Rd Sharma 2020 Solutions for Class 10 Maths Chapter 14 Surface Areas And Volumes are provided here with simple step-by-step explanations. These solutions for Surface Areas And Volumes are extremely popular among Class 10 students for Maths Surface Areas And Volumes Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma 2020 Book of Class 10 Maths Chapter 14 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma 2020 Solutions. All Rd Sharma 2020 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

#### Page No 14.27:

#### Question 1:

How many balls, each of radius I cm, can be made from a solid sphere of lead of radius 8 cm?

#### Answer:

We are given a solid sphere with radius cm.

From this sphere we have to make spherical balls of radius cm.

Let the no. of balls that can be formed be.

We know,

Volume of a sphere =.

So, volume of the bigger solid sphere= …… (a)

Volume of one smaller spherical ball = …… (b)

We know, the volume of the solid sphere should be equal to the sum of the volumes of the n spherical balls formed.

So, using (a) and (b), we get,

Therefore,

Hence, the no. of balls of radius that can be formed out of solid sphere of radius is 512.

#### Page No 14.27:

#### Question 2:

How many spherical bullets each of 5 cm in diameter can be cast from a rectangular block of metal 11dm × 1 m × 5 dm?

#### Answer:

We are given a metallic block of dimension

We know that,

So, the volume of the given metallic block is

We want to know how many spherical bullets can be formed from this volume of the metallic block. It is given that the diameter of each bullet should be 5 cm.

We know,

Volume of a sphere

Here,

Let the no. of bullets formed be *n*.

We know that the sum of the volumes of the bullets formed should be equal to the volume of the metallic block.

Hence the no. of bullets that can be formed is 8400.

#### Page No 14.27:

#### Question 3:

A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of the two of the balls are 1.5 cm and 2 cm respectively. Determine the diameter of the third ball.

#### Answer:

We have one spherical ball of radius 3 cm

So, its volume …… (a)

It is melted and made into 3 balls.

The first ball has radius 1.5 cm

So, its volume …… (b)

The second ball has radius 2 cm

So, its volume …… (c)

We have to find the radius of the third ball.

Let the radius of the third ball be

The volume of this third ball …… (d)

We know that the sum of the volumes of the 3 balls formed should be equal to the volume of the given spherical ball.

Using equations (a), (b), (c) and (d)

Hence the diameter of the third ball should be 5 cm

#### Page No 14.27:

#### Question 4:

2.2 cubic dm of brass is to be drawn into a cylindrical wire 0.25 cm in diameter. Find the length of the wire.

#### Answer:

The brass volume that has to be drawn into a cylindrical wire is given is

We have to make a cylindrical wire out of it with diameter =0.25 cm

So the radius of this wire

We have to find the length of this wire.

Let the length of this wire be

We know that the volume of a cylinder.

We know, the volume of the cylinder should be equal to the volume of the given brass

Therefore, *h *= 448 m

Hence, the length of the cylindrical wire that can be formed is 448 m

#### Page No 14.28:

#### Question 5:

What length of a solid cylinder 2 cm in diameter must be taken to recast into a hollow cylinder of length 16 cm, external diameter 20 cm and thickness 2.5 mm?

#### Answer:

We are given a solid cylinder of, diameter = 2 cm

We have to recast it into a hollow cylinder of length = 16 cm

External Diameter = 20 cm and thickness = 2.5 mm=0.25 cm

We have to find the height of the solid cylinder that can be used to get a hollow cylinder of the desired dimensions.

Volume of a solid cylinder

So,

The volume of the given solid cylinder …… (a)

Here, height has to be found.

Volume of a hollow cylinder

Whereis the external radius and is the internal radius

External radius is given. Thickness of the hollow cylinder is also given.

So, we can find the internal radius of the hollow cylinder.

Thickness=*R*−*r*

So, the volume of the hollow cylinder…… (b)

From (a) and (b) we get,

Hence, the required height of the solid cylinder is *h*=79 cm

#### Page No 14.28:

#### Question 6:

A cylindrical vessel having diameter equal to its height is full of water which is poured into two identical cylindrical vessels with diameter 42 cm and height 21 cm which are filled completely. Find the diameter of the cylindrical vessel.

#### Answer:

A cylindrical vessel whose height is equal to its diameter is given.

It is filled with water.

We know that the volume of a cylinder

In this particular case,

Height is equal to the diameter, that is ,

The volume of cylindrical vessel becomes

The water from this vessel is transferred into two identical cylindrical vessels of

Diameter = 42 cm and, height *h*= 21 cm

Volume of each vessel

We know that the sum of the volumes of the two identical vessels must be equal to the volume of the given cylindrical vessel.

Therefore,

The diameter of the given cylinder is 42 cm

#### Page No 14.28:

#### Question 7:

50 circular plates each of diameter 14 cm and thickness 0.5 cm are placed one above the other to form a right circular cylinder. Find its total surface area.

#### Answer:

We have 50 circular plates, each with diameter = 14 cm

That is, radius = 7 cm and thickness = 0.5 cm

These plates are stacked on top of one another.

So, the total thickness

This is clearly a cylindrical arrangement.

We know,

Total surface area of a cylinder

So, the total surface area of the given arrangement is

#### Page No 14.28:

#### Question 8:

25 circular plates, each of radius 10.5 cm and thickness 1.6 cm, are placed one above the other to form a solid circular cylinder. Find the curved surface area and the volume of the cylinder so formed.

#### Answer:

We have 25 circular plates, each with radius = 10.5 cm and thickness = 1.6 cm

These plates are stacked on top of one another.

So, the total height of the arrangement becomes

Volume of this arrangement

Curved surface area

Hence and

#### Page No 14.28:

#### Question 9:

Find the number of metallic circular discs with 1.5 cm base diameter and of height 0.2 cm to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm .

#### Answer:

Given the diameter of the base of the the circular disc = 1.5 cm

Height = 0.2 cm

Volume of the circular disc = ${\mathrm{\pi r}}^{2}\mathrm{h}=\mathrm{\pi}\times {\left(\frac{1.5}{2}\right)}^{2}\times 0.2=\mathrm{\pi}\times {\left(0.75\right)}^{2}\times 0.2$ ...(i)

Height of the cylinder = 10 cm

Diameter = 4.5 cm

Volume of the cylinder = ${\mathrm{\pi R}}^{2}\mathrm{H}=\mathrm{\pi}{\left(\frac{4.5}{2}\right)}^{2}\times 10=\mathrm{\pi}\times {\left(2.25\right)}^{2}\times 10...\left(\mathrm{ii}\right)$

Now since the circular discs are used to make the cylinder so, let *n* be the number of circular discs required.

$n\times \mathrm{Volume}\mathrm{of}\mathrm{circular}\mathrm{disc}=\mathrm{Volume}\mathrm{of}\mathrm{cylinder}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{Volume}\mathrm{of}\mathrm{cylinder}}{\mathrm{Volume}\mathrm{of}\mathrm{circular}\mathrm{disc}}=n\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{\pi}\times {\left(2.25\right)}^{2}\times 10}{\mathrm{\pi}\times {\left(0.75\right)}^{2}\times 0.2}=n\phantom{\rule{0ex}{0ex}}\Rightarrow n=450$

Hence, 450 metallic circular discs need to be melted to form the right circular cylinder.

#### Page No 14.28:

#### Question 10:

How many spherical lead shots each of diameter 4.2 cm can be obtained from a solid rectangular lead piece with dimension

6cm $\times $ 42 cm $\times $ 21 cm .

#### Answer:

The dimensions of the solid rectangular lead piece is $66\mathrm{cm}\times 42\mathrm{cm}\times 21\mathrm{cm}$.

Diameter of the spherical lead shots = 4.2 cm

Let *n* spherical lead shots be obtained from the rectangular piece.

$n\times \mathrm{volume}\mathrm{of}\mathrm{spherical}\mathrm{lead}\mathrm{shot}=\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{rectangular}\mathrm{lead}\mathrm{piece}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{rectangular}\mathrm{lead}\mathrm{piece}}{\mathrm{volume}\mathrm{of}\mathrm{spherical}\mathrm{lead}\mathrm{shot}}=n\phantom{\rule{0ex}{0ex}}\mathit{\Rightarrow}\frac{66\times 42\times 21}{{\displaystyle \frac{4}{3}}{\mathit{\pi r}}^{\mathit{3}}}\mathit{=}n\phantom{\rule{0ex}{0ex}}\mathit{\Rightarrow}\frac{66\times 42\times 21}{{\displaystyle \frac{\mathrm{4}}{\mathrm{3}}\pi {\left(\frac{\mathrm{4}\mathrm{.}\mathrm{2}}{\mathrm{2}}\right)}^{3}}}\mathit{=}n\phantom{\rule{0ex}{0ex}}\mathit{\Rightarrow}\frac{58212}{38.808}\mathit{=}n\phantom{\rule{0ex}{0ex}}\mathit{\Rightarrow}n\mathit{=}1500\phantom{\rule{0ex}{0ex}}$

Hence, 1500 lead shots can be formed.

DISCLAIMER: There is some error in the question given. Instead of 6 cm, there should be 66 cm.

The result obtained is by taking 66 cm as the dimensions of the rectangular piece.

#### Page No 14.28:

#### Question 11:

How many spherical lead shots of diameter 4 cm can be made out of a solid cube of lead whose edge measures 44 cm .

#### Answer:

Diameter of the spherical lead shots = 4 cm

Edge length of the solid cube (*a*) = 44 cm.

Let *n* be the number of spherical lead shots made out of the solid cube.

$n\times Volumeofthesphericalleadshots=Volumeofthesolidcube\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{Volumeofthesolidcube}{Volumeofthesphericalleadshots}=n\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{a}^{3}}{{\displaystyle \frac{4}{3}}{\mathrm{\pi r}}^{3}}=n\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{44}^{3}}{\frac{4}{3}\mathrm{\pi}\times {\left({\displaystyle \frac{4}{2}}\right)}^{3}}=n\phantom{\rule{0ex}{0ex}}\Rightarrow 2541=n\phantom{\rule{0ex}{0ex}}$

Hence, 2541 spherical lead shots can be made

#### Page No 14.28:

#### Question 12:

Three cubes of a metal whose edges are in the ratios 3 : 4 : 5 are melted and converted into a single cube whose diagonal is $12\sqrt{3}$ cm . Find the edges of the three cubes.

#### Answer:

The three cubes of metal are in the ratio 3 : 4 : 5.

Let the edges of the cubes be 3*x*, 4*x* and 5*x*.

Volume of the three cubes will be

${V}_{1}={\left(3x\right)}^{3}\phantom{\rule{0ex}{0ex}}{V}_{2}={\left(4x\right)}^{3}\phantom{\rule{0ex}{0ex}}{V}_{3}={\left(5x\right)}^{3}$

Diagonal of the single cube = $12\sqrt{3}\mathrm{cm}$

We know diagonal of the cube = $a\sqrt{3}=12\sqrt{3}$

Hence, the side of the cube = 12 cm

Volume of the bigger cube ${V}_{b}={\left(12\right)}^{3}$

Volume of the three cubes = Volume of the single

${\left(3x\right)}^{3}+{\left(4x\right)}^{3}+{\left(5x\right)}^{3}={\left(12\right)}^{3}\phantom{\rule{0ex}{0ex}}\Rightarrow 27{x}^{3}+64{x}^{3}+125{x}^{3}=1728\phantom{\rule{0ex}{0ex}}\Rightarrow 216{x}^{3}=1728\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{3}=\frac{1728}{216}=8\phantom{\rule{0ex}{0ex}}\Rightarrow x=2$

Hence, the edges of the three cubes will be $3\times \left(2\right),4\times \left(2\right),5\times \left(2\right)=6,8,10$ cm

#### Page No 14.28:

#### Question 13:

A solid metallic sphere of radius 10.5 cm is melted and recast into a number of smaller cones, each of radius 3.5 cm and height 3 cm. Find the number of cones so formed.

#### Answer:

Radius of the solid metallic sphere, *r *= 10.5 cm

Radius of the cone, *R* = 3.5 cm

Height of the cone, *H* = 3 cm

Let the number of smaller cones formed be* n. *

Volume of the metallic sphere, ${V}_{s}=\frac{4}{3}\mathrm{\pi}{\left(\mathrm{r}\right)}^{3}=\frac{4}{3}\mathrm{\pi}{\left(10.5\right)}^{3}$

Volume of the cone, ${V}_{c}=\frac{1}{3}\mathrm{\pi}{\left(\mathrm{R}\right)}^{2}\mathrm{H}=\frac{1}{3}\mathrm{\pi}{\left(3.5\right)}^{2}\times 3$

Let the number of cones thus formed be *n*.

$n\times volumeofthecone=volumeofthesphere\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{sphere}\left({\mathrm{V}}_{\mathrm{s}}\right)}{\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{cone}\left({\mathrm{V}}_{\mathrm{c}}\right)}=n\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{\displaystyle \frac{4}{3}}\mathrm{\pi}{\left(10.5\right)}^{3}}{{\displaystyle \frac{1}{3}}\mathrm{\pi}{\left(3.5\right)}^{2}\times 3}=n\phantom{\rule{0ex}{0ex}}\Rightarrow 126=n$

Hence, 126 cones are thus formed.

#### Page No 14.28:

#### Question 14:

The diameter of a metallic sphere is equal to 9 cm. It is melted and drawn into a long wire of diameter 2 mm having uniform cross-section. Find the length of the wire.

#### Answer:

The radius of the metallic sphere is cmmm. Therefore, the volume of the metallic sphere is

Cubic mm

The metallic sphere is melted to produce a long wire of uniform cross section of radius mm. Let the length of the wire be *l* mm. Then, the volume of the wire is

Cubic mm

Since, the volume of the metallic sphere is equal to the volume of the wire, we have

Hence, the length of the wire ismmcm.

Hence

#### Page No 14.28:

#### Question 15:

An iron spherical ball has been melted and recast into smaller balls of equal size. If the radius of each of the smaller balls is 1/4 of the radius of the original ball, how many such balls are made? Compare the surface area, of all the smaller balls combined together with that of the original ball.

#### Answer:

Let the radius of the big metallic ball is 4*r*. Therefore, the volume of the big metallic ball is

The metallic sphere is melted to produce small balls of radius. Then, the volume of each of the small balls is

Since, the volume of the big metallic ball is equal to the sum of the volumes of the small balls, we have the number of produced small balls is

Hence, the number of small balls is

The surface area of the big ball is

The surface area of each of the small ball is

Therefore, the total surface area of the 64 small balls is

Now, we compute the following ratio

Hence, the total surface area of the small balls is equal to four times the surface area of the original big ball.

#### Page No 14.28:

#### Question 16:

A copper sphere of radius 3 cm is melted and recast into a right circular cone of height 3 cm. Find the radius of the base of the cone.

#### Answer:

The radius of the copper sphere is 3cm. Therefore, the volume of the copper sphere is

The copper sphere is melted to produce a right circular cone. The height of the right circular cone is 3cm. Let the base-radius of the right circular cone is *r*. Then, the volume of the right circular cone is

Since, the sphere is melted to recast the cone; the volumes of the sphere and the cone are equal. Hence, we have

Hence, the base-radius of the right circular cone is 6 cm.

#### Page No 14.28:

#### Question 17:

A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m of uniform thickness. Find the thickness of the wire.

#### Answer:

The radius of the copper rod is 0.5 cm and length is 8 cm. Therefore, the volume of the copper rod is

Let the radius of the wire is *r* cm. The length of the wire is 18 m=1800 cm. Therefore, the volume of the wire is

Since, the volume of the copper rod is equal to the volume of the wire; we have

${V}_{1}=V\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{\pi}{r}^{2}\times 1800=\mathrm{\pi}\times {\left(0.5\right)}^{2}\times 8\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{2}=\frac{0.25\times 8}{1800}=\frac{1}{900}\phantom{\rule{0ex}{0ex}}\Rightarrow r=\frac{1}{30}=0.033\mathrm{cm}$

Hence, the radius of the wire is 0.033 cm = 0.33 mm.

So, thickness = $0.33\times 2=0.66\mathrm{mm}$

#### Page No 14.28:

#### Question 18:

The diameters of internal and external surfaces of a hollow spherical shell are 10 cm and 6 cm respectively. If it is melted and recast into a solid cylinder of length of $2\frac{2}{3}$ cm, find the diameter of the cylinder.

#### Answer:

The internal and external radii of the hollow sphere are 3cm and 5cm respectively. Therefore, the volume of the spherical shell is

The spherical shell is melted to recast a solid cylinder of length cm. Let the radius of the solid cylinder is *r* cm. Therefore, the volume of the solid cylinder is

Since, the volume of the hollow spherical shell is equal to the volume of the solid cylinder; we have

Hence, the diameter of the solid cylinder is two times its radius, which is 14 cm.

#### Page No 14.28:

#### Question 19:

How many coins 1.75 cm in diameter and 2 mm thick must be melted to form a cuboid 11 cm × 10 cm × 7 cm?

#### Answer:

The dimension of the cuboid is 11cm10cm7cm. Therefore, the volume of the cuboid is

The radius and thickness of each coin arecm and 2mm = 0.2cm respectively. Therefore, the volume of each coin is

Since, the total volume of the melted coins is same as the volume of the cuboid; the number of required coins is

#### Page No 14.28:

#### Question 20:

The surface area of a solid metallic sphere is 616 cm^{2}. It is melted and recast into a cone of height 28 cm. Find the diameter of the base of the cone so formed (Use π = 22/7).

#### Answer:

The surface area of the metallic sphere is 616 square cm. Let the radius of the metallic sphere is *r*. Therefore, we have

Therefore, the radius of the metallic sphere is 7 cm and the volume of the sphere is

The sphere is melted to recast a cone of height 28 cm. Let the radius of the cone is *R* cm. Therefore, the volume of the cone is

Since, the volumes of the sphere and the cone are same; we have

Hence, the diameter of the base of the cone so formed is two times its radius, which is cm.

#### Page No 14.29:

#### Question 21:

A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied out on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.

#### Answer:

Let the radius of the cone by *r*

Now, Volume cylindrical bucket = Volume of conical heap of sand

$\Rightarrow \mathrm{\pi}{\left(18\right)}^{2}\left(32\right)=\frac{1}{3}\mathrm{\pi}{r}^{2}\left(24\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(18\right)}^{2}\left(32\right)=8{r}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{2}=18\times 18\times 4\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{2}=1296\phantom{\rule{0ex}{0ex}}\Rightarrow r=36\mathrm{cm}$

Let the slant height of the cone be *l*.

Thus , the slant height is given by

$l=\sqrt{{\left(24\right)}^{2}+{\left(36\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{576+1296}\phantom{\rule{0ex}{0ex}}=\sqrt{1872}\phantom{\rule{0ex}{0ex}}=12\sqrt{13}\mathrm{cm}$

**Disclaimer**: The answer given in the book for the slant height is not correct.

#### Page No 14.29:

#### Question 22:

A solid metallic sphere of radius 5.6 cm is melted and solid cones each of radius 2.8 cm and height 3.2 cm are made. Find the number of such cones formed. [CBSE 2014]

#### Answer:

Let the number of such cones formed be *n*

Now, Volume of solid metallic sphere = Volume of *n* solid cones

$\Rightarrow \frac{4}{3}\times \frac{22}{7}\times {\left(5.6\right)}^{3}=n\times \frac{1}{3}\times \frac{22}{7}\times {\left(2.8\right)}^{2}\times 3.2\phantom{\rule{0ex}{0ex}}\Rightarrow 4\times {\left(5.6\right)}^{3}=n\times {\left(2.8\right)}^{2}\times 3.2\phantom{\rule{0ex}{0ex}}\Rightarrow n=28$

#### Page No 14.29:

#### Question 23:

A solid cuboid of iron with dimensions 53 cm ⨯ 40 cm ⨯ 15 cm is melted and recast into a cylindrical pipe. The outer and inner diameters of pipe are 8 cm and 7 cm respectively. Find the length of pipe. [CBSE 2015]

#### Answer:

Volume of solid cuboid of iron = Volume of cylindrical pipe

$\Rightarrow lbh=\mathrm{\pi}h\left({R}^{2}-{r}^{2}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 53\times 40\times 15=\frac{22}{7}\times h\left[{\left(\frac{8}{2}\right)}^{2}-{\left(\frac{7}{2}\right)}^{2}\right]\phantom{\rule{0ex}{0ex}}\Rightarrow 53\times 40\times 15=\frac{22}{7}\times h\left[{4}^{2}-{\left(3.5\right)}^{2}\right]\phantom{\rule{0ex}{0ex}}\Rightarrow 53\times 40\times 15=\frac{22}{7}\times h\times 3.75\phantom{\rule{0ex}{0ex}}\Rightarrow h=2698.18\mathrm{cm}$

**Disclaimer**: The answer given in the book is not correct.

#### Page No 14.29:

#### Question 24:

The diameters of the internal and external surfaces of a hollow spherical shell are 6 cm and 10 cm respectively. If it is melted and recast into a solid cylinder of diameter 14 cm. find the height of the cylinder.

#### Answer:

The internal and external radii of the hollow spherical shell are 3cm and 5cm respectively. Therefore, the volume of the hollow spherical shell is

The hollow spherical shell is melted to recast a cylinder of radius 7cm. Let, the height of the solid cylinder is *h*. Therefore, the volume of the solid cylinder is

Since, the volume of the solid cylinder is same as the volume of the hollow spherical shell, we have

Therefore, the height of the solid cylinder is

#### Page No 14.29:

#### Question 25:

A hollow sphere of internal and external diameters 4 cm and 8 cm respectively is melted into a cone of base diameter 8 cm. Calculate the height of the cone.

#### Answer:

The internal and external radii of the hollow sphere are 2cm and 4cm respectively. Therefore, the volume of the hollow sphere is

The hollow spherical shell is melted to recast a cone of base- radius 4cm. Let, the height of the cone is *h*. Therefore, the volume of the cone is

Since, the volume of the cone is same as the volume of the hollow sphere, we have

Therefore, the height of the cone is

#### Page No 14.29:

#### Question 26:

A hollow sphere of internal and external radii 2 cm and 4 cm respectively is melted into a cone of base radius 4 cm. Find the height and slant height of the cone.

#### Answer:

The internal and external radii of the hollow sphere are 2cm and 4cm respectively. Therefore, the volume of the hollow sphere is

The hollow sphere is melted to produce a right circular cone of base-radius 4cm. Let, the height and slant height of the cone be *h *cm and *l *cm respectively. Then, we have

The volume of the cone is

Since, the volume of the cone and hollow sphere are same, we have

Then, we have

Therefore, the height and the slant height of the cone are 14 cm and 14.56 cm respectively.

#### Page No 14.29:

#### Question 27:

A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radii of two of the balls are 1.5 cm and 2 cm. Find the diameter of the third ball.

#### Answer:

The radius of the big spherical ball is 3cm. Therefore, the volume of the big spherical ball is

cubic cm

The radii of the 1^{st} and 2^{nd} small spherical balls are 1.5 cm and 2 cm respectively. Therefore, the volumes of the 1^{st} and 2^{nd} spherical balls are respectively

cubic cm,

cubic cm

Let, the radius of the 3^{rd} small spherical ball is *r *cm. Then, its volume is

cubic cm

Since, the big spherical ball is melted to produce the three small spherical balls; the volume of the big spherical ball is same as the sum of the volumes of the three small spherical balls. Therefore, we have

Therefore, the diameter of the 3^{rd} ball is

#### Page No 14.29:

#### Question 28:

A path 2 m wide surrounds a circular pond of diameter 40 m. How many cubic metres of gravel are required to grave the path to a depth of 20 cm?

#### Answer:

Diameter of the circular pond is given = 40 m

So, the radius of this pond is 20 m

There is a path surrounding the pond. We are given the thickness of this path as 2 m

We have to grave this path with gravel. The depth of the path is also given 20 cm=0.2 m

This circular path can be viewed as a hollow cylinder of thickness 0.2 m and depth 0.2 m

We know,

Volume of a hollow cylinder

So the volume of the circular path with height 0.2 m

Hence, the volume of gravel required is

#### Page No 14.29:

#### Question 29:

A 16 m deep well with diameter 3.5 m is dug up and the earth from it is spread evenly to form a platform 27.5 m by 7 m. Find the height of the platform.

#### Answer:

Assume the well as a solid right circular cylinder. Then, the radius of the solid right circular cylinder is

m

The well is 16m deep. Thus, the height of the solid right circular cylinder ism.

Therefore, the volume of the solid right circular cylinder is

Let the height of the platform formed be *x *m. The length and the breadth of the platform are *l*=27.5m and *b*=7m respectively. Therefore, the volume of the platform is

Since, the well is spread to form the platform; the volume of the well is equal to the volume of the platform. Hence, we have

Hence, the height of the platform is 0.8 m = 80 cm.

#### Page No 14.29:

#### Question 30:

A well of diameter 2 m is dug 14 m deep. The earth taken out of it is spread evenly all around it to form an embankment of height 40 cm. Find the width of the embankment.

#### Answer:

Assume the well as a solid right circular cylinder. Then, the radius of the solid right circular cylinder is

The well is 14m deep. Thus, the height of the solid right circular cylinder is m.

Therefore, the volume of the solid right circular cylinder is

Since, the embankment is to form around the right circular cylinder. Let the width of the embankment be *x* m. The height of the embankment is *h* = 40 cm = 0.4 m. Therefore, the volume of the platform is

Since, the well is spread to form the platform; the volume of the well is equal to the volume of the platform. Hence, we have

Hence, *x = *5

Hence,

#### Page No 14.29:

#### Question 31:

A well with inner radius 4 m is dug 14 m deep. Earth taken out of it has been spread evenly all around a width of 3 m it to form an embankment. Find the height of the embankment.

#### Answer:

The inner radius of the well is 4m and the height is 14m. Therefore, the volume of the Earth taken out of it is

The inner and outer radii of the embankment are 4m and 4+3=7m respectively. Let the height of the embankment be *h*. Therefore, the volume of the embankment is

Since, the volume of the well is same as the volume of the embankment; we have

Hence, the height of the embankment is

#### Page No 14.29:

#### Question 32:

A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it to a width of 4 m to form an embankment. Find the height of the embankment.

#### Answer:

The inner radius of the well ism and the height is 14m. Therefore, the volume of the Earth taken out of it is

The inner and outer radii of the embankment arem and 4+=m respectively. Let the height of the embankment be *h*. Therefore, the volume of the embankment is

Since, the volume of the well is same as the volume of the embankment; we have

Hence, the height of the embankment is

#### Page No 14.29:

#### Question 33:

Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 9 cm.

#### Answer:

We have the following figure

The length of each side of the cube is 9 cm. We have to find the volume of the largest right circular cone contained in the cube.

The diameter of the base circle is same as the length of the side of the cube. Thus, the diameter of the base circle of the right circular cone is 9 cm. Therefore, the radius of the base of the right circular cone iscm.

From the right angled triangle we have

*h *= 9 cm

Therefore, the volume of the solid right circular cone is

Hence largest volume of cone is

#### Page No 14.29:

#### Question 34:

A cylindrical bucket, 32 cm high and 18 cm of radius of the base, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.

#### Answer:

The height and radius of the cylindrical bucket arecm andcm respectively. Therefore, the volume of the cylindrical bucket is

The bucket is full of sand and is emptied in the ground to form a conical heap of sand of height cm. Let, the radius and slant height of the conical heap be cm and cm respectively. Then, we have

The volume of the conical heap is

Since, the volume of the cylindrical bucket and conical hear are same, we have

Then, we have

Therefore, the radius and the slant height of the conical heap are 36 cm and 43.27 cm respectively.

#### Page No 14.29:

#### Question 35:

Rain water, which falls on a flat rectangular surface of length 6 m and breadth 4 m is transferred into a cylindrical vessel of internal radius 20 cm. What will be the height of water in the cylindrical vessel if a rainfall of 1 cm has fallen?

#### Answer:

The fallen rains are in the form of a cuboid of height 1 cm, length 6 m = 600 cm and breadth 4 m = 400 cm. Therefore, the volume of the fallen rains is

The fallen rains are transferred into a cylindrical vessel of internal radius *r*_{1} = 20 cm. Let, the height of the water in the cylindrical vessel is *h*_{1} cm. Then, the volume of the water in the cylinder is

Since, the volume of the water in the cylinder is same as the volume of the rainfalls, we have

Therefore, the height of the water in the cylinder is 190.9 cm.

#### Page No 14.29:

#### Question 36:

The rain water from a roof of dimensions $22\mathrm{m}\times 20\mathrm{m}$ drains into a cylindrical vessel having diameter of base 2m and height 3.5 m . If the rain water collected from the roof just fills the cylindrical vessel , then find the rainfall in cm .

#### Answer:

The dimension of the roof is $22\mathrm{m}\times 20\mathrm{m}$.

Diameter of the cylinderical vessel = 2 m

Radius of the cylinderical vessel, R = 1 m

Height of the cylinderical vessel, H = 3.5 m

Let the height of the roof be h.

Volume of water thus collected on the roof = $22\times 20\times h$

Volume of the cylinderical vessel = $\mathrm{\pi}{\left(\mathrm{R}\right)}^{2}\mathrm{H}=\mathrm{\pi}{\left(\frac{2}{2}\right)}^{2}\times 3.5=\mathrm{\pi}\times 1\times 3.5=3.5\mathrm{\pi}$

Volume of water collected on the roof = Volume of the cylinderical vessel

$22\times 20\times h=3.5\mathrm{\pi}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{h}=\frac{3.5\mathrm{\pi}}{22\times 20}=0.025\mathrm{m}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{h}=2.5\mathrm{cm}$

#### Page No 14.29:

#### Question 37:

A conical flask is full of water. The flask has base-radius *r *and height *h*. The water is poured into a cylindrical flask of base-radius *mr*. Find the height of water in the cylindrical flask.

#### Answer:

The base-radius and height of the conical flask are *r* and *h* respectively. Let, the slant height of the conical flask is *l*. Therefore, the volume of the water in the conical flask is

The water in the conical flask is poured into a cylindrical flask of base-radius *mr*. Let, the height of the water in the cylindrical flaks is *h*_{1}. Then, the volume of the water in the cylindrical flaks is

Since, the volume of the water in the cylindrical flaks is same as the volume of the water in the conical flaks, we have

Therefore, the height of the water in the cylinder is

#### Page No 14.30:

#### Question 38:

A rectangular tank 15 m long and 11 m broad is required to receive entire liquid contents from a fully cylindrical tank of internal diameter 21 m and length 5 m. Find the least height of the tank that will serve the purpose.

#### Answer:

Suppose height of the rectangular tank is equal to *h*.

Length of the tank = 15 m

Breadth of the tank = 11 m

Further,

length of cylindrical tank = 5 m

Radius of cylindrical tank = $\frac{21}{2}$m

To find out the least height of the tank, equate the volumes of two tanks.

$15\times 11\times h=\mathrm{\pi}{\left(\frac{21}{2}\right)}^{2}\times 5\phantom{\rule{0ex}{0ex}}\Rightarrow h=\frac{22}{7}\times \frac{21}{2}\times \frac{21}{2}\times \frac{5}{15}\times \frac{1}{11}\phantom{\rule{0ex}{0ex}}\Rightarrow h=\frac{21}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow h=10.5$

Hence, the least height of the tank is equal to 10.5.

#### Page No 14.30:

#### Question 39:

A hemispherical bowl of internal radius 9 cm is full of liquid. The liquid is to be filled into cylindrical shaped small bottles each of diameter 3 cm and height 4 cm. How many bottles are necessary to empty the bowl?

#### Answer:

The internal radius of the hemispherical bowl is 9cm. Therefore, the volume of the water in the hemispherical bowl is

The water in the hemispherical bowl is required to transfer into the cylindrical bottles each of radiuscm and height 4cm. Therefore, the volume of each of the cylindrical bottle is

Therefore, the required number of cylindrical bottles is

Hence

#### Page No 14.30:

#### Question 40:

A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical ball dropped into the tub and the level of the water is raised by 6.75 cm. Find the radius of the ball.

#### Answer:

The radius of the cylindrical tub is 12cm. Upon dropping a spherical ball into the tub, the height of the raised water is 6.75cm. Therefore, the volume of the raised water is

Let, the radius of the spherical ball is *r*. Therefore, the volume of the spherical ball is

Since, the volume of the raised water is same as the volume of the spherical ball, we have

Therefore, the radius of the spherical ball is

#### Page No 14.30:

#### Question 41:

500 persons have to dip in a rectangular tank which is 80 m long and 50 m broad. What is the rise in the level of water in the tank, if the average displacement of water by a person is 0.04 m^{3}?

#### Answer:

The average displacement of water by a person is 0.04 cubic m. Hence, the total displacement of water in the rectangular tank by 500 persons is Cubic m.

The length and width of the rectangular tank are 80m and 50m respectively. Upon dipping in the tank, let the height of the raised water is be *h *m. Therefore, the volume of the raised water is

cubic m

Since, the volume of the raised water is same as the volume of the water displaced by 500 persons, we have

Therefore, the water will be raised by

#### Page No 14.30:

#### Question 42:

A cylindrical jar of radius 6 cm contains oil. Iron spheres each of radius 1.5 cm are immersed in the oil. How many spheres are necessary to raise the level of the oil by two centimetres?

#### Answer:

The radius of the cylindrical jar is 6cm. The volume of the oil of height 2cm contained in the jar is

cubic cm

The radius of each small sphere is 1.5cm. Therefore, the volume of each small sphere is

cubic cm

Since, the volume of the raised water is same as the sum of the volumes of the immersed iron spheres, we have the number of immersed sphere is

Therefore, the number of iron spheres is

#### Page No 14.30:

#### Question 43:

A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical from ball of radius 9 cm is dropped into the tub and thus the level of water is raised by *h* cm. What is the value of *h*?

#### Answer:

The radius of the cylindrical tub is 12cm. Upon dropping a spherical ball of radius 9cm into the tub, the height of the raised water is *h *cm. Therefore, the volume of the raised water is

cubic cm

The volume of the spherical ball is

cubic cm

Since, the volume of the raised water is same as the volume of the spherical ball, we have

Therefore, the height of the raised water is

#### Page No 14.30:

#### Question 44:

Metal spheres, each of the radius 2 cm, are packed into a rectangular box of internal dimension 16 cm × 8 cm × 8 cm when 16 spheres are packed the box is filled with preservative liquid. Find the volume of this liquid.

#### Answer:

The radius of each of the metallic sphere is 2cm. Therefore, the volume of each metallic sphere is

The total volume of the 16 spheres is

The internal dimension of the rectangular box is 16cm8cm8cm. Therefore, the volume of the rectangular box is

Therefore, the volume of the liquid is

Hence volume of liquid is

#### Page No 14.30:

#### Question 45:

A vessel in the shape of a cuboid contains some water. If three identical spheres immersed in the water, the level of water is increased by 2 cm. If the area of the base of the cuboid is 160 cm^{2} and its height 12 cm, determine the radius of any of the spheres.

#### Answer:

The area of the base of the cuboid is 160 cm^{2}. After immersing three identical spheres the level of the water is increased by 2 cm. Therefore, the volume of the increased water is

Let the radius of each of the spheres is *r* cm. Then, the volume of each of the sphere is

The total volume of the three spheres is

Since, the volume of the increased water is equal to the total volume of the three spheres; we have

Hence, the radius of each of the sphere is 2.94 cm.

#### Page No 14.30:

#### Question 46:

150 spherical marbles, each of diameter 1.4 cm are dropped in a cylindrical vessel of diameter 7 cm containing some water, which are completely immersed in water. Find the rise in the level of water in the vessel. [CBSE 2014]

#### Answer:

Let the rise in the level of water in the vessel be *h *cm.

Now, Volume of 150 spherical marbles = Volume of water displaced in the vessel

$\Rightarrow 150\times \frac{4}{3}\times \frac{22}{7}\times {\left(\frac{1.4}{2}\right)}^{3}=\frac{22}{7}\times {\left(\frac{7}{2}\right)}^{2}\times h\phantom{\rule{0ex}{0ex}}\Rightarrow 200\times {\left(0.7\right)}^{3}={\left(\frac{7}{2}\right)}^{2}\times h\phantom{\rule{0ex}{0ex}}\Rightarrow h=5.6\mathrm{cm}$

#### Page No 14.30:

#### Question 47:

Sushant has a vessel, of the form of an inverted cone, open at the top, of height 11 cm and radius of top as 2.5 cm and is full of water. Metallic spherical balls each of diameter 0.5 cm are put in the vessel due to which $\left(\frac{2}{5}\right)$^{th} of the water in the vessel flows out. Find how many balls were put in the vessel. Sushant made the arrangement so that the water that flows out irrigates the flower beds. What value has been shown by Sushant?

[CBSE 2014]

#### Answer:

Let the number of the balls be *n*.

Volume of water flows out = Volume of *n* spherical bolls

$\Rightarrow \frac{2}{5}\times \frac{1}{3}\mathrm{\pi}{R}^{2}h=n\times \frac{4}{3}\mathrm{\pi}{r}^{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2}{5}\times {\left(2.5\right)}^{2}\times 11=n\times 4{\left(\frac{0.5}{8}\right)}^{3}\phantom{\rule{0ex}{0ex}}\Rightarrow 27.5=\frac{0.5}{8}n\phantom{\rule{0ex}{0ex}}\Rightarrow n=440\phantom{\rule{0ex}{0ex}}$

#### Page No 14.30:

#### Question 48:

16 glass spheres each of radius 2 cm are packed into a cuboidal box of internal dimensions $16\mathrm{cm}\times 8\mathrm{cm}\times 8\mathrm{cm}$ and then the box is filled with water . Find the volume of the water filled in the box .

#### Answer:

Radius of the glass spheres, r = 2 cm

Dimensions of the cuboidal box = $16\mathrm{cm}\times 8\mathrm{cm}\times 8\mathrm{cm}$

volume of the spheres = ${V}_{s}=\frac{4}{3}\mathrm{\pi}{\left(\mathrm{r}\right)}^{3}=\frac{4}{3}\mathrm{\pi}{\left(2\right)}^{3}$

volume of the cuboidal box = ${V}_{c}=16\times 8\times 8=1024$

Volume of water in the cuboidal box = Volume of the cuboidal box − Volume of the 16 glass spheres

$=1024-16\times \frac{4}{3}\mathrm{\pi}{\left(2\right)}^{3}\phantom{\rule{0ex}{0ex}}=1024-536.6\phantom{\rule{0ex}{0ex}}=487.6{\mathrm{cm}}^{3}$

Hence, the volume of the water in the cuboidal box = 487.6 cm^{3}

#### Page No 14.30:

#### Question 49:

Water flows through a cylindrical pipe , whose inner radius is 1 cm , at the rate of 80 cm /sec in an empty cylindrical tank , the radius of whose base is 40 cm . What is the rise of water level in tank in half an hour ?

#### Answer:

The inner radius of the cylindrical pipe *r* =1 cm.

Rate of flow of water = 80 cm/sec

volume of the water that flows through pipe in 1sec is ${\mathrm{\pi r}}^{2}\times 80=80\mathrm{\pi}{\mathrm{cm}}^{3}$

volume of the water that flows through pipe in half an hour $80\mathrm{\pi}\times 30\times 60=144000\mathrm{\pi}{\mathrm{cm}}^{3}$

radius of the base of the cylindrical tank is R = 40 cm

let the water level in the cylinderical tank after half an hour be *h*.

volume of the raised water = $\mathrm{\pi}{\left(\mathrm{R}\right)}^{2}\mathrm{h}=\mathrm{\pi}{\left(40\right)}^{2}\mathrm{h}$

volume of the raised water in tank = volume of the water that flows through pipe

$\Rightarrow \mathrm{\pi}{\left(40\right)}^{2}\mathrm{h}=144000\mathrm{\pi}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{h}=\frac{144000}{1600}=90\mathrm{cm}$

Thus water level will rise by 90 cm in half an hour.

#### Page No 14.30:

#### Question 50:

Water in a canal 1.5 m wide and 6 m deep is flowing with a speed of 10km/hr. How much area will it irrigate in 30 minutes if 8 cm of standing water is desired?

#### Answer:

The canal is 1.5 m wide and 6 m deep. The water is flowing in the canal at 10 km/hr. Hence, in 30 minutes, the length of the flowing standing water is

Therefore, the volume of the flowing water in 30 min is

Thus, the irrigated area in 30 min of 8 cm=0.08 m standing water is

#### Page No 14.30:

#### Question 51:

A farmer runs a pipe of internal diameter 20 cm from the canal into a cylindrical tank in his field which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?

#### Answer:

The internal radius of the pipe is 10 cm=0.1 m. The water is flowing in the pipe at 3km/hr = 3000m/hr. Let the cylindrical tank will be filled in *t* hours. Therefore, the length of the flowing water in *t* hours is

Therefore, the volume of the flowing water is

The radius of the cylindrical tank is 5 m and the height is 2 m. Therefore, the volume of the cylindrical tank is

Since, we have considered that the tank will be filled in *t* hours; therefore the volume of

the flowing water in *t* hours is same as the volume of the cylindrical tank. Hence, we have

Hence, the tank will be filled in

#### Page No 14.31:

#### Question 52:

A cylindrical tank full of water is emptied by a pipe at the rate of 225 litres per minute. How much time will it take to empty half the tank, if the diameter of its base is 3 m and its height is 3.5 m? [Use $\mathrm{\pi}=\frac{22}{7}$] [CBSE 2014]

#### Answer:

Volume of cylidrical tank = $\frac{22}{7}\times {\left(\frac{3}{2}\right)}^{2}\times 3.5$ = 24.75 m^{3 }

Now, 1 m^{3} = 1000 L

∴ 24.75 m^{3} = 24750 L

Half the capacity of tank = 12375 L

Time taken by the pipe to empty 225 litres = 1 minute

Time taken by the pipe to empty 1 litre = $\frac{1}{225}$ minutes

Time taken by the pipe to empty 12375 litres = $\frac{1}{225}\times 12375=55\mathrm{minutes}$

#### Page No 14.31:

#### Question 53:

Water is flowing at the rate of 2.52 km/h through a cylindrical pipe into a cylindrical tank, the radius of the base is 40 cm. If the increase in the level of water in the tank, in half an hour is 3.15 m, find the internal diameter of the pipe. [CBSE 2015]

#### Answer:

Increase in the level of water in half an hour, *h = *3.15* *m* = *315 cm

Radius of the water tank, *r = *40 cm

Volume of water that falls in the tank in half an hour = π*r*^{2}*h*

= π × (40)^{2} × 315

= 5,04,000 π cm^{3}

Rate of flow of water = 2.52 km/h

Length of water column in half an hour = 2.52 ÷ 2 = 1.26 km = 1,26,000 cm

Let the internal diameter of the cylindrical pipe be* d.*

Volume of the water that flows through the pipe in half an hour = $\mathrm{\pi}{\left(\frac{d}{2}\right)}^{2}\times 126000$

We know

Volume of the water that flows through the pipe in half an hour = Volume of water that falls in the tank in half an hour

$\Rightarrow \mathrm{\pi}{\left(\frac{d}{2}\right)}^{2}\times 126000=504000\mathrm{\pi}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(\frac{d}{2}\right)}^{2}=4\phantom{\rule{0ex}{0ex}}\Rightarrow d=4\mathrm{cm}$

Thus, the internal diameter of the pipe is 4 cm.

#### Page No 14.31:

#### Question 54:

Water flows at the rate of 15 km/hr through a pipe of diameter 14 cm into a cuboidal pond which is 50 m long and 44 m wide . In what time will the level of water in the pond rise by 21 cm.

#### Answer:

Let the level of water in the pond rises by 21 cm in *t* hours.

Speed of water = 15 km/hr = 15000 m/hr

$\mathrm{Diameter}\mathrm{of}\mathrm{the}\mathrm{pipe}=14\mathrm{cm}=\frac{14}{100}\mathrm{m}\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{pipe},\mathrm{r}=\frac{14}{2\times 100}=\frac{7}{100}\mathrm{m}$

Volume of water flowing out of the pipe in 1 hour

$={\mathrm{\pi r}}^{2}\mathrm{h}\phantom{\rule{0ex}{0ex}}=\mathrm{\pi}\times {\left(\frac{7}{100}\right)}^{2}\times 15000{\mathrm{m}}^{3}\phantom{\rule{0ex}{0ex}}=231{\mathrm{m}}^{3}$

∴ Volume of water flowing out of the pipe in* t* hours = 231*t* m^{3}.

Volume of water in the cuboidal pond

$=50\mathrm{m}\times 44\mathrm{m}\times \frac{21}{100}\mathrm{m}\phantom{\rule{0ex}{0ex}}=462{\mathrm{m}}^{3}$

Volume of water flowing out of the pipe in *t* hours = Volume of water in the cuboidal pond

∴ 231*t* = 462

Thus, the water in the pond rise by 21 cm in 2 hours.

#### Page No 14.31:

#### Question 55:

A canal is 300 cm wide and 120 cm deep. The water in the canal is flowing with a speed of 20 km/hr. How much area will it irrigate in 20 minutes if 8 cm of standing water is desired ?

#### Answer:

Width of the canal = 300 cm = 3 m

Depth of the canal = 120 cm = 1.2 m

Speed of water flow = 20 km/h = 20000 m/h

Distance covered by water in 1 hour or 60 min = 20000 m

So, distance covered by the water in 20 min = $\frac{20}{60}\times 20000=\frac{20000}{3}m$

Amount of water irrigated in 20 mintues

$=3\times 1.2\times \frac{20000}{3}=24000$ m^{3}

Area irrigated by this water if 8 cm of standing water is desired will be

$\frac{24000}{{\displaystyle \frac{8}{100}}}=300000{\mathrm{m}}^{2}$

So, area irrigated will be 300000 m^{2} or 30 hectors.

#### Page No 14.31:

#### Question 56:

The sum of the radius of base and height of a solid right circular cylinder is 37 cm . If the total surface area of the solid cylinder is 1628 cm^{2} , find the volume of cylinder. (Use $\mathrm{\pi}=$22 / 7 )

#### Answer:

Let the radius of the base of the cylinder be *r* cm

Let the height be *h* cm.

Now given that *r + h* = 37 cm

Total surface area = 1628 cm^{2 }...(i)

$Totalsurfaceareaofthecylinder=2{\mathrm{\pi r}}^{2}+2\mathrm{\pi rh}\phantom{\rule{0ex}{0ex}}=2\mathrm{\pi r}\left(\mathrm{r}+\mathrm{h}\right)\phantom{\rule{0ex}{0ex}}=2\mathrm{\pi r}\times 37\phantom{\rule{0ex}{0ex}}=74\mathrm{\pi r}...\left(\mathrm{ii}\right)$

From equation (i) and (ii) we get

$74\mathrm{\pi r}=1628\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{r}=\frac{1628}{74\mathrm{\pi}}=7{\mathrm{cm}}^{3}$

Thus, the height will be 37 − 7 = 30 cm

Thus, the volume of the cylinder = ${\mathrm{\pi r}}^{2}\mathrm{h}=\mathrm{\pi}{\left(7\right)}^{2}\times 30=4620{\mathrm{cm}}^{3}$

Hence the volume is 4620 cm^{3}

#### Page No 14.31:

#### Question 57:

A tent of height 77 dm is in the form a right circular cylinder of diameter 36 m and height 44 dm surmounted by a right circular cone. Find the cost of the canvas at Rs 3.50 per m^{2}. [Use π = 22/7]

#### Answer:

The height of the tent is 77dm = 7.7m. The height of the upper portion of the tent is

44dm = 4.4m. Therefore, the height of the cylindrical part isdm = 3.3m. The radius of the cylindrical part is m.

Let the slant height of the cone part is *l* m. Then, we have

Therefore, the slant height of the cone part is 18.3 m.

The curved surface area of the cylindrical part is

The curved surface area of the cone part is

Therefore, the total curved surface area of the tent is

The cost of canvas per m^{2} is Rs 3.50. Hence, the total cost for canvas in Rs is

Hence total cost is

#### Page No 14.31:

#### Question 58:

The largest sphere is to be curved out of a right circular cylinder of radius 7 cm and height 14 cm. Find the volume of the sphere.

#### Answer:

The radius of the right circular cylinder is 7cm and the height is 14cm. Therefore, the radius of the largest sphere curved out from the cylinder is the minimum of the radius and half the height of the cylinder, which is 7cm. Therefore, the volume of the sphere is

#### Page No 14.31:

#### Question 59:

A right angled triangle whose sides are 3 cm, 4 cm and 5 cm is revolved about the sides containing the right angle in two ways. Find the difference in volumes of the two cones so formed. Also, find their curved surfaces.

#### Answer:

We consider the following figure as follows

Let the angle B is right angle and the sides of the triangle are AB* *= 4cm, BC* *= 3cm,

AC* *= 5cm.

When the triangle is revolved about the side AB, then the base-radius, height and slant height of the produced cone becomes BC*, *AB and AC respectively. Therefore, the volume of the produced cone is

In this case, the curved surface area of the cone is

When the triangle is revolved about the side BC, then the base-radius, height and slant height of the produced cone becomes AB*, *BC and AC respectively. Therefore, the volume of the produced cone is

In this case, the curved surface area of the cone is

Therefore, the difference between the volumes of the two cones so formed is

Hence the difference between the volumes is

And surface areas are

#### Page No 14.31:

#### Question 60:

A 5 m wide cloth is used to make a conical tent of base diameter 14 m and height 24 m. Find the cost of cloth used at the rate of Rs 25 per metre. [Use π = 22/7] [CBSE 2014]

#### Answer:

Let the slant height of the cone be *l*.

Thus , the slant height is given by

$l=\sqrt{{\left(\frac{14}{2}\right)}^{2}+{\left(24\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{49+576}\phantom{\rule{0ex}{0ex}}=\sqrt{625}\phantom{\rule{0ex}{0ex}}=25\mathrm{m}$

Now, the curved surface area of the tent is given by

$\frac{22}{7}\times \frac{14}{2}\times 25\phantom{\rule{0ex}{0ex}}=550{\mathrm{m}}^{2}$

The curved surface area will be equal to the area of the cloth

$\Rightarrow 550=\mathrm{Length}\times 5\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{Length}=110\mathrm{m}$

Now, the cost of cloth is given by

110 ⨯ 25

= Rs 2750

#### Page No 14.31:

#### Question 61:

The volume of a hemi-sphere is 2425$\frac{1}{2}$ cm^{3}. Find its curved surface area. (Use π = 22/7)

#### Answer:

Let the radius of the hemisphere be *r *cm.

Volume of hemisphere = $2425\frac{1}{2}{\mathrm{cm}}^{3}$

$\Rightarrow \frac{2}{3}\mathrm{\pi}{r}^{3}=\frac{4851}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2}{3}\times \frac{22}{7}{r}^{3}=\frac{4851}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{3}=\frac{4851\times 3\times 7}{2\times 2\times 22}\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{3}=\frac{441\times 21}{2\times 2\times 2}\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{3}={\left(\frac{21}{2}\right)}^{3}\phantom{\rule{0ex}{0ex}}\Rightarrow r=\frac{21}{2}\mathrm{cm}$

Now, the curved surface area of hemisphere is given by

$2\mathrm{\pi}{r}^{2}\phantom{\rule{0ex}{0ex}}=2\times \frac{22}{7}\times {\left(\frac{21}{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}=693{\mathrm{cm}}^{2}$

#### Page No 14.31:

#### Question 62:

The difference between the outer and inner curved surface areas of a hollow right circular cylinder 14 cm long is 88 cm^{2}. If the volume of metal used in making the cylinder is 176 cm^{3}, find the outer and inner diameters of the cylinder. (Use π = 22/7)

#### Answer:

The height of the hollow cylinder is 14 cm. Let the inner and outer radii of the hollow cylinder are *r* cm and *R* cm respectively. The difference between the outer and inner surface area of the hollow cylinder is

By the given condition, this difference is 88 square cm. Hence, we have

The volume of the metal used in making the cylinder is

By the given condition, the volume of the metal is 176 cubic cm. Hence, we have

Hence, we have two equations with unknowns *R* and *r*

Adding the two equations, we have

Then from the second equation, we have

Therefore, the outer and inner diameters of the hollow cylinder are 5cm and 3cm respectively.

#### Page No 14.31:

#### Question 63:

The internal and external diameters of a hollow hemispherical vessel are 21 cm and 25.2 cm respectively. The cost of painting 1 cm^{2} of the surface is 10 paise. Find the total cost to paint the vessel all over.

#### Answer:

We are given the following hemi hollow sphere

The internal and external radii of the hollow hemispherical vessel arecm and cm respectively. Therefore, the total surface area of the hollow hemispherical vessel is

The cost of painting 1 square cm is 10 paise. Therefore the total cost of painting the vessel all over is

Hence total cost of painting is rupees

#### Page No 14.31:

#### Question 64:

Prove that the surface area of a sphere is equal to the curved surface area of the circumscribed cylinder.

#### Answer:

We have the following figure to visualize the situation

Let the radius of the sphere is *r*. Therefore, the surface area of the sphere is

The circumscribed cylinder of the sphere must have radius *r* cm and height 2*r* cm. Therefore, the curved surface area of the cylinder is

Hence, *S* and *S*_{1} are same. Thus the proof is complete.

#### Page No 14.31:

#### Question 65:

If the total surface area of a solid hemisphere is 462 cm^{2}, find its volume (Take π = 22/7 ) [CBSE 2014]

#### Answer:

Let the radius of the hemisphere be r cm.

Total surface area of hemisphere = 462 cm^{2}

$\Rightarrow 3\mathrm{\pi}{r}^{2}=462\phantom{\rule{0ex}{0ex}}\Rightarrow 3\times \frac{22}{7}\times {\left(r\right)}^{2}=462\phantom{\rule{0ex}{0ex}}\Rightarrow {r}^{2}=49\phantom{\rule{0ex}{0ex}}\Rightarrow r=7\mathrm{cm}$

Now, the volume of hemisphere is given by

$\frac{2}{3}\mathrm{\pi}{r}^{3}\phantom{\rule{0ex}{0ex}}=\frac{2}{3}\times \frac{22}{7}{\left(7\right)}^{3}\phantom{\rule{0ex}{0ex}}=\frac{2156}{3}\phantom{\rule{0ex}{0ex}}=718\frac{2}{3}{\mathrm{cm}}^{3}$

#### Page No 14.31:

#### Question 66:

Water flows at the rate of 10 m / minute through a cylindrical pipe 5 mm in diameter . How long would it take to fill a conical vessel whose diameter at the base is 40 cm and depth 24 cm.

#### Answer:

Diameter of the pipe = 5 mm = 0.5 cm

Radius of the pipe = $\frac{0.5}{2}\mathrm{cm}$ = $\frac{1}{4}cm$

Rate of flow of water through the pipe = 10 m/min = 1000 cm/min

Volume of water that flows out through the pipe in 1 min = ${\mathrm{\pi r}}^{2}\mathrm{h}=\mathrm{\pi}\times {\left(\frac{1}{4}\right)}^{2}\times 1000{\mathrm{cm}}^{3}$

Volume of water flowing out through the pipe in t min = $\mathrm{\pi}{\left(\frac{1}{4}\right)}^{2}\times 1000t$

Diameter of the conical vessel = 40 cm

Radius = 20 cm

Height or depth = 24 cm

Volume of the conical vessel = $\frac{1}{3}{\mathrm{\pi R}}^{2}\mathrm{H}=\frac{1}{3}\mathrm{\pi}{\left(20\right)}^{2}\times 24=3200\mathrm{\pi}$

Time required to fill the vessel = $\frac{\mathrm{capacity}\mathrm{of}\mathrm{the}\mathrm{vessel}}{\mathrm{volume}\mathrm{of}\mathrm{water}\mathrm{flowing}\mathrm{per}\mathrm{min}}$

$t=\frac{3200\mathrm{\pi}}{\mathrm{\pi}{\left({\displaystyle \frac{1}{4}}\right)}^{2}\times 1000}\phantom{\rule{0ex}{0ex}}t=51.2$

So, the time required is 51.2 min = 51 min 12 sec

#### Page No 14.31:

#### Question 67:

A solid right circular cone of height 120 cm and radius 60 cm is placed in a right circular cylinder full of water of height 180 cm such that it touches the bottom . Find the volume of water left in the cylinder , if the radius of the cylinder is equal to the radius of te cone .

#### Answer:

Height of the cone, *h* = 120 cm

Radius of the cone, *r* = 60 cm

Height of the cylinder, H = 180 cm

Radius of the cylinder, R = 60 cm

Volume of the cylinder = ${\mathrm{\pi R}}^{2}\mathrm{H}=\mathrm{\pi}{\left(60\right)}^{2}\times 180{\mathrm{cm}}^{3}$

Volume of the cone = $\frac{1}{3}{\mathrm{\pi r}}^{2}\mathrm{h}=\frac{1}{3}\mathrm{\pi}{\left(60\right)}^{2}\times 120$

Volume of water left in the cylinder = Volume of cylinder − volume of the cone

$=\mathrm{\pi}{\left(60\right)}^{2}\times 180-\frac{1}{3}\mathrm{\pi}{\left(60\right)}^{2}\times 120\phantom{\rule{0ex}{0ex}}=\mathrm{\pi}{\left(60\right)}^{2}\left[180-40\right]\phantom{\rule{0ex}{0ex}}=\mathrm{\pi}\times 3600\left[140\right]\phantom{\rule{0ex}{0ex}}=1584000{\mathrm{cm}}^{3}\phantom{\rule{0ex}{0ex}}=1.584{\mathrm{m}}^{3}$

#### Page No 14.32:

#### Question 68:

A heap of rice in the form of a cone of diameter 9 m and height 3.5 m. Find the volume of rice. How much canvas cloth is required to cover the heap ?

#### Answer:

The heap of rice is in the form of a cone.

Diameter, d = 9 m

radius, r = $\frac{9}{2}\mathrm{m}$

height, h = 3.5 m

$Volume,V=\frac{1}{3}{\mathrm{\pi r}}^{2}\mathrm{h}\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\mathrm{\pi}{\left(\frac{9}{2}\right)}^{2}\times 3.5\phantom{\rule{0ex}{0ex}}=74.25{\mathrm{m}}^{3}$

Thus, volume of rice = 74.25 m^{3}

The canvas cloth required to cover the heap will be the curved surface area of the cone

$l=\sqrt{{h}^{2}+{r}^{2}}\phantom{\rule{0ex}{0ex}}l=\sqrt{3.{5}^{2}+{\left(\frac{9}{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}l=\sqrt{12.25+20.25}\phantom{\rule{0ex}{0ex}}l=5.7m$

$CSA=\mathrm{\pi rl}\phantom{\rule{0ex}{0ex}}=\mathrm{\pi}\times \left(\frac{9}{2}\right)\times 5.7\phantom{\rule{0ex}{0ex}}=80.62{\mathrm{m}}^{2}$

Hence, the canvas cloth required to cover the heap will be 80.62 m^{2}

#### Page No 14.32:

#### Question 69:

A cylindrical bucket of height 32 cm and base radius 18 cm is filled with sand . This bucket is emptied on the ground and a conical heap of sand is formed , If the height of the conical heap is 24 cm , find the radius and slant height of the heap.

#### Answer:

The cylinderical bucket has the height H = 32 cm

radius, R = 18 cm

Volume of the cylinderical bucket will be

$V=\mathrm{\pi}{\left(\mathrm{R}\right)}^{2}\mathrm{H}=\mathrm{\pi}{\left(18\right)}^{2}\times 32$

Height of the conical heap, h = 24 cm

Volume of cylinderical bucket = Volume of the conical heap

$\mathrm{\pi}{\left(\mathrm{R}\right)}^{2}\mathrm{H}=\frac{1}{3}{\mathrm{\pi r}}^{2}\mathrm{h}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(18\right)}^{2}\times 32=\frac{1}{3}{\mathrm{r}}^{2}\left(24\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{\left(18\right)}^{2}\times 32\times 3}{24}={\mathrm{r}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 1296={\mathrm{r}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 36=\mathrm{r}$

Hence, the radius of the conical heap = 36 cm

$\mathrm{Slant}\mathrm{height}l=\sqrt{{h}^{2}+{r}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow l=\sqrt{{24}^{2}+{36}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow l=\sqrt{576+1296}\phantom{\rule{0ex}{0ex}}\Rightarrow l=\sqrt{1872}\phantom{\rule{0ex}{0ex}}\Rightarrow l=43.2cm\phantom{\rule{0ex}{0ex}}$

Thus, the slant height l = 43.27 cm

#### Page No 14.32:

#### Question 70:

A hemispherical bowl of internal radius 9 cm is full of liquid . The liquid is to be filled into cylindrical shaped bottles each of radius 1.5 cm and height 4 cm . How many bottles are needed to empty the bowl ?

#### Answer:

The radius of the hemispherical bowl, R = 9 cm

Radius of the cylinderical bottles, r = 1.5 cm

Height of the bottles, h = 4 cm

Let the number of bottles required be n.

Volume of the hemispherical bowl = n × Volume of the cylinderical bottles

$\frac{VolumeofthehemisphericalBowl}{Volumeofthecylindericalbottles}=n\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{\displaystyle \frac{2}{3}{\mathrm{\pi R}}^{3}}}{{\mathrm{\pi r}}^{2}\mathrm{h}}=n\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\frac{2}{3}{{\displaystyle \left(9\right)}}^{3}}{{{\displaystyle \left(1.5\right)}}^{2}{\displaystyle \left(4\right)}}=n\phantom{\rule{0ex}{0ex}}\Rightarrow 54=n$

Hence, the 54 bottles are required.

#### Page No 14.32:

#### Question 71:

A factory manufactures 120,000 pencils daily . The pencil are cylindrical in shape each of length 25 cm and circumference of base as 1.5 cm . Determine the cost of colouring the curved surfaces of the pencils manufactured in one day at ₹0.05 per dm^{2}.

#### Answer:

Length of the pencil, h = 25 cm

circumference of the base = 1.5 cm

Curved surface area of the pencil which needs to be painted will be

$CSA=circumference\times height\phantom{\rule{0ex}{0ex}}=1.5\times 25c{m}^{2}\phantom{\rule{0ex}{0ex}}=37.5c{m}^{2}$

= 0.375 dm^{2}

Pencils manufactured in one day = 120000

So, the total area to be painted will be $120000\times 0.375d{m}^{2}=45000d{m}^{2}\phantom{\rule{0ex}{0ex}}$

Cost of painting this area will be $45000\times 0.05=Rs2250$

#### Page No 14.32:

#### Question 72:

The $\frac{3}{4}$th part of a conical vessel of internal radius 5 cm and height 24 cm is full of water. The water is emptied into a cylindrical vessel with internal radius 10 cm. Find the height of water in cylindrical vessel.

#### Answer:

Radius of conical vessel *r* = 5 cm

Height of conical vessel *h* = 24 cm

The volume of water = $\frac{3}{4}\times $volume of conical vessel.

$=\frac{3}{4}\times \frac{1}{3}\pi {r}^{2}h\phantom{\rule{0ex}{0ex}}=\frac{3}{4}\times \frac{1}{3}\pi \times 25\times 24\phantom{\rule{0ex}{0ex}}=150\mathrm{\pi}\phantom{\rule{0ex}{0ex}}$

Let *h*' be the height of cylindrical vessel, which filled by the water of conical vessel,

Radius of cylindrical vessel = 10 cm

Clearly,

Volume of cylindrical vessel = volume of water

$\pi {\left(10\right)}^{2}h=150\pi \phantom{\rule{0ex}{0ex}}\Rightarrow h=\frac{150\pi}{100\pi}\phantom{\rule{0ex}{0ex}}\Rightarrow h=1.5\mathrm{cm}$

Thus, the height of cylindrical vessel is 1.5 cm.

#### Page No 14.60:

#### Question 1:

A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of cylinder is 24 m. The height of the cylindrical portion is 11 m while the vertex of the cone is 16 m above the ground. Find the area of canvas required for the tent.

#### Answer:

We have a right circular cylinder surmounted by a cone.

Diameter of cylinder = 24 m, Height if cylindrical portion = 11 m and the vertex of the cone is 16 meters above the ground. We have to find the area of canvas required for the tent.

Suppose curved area of the cone portion is.

From the above figure the slant height of the top is given by

Now, Let us suppose that the curved area of cylinder is

Therefore, the area of canvas is given by

Hence,

#### Page No 14.60:

#### Question 2:

A rocket is in the form of a circular cylinder closed at the lower end with a cone of the same radius attached to the top. The cylinder is of radius 2.5 m and height 21 m and the cone has the slant height 8 m. Calculate the total surface area and the volume of the rocket.

#### Answer:

Given:

Radius of the cylinder, height of the cylinder, , slant height of the cone.We have to find total surface area and volume of the rocket

Let us assume that the area of the cone is.

The area of the cylinder is given by

Total areais

Now, we are going to find the volume of the rocket *V*.

Volume of the cone is given by

Volume of the cylinder is

Total volume of the cone is given by

Hence, the area and the volume of the rocket is

#### Page No 14.60:

#### Question 3:

A tent of height 77 dm is in the form of a right circular cylinder of diameter 36 m and height 44 dm surmounted by a right circular cone. Find the cost of the canvas at Rs. 3.50 per m^{2}

(Use π = 22/7).

#### Answer:

Given:

Height of the tent h = 77 dm = 7.7 m, diameter of cylinder

Height of the cylinder h_{1} = 44 dm = 4.4 m, height of cone h_{2} = 33 dm = 3.3 m

We have the following diagram

Radius

The curved area of cylinder is given by

${S}_{1}=2\mathrm{\pi rh}=2\times \frac{22}{7}\times 18\times 4.4\phantom{\rule{0ex}{0ex}}=497.82{\mathrm{m}}^{2}$

The slant height of the cone is

$l=\sqrt{{h}^{2}+{r}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{3.{3}^{2}+{18}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{334.89}\phantom{\rule{0ex}{0ex}}=18.3m$

The curved area of the cone is given by

${S}_{2}=\mathrm{\pi rl}=\frac{22}{7}\times 18\times 18.3\phantom{\rule{0ex}{0ex}}=1035.25{\mathrm{m}}^{2}$

The total area of the canvas required is given as

S = S_{1} + S_{2}

= 497.82 + 1035.25

=1533.07 m^{2}

Therefore the cost of the canvas at the rate of Rs 3.5 per square meter is given by

$1533.07\times 3.5\phantom{\rule{0ex}{0ex}}=Rs5365.745$

Hence the cost of the canvas is Rs 5365.745

#### Page No 14.60:

#### Question 4:

A toy is in the form of a cone surmounted on a hemisphere. The diameter of the base and the height of the cone are 6 cm and 4 cm, respectively. Determine the surface area of the toy. (Use π = 3.14)

#### Answer:

Given that, a toy is in the form of a cone surmounted on the hemisphere.

Diameter of the baseand the height of the cone, then we have to find the surface area of the toy.

We have the following figure

The radius of the base is

From the above figure, the slant height of the cone is

We know that when the surface area of the cone is, then

The surface area of the hemisphere is

Therefore the surface area of the toy is

Hence,

#### Page No 14.60:

#### Question 5:

A solid is in the form of a right circular cylinder, with a hemisphere at one end and a cone at the other end. The radius of the common base is 3.5 cm and the heights of the cylindrical and conical portions are 10 cm. and 6 cm, respectively. Find the total surface area of the solid. (Use π = 22/7)

#### Answer:

We have the following diagram

For cone, we have

Curved surface area of the cone is given as

For cylindrical part, we have

Curved surface area of the cylinder is

The surface area of the hemisphere is

Total surface area of the solid is given by

Hence the total surface area of the solid is

#### Page No 14.60:

#### Question 6:

A toy is in the shape of a right circular cylinder with a hemisphere on one end and a cone on the other. The radius and height of the cylindrical part are 5 cm and 13 cm respectively. The radii of the hemispherical and conical parts are the same as that of the cylindrical part. Find the surface area of the toy if the total height of the toy is 30 cm.

#### Answer:

We have the following diagram

For cylindrical part, we have

Therefore, the curved surface area of the cylinder is given by

For conical part, we have

Therefore, the curved surface area of the conical part is

For hemisphere, we have

Therefore the surface area of the hemisphere is

The total surface area of the toy is

Hence, total surface area of the toy is

#### Page No 14.60:

#### Question 7:

A cylindrical tub of radius 5 cm and length 9.8 cm is full of water. A solid in the form of a right circular cone mounted on a hemisphere is immersed in the tub. If the radius of the hemisphere is immersed in the tub. If the radius of the hemi-sphere is 3.5 cm and height of the cone outside the hemisphere is 5 cm, find the volume of the water left in the tub (Take π = 22/7)

#### Answer:

To find the volume of the water left in the tube, we have to subtract the volume of the hemisphere and cone from volume of the cylinder.

For right circular cylinder, we have

The volume of the cylinder is

For hemisphere and cone, we have

Therefore the total volume of the cone and hemisphere is

The volume of the water left in the tube is

Hence, the volume of the water left in the tube is

#### Page No 14.60:

#### Question 8:

A circus tent has cylindrical shape surmounted by a conical roof. The radius of the cylindrical base is 20 m. The heights of the cylindrical and conical portions are 4.2 m and 2.1 m respectively. Find the volume of the tent.

#### Answer:

Given that:

Radius of the cylindrical base

Height of the cylindrical portion

Height of the conical portion

The volume of the cylinder is given by the following formula

The volume of the conical portion is

Therefore, the total volume of the circus tent is

Hence, the volume of the circus tent is

#### Page No 14.60:

#### Question 9:

A petrol tank is a cylinder of base diameter 21 cm and length 18 cm fitted with conical ends each of axis length 9 cm. Determine the capacity of the tank.

#### Answer:

To find the total capacity of the tank, we have to add the volume of the cylinder and cone.

Diameter of the cylinder,

Radius of the cylinder, $r=\frac{d}{2}=\frac{21}{2}\mathrm{cm}$

Height of the cylinder, ${h}_{1}=18\mathrm{cm}$

Also, radius of cone, *r* = $\frac{21}{2}\mathrm{cm}$

Height of the cone, ${h}_{2}=9\mathrm{cm}$

Now,

Total capacity of the tank

= Volume of the cylinder + Volume of 2 cones

$=\mathrm{\pi}{r}^{2}{h}_{1}+2\times \frac{1}{3}\mathrm{\pi}{r}^{2}{h}_{2}\phantom{\rule{0ex}{0ex}}=\mathrm{\pi}{r}^{2}\left({h}_{1}+\frac{2}{3}{h}_{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{22}{7}\times {\left(\frac{21}{2}\right)}^{2}\times \left(18+\frac{2}{3}\times 9\right)\phantom{\rule{0ex}{0ex}}=\frac{22}{7}\times {\left(\frac{21}{2}\right)}^{2}\times 24\phantom{\rule{0ex}{0ex}}=8316{\mathrm{cm}}^{3}$

Hence the total capacity of the tank is 8316 cm^{3}.

#### Page No 14.60:

#### Question 10:

A conical hole is drilled in a circular cylinder of height 12 cm and base radius 5 cm. The height and the base radius of the cone are also the same. Find the whole surface and volume of the remaining cylinder.

#### Answer:

Given that:

We have the following diagram

Slant height of cone is given by

The total surface area of the remaining part is given by

The volume of the remaining part is given by

Hence,

#### Page No 14.60:

#### Question 11:

A tent is in the form of a cylinder of diameter 20 m and height 2.5 m, surmounted by a cone of equal base and height 7.5 m. Find the capacity of the tent and the cost of the canvas at Rs 100 per square metre.

#### Answer:

Given that:

Radius of the base

Height of the cylinder

Height of the cone

Slant height of the cone

The total capacity of the tent is given by

The total area of canvas required for the tent is

Therefore, the total cost of the canvas is

Hence, the total capacity and cost is

#### Page No 14.61:

#### Question 12:

A boiler is in the form of a cylinder 2 m long with hemispherical ends each of 2 metre diameter. Find the volume of the boiler.

#### Answer:

Given that:

Height of the cylinder

Radius of the cylinder and hemisphere are same and is given by

The volume of the cylinder is cylinder is

There are two hemispheres at each ends of the cylinder, therefore the volume of the two hemispheres is

Therefore, the total volume of the boiler is given by

Hence the volume of the boiler is

#### Page No 14.61:

#### Question 13:

A vessel is a hollow cylinder fitted with a hemispherical bottom of the same base. The depth of the cylinder is $\frac{14}{3}$ m and the diameter of hemisphere is 3.5 m. Calculate the volume and the internal surface area of the solid.

#### Answer:

Given that:

Radius of the same base

Height of the cylinder

The volume of the vessel is given by

The internal surface area of the solid is

Hence, the volume of the vessel and internal surface area of the solid is

#### Page No 14.61:

#### Question 14:

A solid is composed of a cylinder with hemispherical ends. If the whole length of the solid is 104 cm and the radius of each of the hemispherical ends is 7 cm, find the cost of polishing its surface at the rate of Rs 10 per dm^{2} .

#### Answer:

We have a solid composed of cylinder with hemispherical ends.

Radius of the two curved surfaces

Height of cylinder is *h*.

Total height of the body

So, total surface area is given by,

Change the units of curved surface area as,

Cost of polishing the surface is Rs 10 per.

So total cost,

#### Page No 14.61:

#### Question 15:

A cylindrical vessel of diameter 14 cm and height 42 cm is fixed symmetrically inside a similar vessel of diameter 16 cm and height 42 cm. The total space between the two vessels is filled with cork dust for heat insulation purposes. How many cubic centimeters of cork dust will be required?

#### Answer:

We have to find the volume of cork dust filled between the two vessels.

Radius of outer vessel

Radius of inner vessel

Height of the cylinder

So, volume of cork dust filled between the two vessels,

Volume of cork dust filled between the two vessels is 1980.

#### Page No 14.61:

#### Question 16:

A cylindrical road roller made of iron is 1 m long, Its internal diameter is 54 cm and the thickness of the iron sheet used in making the roller is 9 cm. Find the mass of the roller, if 1 cm^{3} of iron has 7.8 gm mass. (Use π = 3.14)

#### Answer:

We have to find the mass of the roller.

Radius of inner cylinder

Radius of outer cylinder

Length of the cylinder

So, volume of iron,

It is given that, 1 of iron has a mass of 7.8 gm.

So the mass of iron used,

#### Page No 14.61:

#### Question 17:

A vessel in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

#### Answer:

We have to find the inner surface area of a vessel which is in the form of a hemisphere mounted by a hollow cylinder.

Radius of hemisphere and cylinder

Total height of vessel

So, the inner surface area of a vessel,

#### Page No 14.61:

#### Question 18:

A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

#### Answer:

We have to find the total surface area of a toy which is a cone surmounted on a hemisphere.

Radius of hemisphere and the base of the cone

Height of the cone,

h = 15.5 − 3.5 = 12 cm

$\mathrm{slant}\mathrm{height}\left(l\right)=\sqrt{{\mathrm{h}}^{2}+{\mathrm{r}}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{12}^{2}+3.{5}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{156.25}\phantom{\rule{0ex}{0ex}}=12.5\mathrm{cm}$

So, total surface area of toy,

$S=\mathrm{\pi rl}+2{\mathrm{\pi r}}^{2}\phantom{\rule{0ex}{0ex}}=\mathrm{\pi r}\left(\mathrm{l}+2\mathrm{r}\right)\phantom{\rule{0ex}{0ex}}=\frac{22}{7}\times 3.5\left(12.5+2\times 3.5\right)\phantom{\rule{0ex}{0ex}}=214.5{\mathrm{cm}}^{2}$

#### Page No 14.61:

#### Question 19:

The difference between outside and inside surface areas of cylindrical metallic pipe 14 cm long is 44 m^{2}. If the pipe is made of 99 cm^{3} of metal, find the outer and inner radii of the pipe.

#### Answer:

We have to find the outer and inner radius of a hollow pipe.

Radius of inner pipe be

Radius of outer cylinder be

Length of the cylinder

Difference between the outer and the inner surface area is 44

So,

So,

…… (1)

So, volume of metal used is 99 , so,

Use equation (1) in the above to get,

Therefore,

…… (2)

Solve equation (1) and (2) to get,

#### Page No 14.61:

#### Question 20:

A right circular cylinder having diameter 12 cm and height 15 cm is full ice-cream. The ice-cream is to be filled in cones of height 12 cm and diameter 6 cm having a hemispherical shape on the top. Find the number of such cones which can be filled with ice-cream.

#### Answer:

We have to find the number of cones which can be filled using the ice cream in the cylindrical vessel.

Radius of the cylinder

Height of cylinder

Radius of cone and the hemisphere on it

Height of cone

Let ‘*n*’ number of cones filled. So we can write it as,

So,

Now put the values to get,

Therefore,

#### Page No 14.61:

#### Question 21:

A solid iron pole having cylindrical portion 110 cm high and of base diameter 12 cm is surmounted by a cone 9 cm high. Find the mass of the pole, given that the mass of 1 cm^{3} of iron is 8 gm.

#### Answer:

We have to find the mass of a pole having a cylindrical base surmounted by a cone.

Radius of cone and cylinder

Height of cylinder

Height of cone

So volume of the pole is,

Put the values to get,

Mass of 1 of iron is 8 gm.

Therefore mass of the iron,

#### Page No 14.61:

#### Question 22:

A solid toy is in the form of a hemisphere surmounted by a right circular cone. height of the cone is 2 cm and the diameter of the base is 4 cm. If a right circular cylinder circumscribes the toy, find how much more space it will cover.

#### Answer:

We have to find the remaining volume of the cylinder when the toy is inserted into it. The toy is a hemisphere surmounted by a cone.

Radius of cone, cylinder and hemisphere

Height of cone

Height of the cylinder

So the remaining volume of the cylinder when the toy is inserted into it,

Put the values to get,

#### Page No 14.61:

#### Question 23:

A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottoms. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.

#### Answer:

We have to find the remaining volume of water left in the cylinder when the solid is inserted into it. The solid is a hemisphere surmounted by a cone.

Radius of cone, cylinder and hemisphere

Height of cone

Height of the cylinder

So the remaining volume of water left in the cylinder when the solid is inserted into it,

Put the values to get,

#### Page No 14.61:

#### Question 24:

A cylindrical vessel with internal diameter 10 cm and height 10.5 cm is full of water. A solid cone of base diameter 7 cm and height 6 cm is completely immersed in water. Find the value of water (i) displaced out of the cylinder (ii) left in the cylinder. (Take π 22/7)

#### Answer:

We have a cylindrical vessel in which a cone is inserted. We have,

Radius of the cylinder

Radius of cone

Height of cylinder

Height of cone

(i) We have to find the volume of water displaced from the cylinder when cone is inserted.

So,

So volume of water displaced,

(ii) We have to find the volume of water remaining in the cylinder.

So volume of the water left in the cylinder,

#### Page No 14.61:

#### Question 25:

A hemispherical depression is cut out from one face of a cubical wooden block of edge 21 cm, such that the diameter of the hemisphere is equal to the edge of the cube. Determine the volume and total surface area of the remaining block.

#### Answer:

We have to find the remaining volume and surface area of a cubical box when a hemisphere is cut out from it.

Edge length of cube

Radius of hemisphere

Therefore volume of the remaining block,

So,

So, remaining surface area of the box,

Therefore,

Put the values to get the remaining surface area of the box,

#### Page No 14.61:

#### Question 26:

A toy is in the form of a hemisphere surmounted by a right circular cone of the same base radius as that of the hemisphere. If the radius of the base of the cone is 21 cm and its volume is 2/3 of the volume of hemisphere, calculate the height of the cone and the surface area of the toy.

(Use $\mathrm{\pi}=22/7$).

#### Answer:

**Solution:**

Let the height of the conical part be *h*.

Radius of the cone = Radius of the hemisphere = *r* = 21 cm

The toy can be diagrammatically represented as

Volume of the cone =

Volume of the hemisphere =

According to given information:

Volume of the cone × Volume of the hemisphere

Thus, surface area of the toy = Curved surface area of cone + Curved surface area of hemisphere

#### Page No 14.62:

#### Question 27:

A solid is in the shape of a cone surmounted on a hemisphere, the radius of each of them is being 3.5 cm and the total height of solid is 9.5 cm. Find the volume of the solid. (Use π = 22/7).

#### Answer:

Height of cone = 9.5 − 3.5 = 6 cm

Volume of the solid = Volume of cone + Volume of hemisphere

$=\frac{1}{3}\mathrm{\pi}{r}^{2}h+\frac{2}{3}\mathrm{\pi}{r}^{3}\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\times \frac{22}{7}\times {\left(3.5\right)}^{2}\times 6+\frac{2}{3}\times \frac{22}{7}\times {\left(3.5\right)}^{3}\phantom{\rule{0ex}{0ex}}=77+89.83\phantom{\rule{0ex}{0ex}}=166.83{\mathrm{cm}}^{3}$

#### Page No 14.62:

#### Question 28:

An wooden toy is made by scooping out a hemisphere of same radius from each end of a solid cylinder. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the volume of wood in the toy. (Use π = 22/7).

#### Answer:

Volume of wood in the toy = Volume of cylinder − 2(Volume of hemisphere)

$=\mathrm{\pi}{r}^{2}h-2\times \frac{2}{3}\mathrm{\pi}{r}^{3}\phantom{\rule{0ex}{0ex}}=\frac{22}{7}\times {\left(3.5\right)}^{2}\times 10-2\times \frac{2}{3}\times \frac{22}{7}\times {\left(3.5\right)}^{3}\phantom{\rule{0ex}{0ex}}=385-179.67\phantom{\rule{0ex}{0ex}}=205.33{\mathrm{cm}}^{3}$

#### Page No 14.62:

#### Question 29:

The largest possible sphere is carved out of a wooden solid cube of side 7 cm. Find the volume of the wood left. (Use $\mathrm{\pi}=\frac{22}{7}$). [CBSE 2014]

#### Answer:

The radius of the largest possible sphere is carved out of a wooden solid cube is equal to the half of the side of the cube.

Radius of the sphere = $\frac{7}{2}=3.5$

Volume of the wood left = Volume of cube − Volume of sphere

$={\left(\mathrm{Side}\right)}^{3}-\frac{4}{3}\mathrm{\pi}{r}^{3}\phantom{\rule{0ex}{0ex}}={7}^{3}-\frac{4}{3}\times \frac{22}{7}\times {\left(3.5\right)}^{3}\phantom{\rule{0ex}{0ex}}=343-179.67\phantom{\rule{0ex}{0ex}}=163.33{\mathrm{cm}}^{3}$

#### Page No 14.62:

#### Question 30:

From a solid cylinder of height 2.8 cm and diameter 4.2 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid. (Take $\mathrm{\pi}=\frac{22}{7}$). [CBSE 2014]

#### Answer:

Total surface area of the remaining solid

= CSA of cylindrical part + CSA of conical part + Area of cylindrical base

$=2\mathrm{\pi}rh+\mathrm{\pi}rl+\mathrm{\pi}{r}^{2}\phantom{\rule{0ex}{0ex}}=2\times \frac{22}{7}\times \frac{4.2}{2}\times 2.8+\frac{22}{7}\times \frac{4.2}{2}\times \sqrt{{\left(\frac{4.2}{2}\right)}^{2}+{\left(2.8\right)}^{2}}+\frac{22}{7}\times {\left(\frac{4.2}{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}=2\times \frac{22}{7}\times 2.1\times 2.8+\frac{22}{7}\times 2.1\times \sqrt{{\left(2.1\right)}^{2}+{\left(2.8\right)}^{2}}+\frac{22}{7}\times {\left(2.1\right)}^{2}\phantom{\rule{0ex}{0ex}}=2\times \frac{22}{7}\times 2.1\times 2.8+\frac{22}{7}\times 2.1\times 3.5+\frac{22}{7}\times {\left(2.1\right)}^{2}\phantom{\rule{0ex}{0ex}}=36.96+23.1+13.86\phantom{\rule{0ex}{0ex}}=73.92{\mathrm{cm}}^{2}$

The total surface area of the remaining solid is 73.92 cm^{2}.

#### Page No 14.62:

#### Question 31:

The largest cone is curved out from one face of solid cube of side 21 cm. Find the volume of the remaining solid. [CBSE 2015]

#### Answer:

The radius of the largest possible cone is carved out of a solid cube is equal to the half of the side of the cube.

Also, the height of the cone is equal to the side of the cube.

Radius of the cone = $\frac{21}{2}=10.5$ cm

Volume of the remaining solid = Volume of cube − Volume of cone

$={\left(\mathrm{Side}\right)}^{3}-\frac{1}{3}\mathrm{\pi}{r}^{2}h\phantom{\rule{0ex}{0ex}}={\left(21\right)}^{3}-\frac{1}{3}\times \frac{22}{7}\times {\left(10.5\right)}^{2}\times 21\phantom{\rule{0ex}{0ex}}=9261-2425.5\phantom{\rule{0ex}{0ex}}=6835.5{\mathrm{cm}}^{3}$

**Disclaimer**: The answer given in the book is not correct.

#### Page No 14.62:

#### Question 32:

A solid wooden toy is in the form of a hemisphere surmounted by a cone of same radius. The radius of hemisphere is 3.5 cm and the total wood used in the making of toy is $166\frac{5}{6}$ cm^{3}. Find the height of the toy. Also, find the cost of painting the hemispherical part of the toy at the rate of Rs 10 per cm^{2} . (Take$\mathrm{\pi}=\frac{22}{7}$). [CBSE 2015]

#### Answer:

Volume of solid wooden toy = $166\frac{5}{6}$ cm^{3}

⇒ Volume of cone + Volume of hemisphere = $166\frac{5}{6}$

$\Rightarrow \frac{1}{3}\mathrm{\pi}{r}^{2}h+\frac{2}{3}\mathrm{\pi}{r}^{3}=\frac{1001}{6}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{3}\times \frac{22}{7}\times {\left(3.5\right)}^{2}\times h+\frac{2}{3}\times \frac{22}{7}\times {\left(3.5\right)}^{3}=\frac{1001}{6}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{3}\times \frac{22}{7}\times {\left(3.5\right)}^{2}\left(h+7\right)=\frac{1001}{6}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{38.5}{3}\left(h+7\right)=\frac{1001}{6}\phantom{\rule{0ex}{0ex}}\Rightarrow h+7=13\phantom{\rule{0ex}{0ex}}\Rightarrow h=6\mathrm{cm}$

Height of the solid wooden toy

= Height of cone + Radius of hemisphere

= 6 + 3.5

= 9.5 cm

Now, curved surface area of hemisphere = $2\mathrm{\pi}{r}^{2}$

$=2\times \frac{22}{7}\times {\left(3.5\right)}^{2}\phantom{\rule{0ex}{0ex}}=77{\mathrm{cm}}^{2}$

Cost of painting the hemispherical part of the toy = 10 ⨯ 77 = Rs 770

#### Page No 14.62:

#### Question 33:

In the given figure, from a cuboidal solid metalic block, of dimensions 15 cm ⨯ 10 cm ⨯ 5 cm, a cylindrical hole of diameter 7 cm is drilled out. Find the surface area of the remaining block. (Take $\mathrm{\pi}=\frac{22}{7}$). [CBSE 2015]

#### Answer:

Surface area of the remaining block

= Total Surface area of cubic block + Curved Surface area of cylinder − 2(Area of circular base)

$=2\left(lb+bh+lh\right)+2\mathrm{\pi}rh-2\mathrm{\pi}{r}^{2}\phantom{\rule{0ex}{0ex}}=2\left(15\times 10+10\times 5+15\times 5\right)+2\times \frac{22}{7}\times \frac{7}{2}\times 5-2\times \frac{22}{7}\times {\left(\frac{7}{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}=2\times 275+110-77\phantom{\rule{0ex}{0ex}}=583{\mathrm{cm}}^{2}$

#### Page No 14.62:

#### Question 34:

A building is in the form of a cylinder surmounted by a hemi-spherical vaulted dome and contains $41\frac{19}{21}{\mathrm{m}}^{3}$ of air. If the internal diameter of dome is equal to its total height above the floor , find the height of the building ?

#### Answer:

let the total height of the building be H m.

let the radius of the base be r m. Therefore the radius of the hemispherical dome is r m.

Now given that internal diameter = total height

$\Rightarrow 2r=H$

Total height of the building = height of the cylinder +radius of the dome

⇒ H = h + r

⇒ 2r = h + r

⇒ r = h

Volume of the air inside the building = volume of the cylinder+ volume of the hemisphere

$\Rightarrow 41\frac{19}{21}={\mathrm{\pi r}}^{2}\mathrm{h}+\frac{2}{3}{\mathrm{\pi r}}^{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{880}{21}={\mathrm{\pi h}}^{2}\mathrm{h}+\frac{2}{3}{\mathrm{\pi h}}^{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{880}{21}={\mathrm{\pi h}}^{3}\left(1+\frac{2}{3}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{880}{21}={\mathrm{\pi h}}^{3}\left(\frac{5}{3}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{h}=2\mathrm{m}$

Hence, height of the building H = 2 × 2 = 4m

#### Page No 14.62:

#### Question 35:

A pen stand made of wood is in the shape of a cuboid with four conical depression and a cubical depression to hold the pens and pins , respectively . The dimension of the cuboid are $10\mathrm{cm}\times 5\mathrm{cm}\times 4\mathrm{cm}$. The radius of each of the conical depression is 0.5 cm and the depth is 2.1 cm . The edge of the cubical depression is 3 cm . Find the volume of the wood in the entire stand.

#### Answer:

The dimensions of the cuboid = 10 cm × 5 cm × 4 cm

Volume of the total cuboid = 10 cm × 5 cm × 4 cm = 200 cm^{3}

Radius of the conical depressions, r = 0.5 cm

Depth, h = 2.1 cm

Volume of the conical depression = $\frac{1}{3}{\mathrm{\pi r}}^{2}\mathrm{h}=\frac{1}{3}\mathrm{\pi}{\left(0.5\right)}^{2}\left(2.1\right)=0.5495$ cm^{3}

Edge of cubical depression, a = 3 cm

Volume of the cubical depression = ${a}^{3}={3}^{3}=27c{m}^{3}$

Volume of wood used to make the entire stand = Volume of the total cuboid − volume of conical depression − volume of cubical depression

$=200-4\times 0.5495-27\phantom{\rule{0ex}{0ex}}=170.802c{m}^{3}\phantom{\rule{0ex}{0ex}}$

#### Page No 14.62:

#### Question 36:

A building is in the form of a cylinder surrounded by a hemispherical dome. The base diameter of the dome is equal to $\frac{2}{3}$ of the total height of the building . Find the height of the building , if it contains $67\frac{1}{21}{m}^{3}$ of air.

#### Answer:

Let the radius of the dome be r.

Diameter be d.

Let the height of the building be H.

Given $d=\frac{2}{3}H$

$\Rightarrow 2r=\frac{2}{3}H\phantom{\rule{0ex}{0ex}}\Rightarrow r=\frac{H}{3}\phantom{\rule{0ex}{0ex}}\Rightarrow 3r=H$

Also, h + r = H

$\Rightarrow 3r=h+r\phantom{\rule{0ex}{0ex}}\Rightarrow 2r=h\phantom{\rule{0ex}{0ex}}\Rightarrow r=\frac{h}{2}$

Volume of air = Volume of air in the cylinder + Volume of air int he hemispherical dome

$\Rightarrow {\mathrm{\pi r}}^{2}\mathrm{h}+\frac{2}{3}{\mathrm{\pi r}}^{3}=67\frac{1}{21}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{\pi}{\left(\frac{\mathrm{h}}{2}\right)}^{2}\mathrm{h}+\frac{2}{3}\mathrm{\pi}{\left(\frac{h}{2}\right)}^{3}=\frac{1408}{21}\phantom{\rule{0ex}{0ex}}\Rightarrow {h}^{3}=\frac{11264}{176}=64\phantom{\rule{0ex}{0ex}}\Rightarrow h=4m\phantom{\rule{0ex}{0ex}}$

Hence, the radius will be $r=\frac{h}{2}=\frac{4}{2}=2m$

Height of the building, H = 3r = $3\times 2=6m$

#### Page No 14.62:

#### Question 37:

A solid toy s in the form of a hemisphere surrounded by a right circular cone . The height of cone is 4 cm and the diameter of the base is 8 cm . Determine the volume of the toy. If a cube circumscribes the toy , then find the difference of the volumes of cube and the toy .

#### Answer:

The height of the cone, h = 4 cm

Diameter of the base, d = 8 cm

Radius of the cone, r = 4 cm

lateral side will be

$l=\sqrt{{h}^{2}+{r}^{2}}\phantom{\rule{0ex}{0ex}}l=\sqrt{{\left(4\right)}^{2}+{\left(4\right)}^{2}}=4\sqrt{2}\phantom{\rule{0ex}{0ex}}$

Volume of the toy = volume of hemisphere + volume of cone

$\Rightarrow V=\frac{2}{3}{\mathrm{\pi r}}^{3}+\frac{1}{3}{\mathrm{\pi r}}^{2}\mathrm{h}\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{\pi r}}^{2}}{3}\left[2\mathrm{r}+\mathrm{h}\right]\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi}{\left(4\right)}^{2}}{3}\left[2\times 4+4\right]\phantom{\rule{0ex}{0ex}}=201.14{\mathrm{cm}}^{3}$

When the cube circumscribes the toy, then

Volume of the cube = ${a}^{3}={\left(8\right)}^{3}=512c{m}^{3}$

$\mathrm{Volume}\mathrm{of}\mathrm{cube}-\mathrm{volume}\mathrm{of}\mathrm{toy}=512-201.14=310.86c{m}^{3}$

Total surface area of the toy = curved surface area of the hemisphere + curved surface area of the cone

$=2\mathrm{\pi}{r}^{2}+\mathrm{\pi}rl\phantom{\rule{0ex}{0ex}}=2\mathrm{\pi}{\left(4\right)}^{2}+\mathrm{\pi}\times 4\times 4\sqrt{2}\phantom{\rule{0ex}{0ex}}=\mathrm{\pi}{\left(4\right)}^{2}\left[2+\sqrt{2}\right]\phantom{\rule{0ex}{0ex}}=171.68{\mathrm{cm}}^{2}$

#### Page No 14.63:

#### Question 38:

A circus tent is in the shape of cylinder surmounted by a conical top of same diameter. If their common diameter is 56 m, the height of the cylindrical part is 6 m and the total height of the tent above the ground is 27 m, find the area of the canvas used in making the tent.

#### Answer:

Total height of the tent above the ground = 27 m

Height of the cylinderical part, ${h}_{1}$= 6 m

Height of the conical part, ${h}_{2}$ = 21 m

Diameter = 56 m

Radius = 28 m

Curved surface area of the cylinder, CSA_{1} = $2\mathrm{\pi}r{h}_{1}=2\mathrm{\pi}\times 28\times 6=336\mathrm{\pi}$

Curved surface area of the cylinder, CSA_{2} will be

$\mathrm{\pi}rl=\mathrm{\pi}r\left(\sqrt{{h}^{2}+{r}^{2}}\right)=\mathrm{\pi}\times 28\times \left(\sqrt{{21}^{2}+{28}^{2}}\right)=28\mathrm{\pi}\left(\sqrt{441+784}\right)\phantom{\rule{0ex}{0ex}}=28\mathrm{\pi}\times 35\phantom{\rule{0ex}{0ex}}=980\mathrm{\pi}$

Total curved surface area = CSA of cylinder + CSA of cone

= CSA_{1 }+ CSA_{2}

$=336\mathrm{\pi}+980\mathrm{\pi}\phantom{\rule{0ex}{0ex}}=1316\mathrm{\pi}\phantom{\rule{0ex}{0ex}}=4136{\mathrm{m}}^{2}$

Thus, the area of the canvas used in making the tent is 4136 m^{2 }.

#### Page No 14.78:

#### Question 1:

A bucket has top and bottom diameter of 40 cm and 20 cm respectively. Find the volume of the bucket if its depth is 12 cm. Also, find the cost of tin sheet used for making the bucket at the rate of Rs. 1.20 per dm^{2 }. (Use π = 3.14)

#### Answer:

The radii of the top and bottom circles are *r*_{1}* = *20 cm and *r*_{2} = 10 cm respectively. The height of the bucket is *h *= 12 cm. Therefore, the volume of the bucket is

The slant height of the bucket is

The total surface area of the bucket is

The total cost of tin sheet used for making the bucket is

#### Page No 14.78:

#### Question 2:

A frustum of a right circular cone has a diameter of base 20 cm, of top 12 cm, and height 3 cm. Find the area of its whole surface and volume.

#### Answer:

The radii of the bottom and top circles are *r*_{1}* = *10 cm and *r*_{2} = 6 cm respectively. The height of the frustum cone is *h *= 3 cm. Therefore, the volume of the bucket is

Hence

The slant height of the bucket is

cm

The total surface area of the frustum cone is

Square cm

Square cm

Hence

#### Page No 14.78:

#### Question 3:

The slant height of the frustum of a cone is 4 cm and the perimeters of its circular ends are 18 cm and 6 cm. Find the curved surface of the frustum.

#### Answer:

The slant height of the frustum of the cone is *l *= 4 cm. The perimeters of the circular ends are 18 cm and 6 cm. Let the radii of the bottom and top circles are *r*_{1} cm and *r*_{2} cm respectively. Then, we have

The curved surface area of the frustum cone is

= 48 Square cm

#### Page No 14.78:

#### Question 4:

The perimeters of the ends of a frustum of a right circular cone are 44 cm and 33 cm. If the height of the frustum be 16 cm, find its volume, the slant surface and the total surface.

#### Answer:

The height of the frustum of the cone is *h *= 16 cm. The perimeters of the circular ends are 44 cm and 33 cm. Let the radii of the bottom and top circles are *r*_{1} cm and *r*_{2} cm respectively. Then, we have

The slant height of the bucket is

The curved/slant surface area of the frustum cone is

Hence

The volume of the frustum of the cone is

Hence

The total surface area of the frustum cone is

= 860.25 Square cm

Hence

#### Page No 14.78:

#### Question 5:

If the radii of the circular ends of a conical bucket which is 45 cm high be 28 cm and 7 cm, find the capacity of the bucket. (Use π = 22/7).

#### Answer:

The height of the conical bucket is *h *= 45 cm. The radii of the bottom and top circles are *r*_{1} = 28cm and *r*_{2} =7cm respectively.

The volume/capacity of the conical bucket is

Hence

#### Page No 14.78:

#### Question 6:

The height of a cone is 20 cm. A small cone is cut off from the top by a plane parallel to the base. If its volume be 1/125 of the volume of the original cone, determine at what height above the base the section is made.

#### Answer:

We have the following situation as shown in the figure

V

Let VAB be a cone of height *h*_{1}_{ }*= *VO_{1}* *=20cm. Then from the symmetric triangles VO _{1}A and VOA_{1}, we have

It is given that, volume of the cone VA_{1}O istimes the volume of the cone VAB. Hence, we have

Hence, the height at which the section is made is 20 − 4 = 16 cm.

#### Page No 14.78:

#### Question 7:

If the radii of the circular ends of a bucket 24 cm high are 5 cm and 15 cm respectively, find the surface area of the bucket.

#### Answer:

The height of the conical bucket is *h *= 24 cm. The radii of the bottom and top circles are *r*_{1} = 15cm and *r*_{2} = 5cm respectively.

The slant height of the bucket is

The curved surface area of the bucket is

cm^{2}

Hence the curved surface area of the bucket is

#### Page No 14.79:

#### Question 8:

The radii of the circular bases of a frustum of a right circular cone are 12 cm and 3 cm and the height is 12 cm. Find the total surface area and the volume of the frustum.

#### Answer:

The height of the frustum cone is *h *= 12 cm. The radii of the bottom and top circles are *r*_{1} = 12cm and *r*_{2} = 3cm respectively.

The slant height of the frustum cone is

The total surface area of the frustum cone is

Hence

The volume of the frustum cone is

Hence

#### Page No 14.79:

#### Question 9:

A tent consists of a frustum of a cone capped by a cone. If the radii of the ends of the frustum be 13 m and 7 m , the height of the frustum be 8 m and the slant height of the conical cap be 12 m, find the canvas required for the tent.

(Take : π = 22/7)

#### Answer:

The height of the frustum cone is *h *= 8 m. The radii of the end circles of the frustum are *r*_{1} = 13m and *r*_{2} =7m.

The slant height of the frustum cone is

The curved surface area of the frustum is

The base-radius of the upper cap cone is 7m and the slant height is 12m. Therefore, the curved surface area of the upper cap cone is

Hence, the total canvas required for the tent is

Hence total canvas is

#### Page No 14.79:

#### Question 10:

A milk container of height 16 cm is made of metal sheet in the form of a frustum of a cone with radii of its lower and upper ends as 8 cm and 20 cm respectively . Find the cost of milk at the rate of ₹44 per litre which the container can hold.

#### Answer:

Radius, r_{1} = 8 cm and r_{2} = 20 cm

Height, h = 16 cm

Volume of milk that the container can hold

$V=\frac{1}{3}\mathrm{\pi h}\left({\mathrm{r}}_{1}^{2}+{\mathrm{r}}_{1}{\mathrm{r}}_{2}+{\mathrm{r}}_{2}^{2}\right)\phantom{\rule{0ex}{0ex}}\mathrm{V}=\frac{1}{3}\mathrm{\pi}\times 16\left({8}^{2}+8\times 20+{20}^{2}\right)\phantom{\rule{0ex}{0ex}}\mathrm{V}=10459.42{\mathrm{cm}}^{3}=10.45942\mathrm{liters}$

Cost of milk will be $10.45942\times 44=Rs460.21$

#### Page No 14.79:

#### Question 11:

A bucket is in the form of a frustum of a cone of height 30 cm with radii of its lower and upper ends as 10 cm and 20 cm respectively. Find the capacity and surface area of the bucket. Also, find the cost of milk which can completely fill the container , at thr rate of ₹25 per litre. (Use $\mathrm{\pi}=3.14).$

#### Answer:

Height of the bucket, h = 30 cm

Radii r_{1} = 10 cm and r_{2} = 20 cm

Capacity of the bucket,

$V=\frac{1}{3}\mathrm{\pi h}\left({\mathrm{r}}_{1}^{2}+{\mathrm{r}}_{1}{\mathrm{r}}_{2}+{\mathrm{r}}_{2}^{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\mathrm{\pi}\times 30\left({10}^{2}+10\times 20+{20}^{2}\right)\phantom{\rule{0ex}{0ex}}=21980{\mathrm{cm}}^{3}\phantom{\rule{0ex}{0ex}}=21.980\mathrm{liters}$

$l=\sqrt{{h}^{2}+{\left({r}_{2}-{r}_{1}\right)}^{2}}\phantom{\rule{0ex}{0ex}}l=\sqrt{{30}^{2}+{\left(20-10\right)}^{2}}=10\sqrt{10}$

Surface area of the bucket

$S=CSA+areaofthebase\phantom{\rule{0ex}{0ex}}S=\mathrm{\pi}\left({\mathrm{r}}_{1}+{\mathrm{r}}_{2}\right)\mathrm{l}+{{\mathrm{\pi r}}_{1}}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{S}=\mathrm{\pi}\left(10+20\right)10\sqrt{10}+\mathrm{\pi}{\left(10\right)}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{S}=2978.86+314=3292.86{\mathrm{cm}}^{2}$

Cost of milk which can completely fill the container at Rs 25/litre

= 21.980 × 25

= Rs 549.50

#### Page No 14.79:

#### Question 12:

A bucket is in the form of a frustum of a cone with a capacity of 12308.8 cm^{3} of water. The radii of the top and bottom circular ends are 20 cm and 12 cm respectively. Find the height of the bucket and the area of the metal sheet used in its making. (Use π = 3.14).

#### Answer:

Let the depth of the bucket is *h *cm. The radii of the top and bottom circles of the frustum bucket are *r*_{1} =20cm and *r*_{2} =12cm respectively.

The volume/capacity of the bucket is

Given that the capacity of the bucket isCubic cm. Thus, we have

Hence, the height of the bucket is

The slant height of the bucket is

The surface area of the used metal sheet to make the bucket is

Hence

#### Page No 14.79:

#### Question 13:

A bucket made of a aluminium sheet is of height 20 cm and its upper and lower ends are of radius 25 cm and 10 cm respectively. Find the cost of making the bucket if the aluminium sheet costs Rs 70 per 100 cm^{2} . (Use π = 3.14).

#### Answer:

The height of the bucket is 20cm. The radii of the upper and lower circles of the bucket are *r*_{1} =25cm and *r*_{2} =10cm respectively.

The slant height of the bucket is

The surface area of the used aluminium sheet to make the bucket is

Therefore, the total cost of making the bucket is

Hence the total cost is

#### Page No 14.79:

#### Question 14:

The radii of the circular ends of a solid frustum of a cone are 33 cm and 27 cm and its slant height is 10 cm. Find its total surface area.

#### Answer:

The slant height of the frustum of a cone is *l=*10cm. The radii of the upper and lower circles of the bucket are *r*_{1} =33cm and *r*_{2} =27cm respectively.

The total surface area of the frustum of the cone is

Hence total surface area is

#### Page No 14.79:

#### Question 15:

A bucket made up of a metal sheet is in the form of a frustum of a cone of height 16 cm with diameters of its lower and upper ends as 16 cm and 40 cm respectively. Find the volume of the bucket. Also, find the cost of the bucket if the cost of metal sheet used in Rs 20 per 100 cm^{2} .

(Use π = 3.14)

#### Answer:

The height of the bucket is *h*=16cm. The radii of the upper and lower circles of the bucket are *r*_{1} =20 cm and *r*_{2} = 8 cm respectively.

The slant height of the bucket is

The volume of the bucket is

Hence the volume of the bucket is

The surface area of the used metal sheet to make the bucket is

Therefore, the total cost of making the bucket is

#### Page No 14.79:

#### Question 16:

A solid is in the shape of a frustum of a cone, The diameters of the two circular ends are 60 cm and 36 cm and the height is 9 cm. Find the area of its whole surface and the volume.

#### Answer:

The height of the frustum of a cone is *h*=9cm. The radii of the upper and lower circles of the frustum of the cone are *r*_{1} =30cm and *r*_{2} =18cm respectively.

The slant height of the frustum of the cone is

The volume of the frustum of the cone is

The total surface area of the frustum of the cone is

#### Page No 14.79:

#### Question 17:

A milk container is made of metal sheet in the shape of frustum of cone whose volume is 10459$\frac{3}{7}$cm^{3} . The radii of its lower and upper circular ends are 8 cm and 20 cm respectively. Find the cost of metal sheet used in making the container at the rate of Rs. 1.40 per cm^{2} .

(Use π = 22/7)

#### Answer:

Let the depth of the container is *h *cm. The radii of the top and bottom circles of the container are *r*_{1} =20cm and *r*_{2} =8cm respectively.

The volume/capacity of the container is

Given that the capacity of the bucket is. Thus, we have

cm

Hence, the height of the container is 16 cm.

The slant height of the container is

The surface area of the used metal sheet to make the container is

The cost to make the container is

#### Page No 14.79:

#### Question 18:

A solid cone of base radius 10 cm is cut into two parts through the mid-point of its height, by a plane parallel to its base. Find the ratio in the volumes of two parts of the cone.

#### Answer:

Let the height of the cone be *H*.

Now, the cone is divided into two parts by the parallel plane

∴ OC = CA

Now, In ∆OCD and OAB

∠OCD = OAB (Corresponding angles)

∠ODC = OBA (Corresponding angles)

By AA-similarity criterion ∆OCD ∼ ∆OAB

$\therefore \frac{\mathrm{CD}}{\mathrm{AB}}=\frac{\mathrm{OC}}{\mathrm{OA}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{CD}}{10}=\frac{H}{2\times H}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{CD}=5\mathrm{cm}$

$\frac{\mathrm{Volume}\mathrm{of}\mathrm{first}\mathrm{part}}{\mathrm{Volume}\mathrm{of}\mathrm{second}\mathrm{part}}=\frac{{\displaystyle \frac{1}{3}}\mathrm{\pi}{\left(\mathrm{CD}\right)}^{2}\left(\mathrm{OC}\right)}{{\displaystyle \frac{1}{3}}\mathrm{\pi CA}\left[{\left(\mathrm{AB}\right)}^{2}+\left(\mathrm{AB}\right)\left(\mathrm{CD}\right)+{\mathrm{CD}}^{2}\right]}\phantom{\rule{0ex}{0ex}}=\frac{{\displaystyle {\left(5\right)}^{2}}}{{\displaystyle \left[{\left(10\right)}^{2}+\left(10\right)\left(5\right)+{5}^{2}\right]}}\phantom{\rule{0ex}{0ex}}=\frac{25}{100+50+25}\phantom{\rule{0ex}{0ex}}=\frac{25}{175}\phantom{\rule{0ex}{0ex}}=\frac{1}{7}$

#### Page No 14.79:

#### Question 19:

A bucket open at the top, and made up of a metal sheet is in the form of a frustum of a cone. The depth of the bucket is 24 cm and the diameters of its upper and lower circular ends are 30 cm and 10 cm respectively. Find the cost of metal sheet used in it at the rate of Rs. 10 per 100 cm^{2} .

(Use π = 3.14).

#### Answer:

The slant height of the bucket is given by

$l=\sqrt{{h}^{2}+{\left(R-r\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(24\right)}^{2}+{\left(15-5\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{576+100}\phantom{\rule{0ex}{0ex}}=\sqrt{676}\phantom{\rule{0ex}{0ex}}=26\mathrm{cm}$

Surface area of bucket

= Curved surface area of bucket + Area of the smaller circlular base

$=\mathrm{\pi}l\left(R+r\right)+\mathrm{\pi}{r}^{2}\phantom{\rule{0ex}{0ex}}=3.14\times 26\times \left(15+5\right)+3.14\times 5\times 5\phantom{\rule{0ex}{0ex}}=1632.8+78.5\phantom{\rule{0ex}{0ex}}=1711.3{\mathrm{cm}}^{2}$

Cost of metal sheet used = $\frac{10}{100}\times 1711.3=\mathrm{Rs}171.13$

#### Page No 14.79:

#### Question 20:

In the given figure, from the top of a solid cone of height 12 cm and base radius 6 cm, a cone of height 4 cm is removed by a plane parallel to the base. Find the total surface area of the remaining solid. (Use $\mathrm{\pi}=\frac{22}{7}\mathrm{and}\sqrt{5}=2.236$). [CBSE 2015]

#### Answer:

Now, In ∆OCD and OAB

∠OCD = OAB (Corresponding angles)

∠ODC = OBA (Corresponding angles)

By AA-similarity criterion ∆OCD ∼ ∆OAB

$\therefore \frac{\mathrm{CD}}{\mathrm{AB}}=\frac{\mathrm{OC}}{\mathrm{OA}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{CD}}{6}=\frac{4}{12}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{CD}=2\mathrm{cm}$

The slant height of the bucket is given by

$l=\sqrt{{h}^{2}+{\left(R-r\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{{\left(8\right)}^{2}+{\left(6-2\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{64+16}\phantom{\rule{0ex}{0ex}}=\sqrt{80}\phantom{\rule{0ex}{0ex}}=4\sqrt{5}\mathrm{cm}$

Surface area of the remaining solid

= Curved surface area of figure + Area of the smaller circle + Area of the larger circle

$=\mathrm{\pi}l\left(R+r\right)+\mathrm{\pi}{R}^{2}+\mathrm{\pi}{r}^{2}\phantom{\rule{0ex}{0ex}}=\frac{22}{7}\times 4\sqrt{5}\times \left(6+2\right)+\frac{22}{7}\times 6\times 6+\frac{22}{7}\times 2\times 2\phantom{\rule{0ex}{0ex}}=\frac{22}{7}\times 4\sqrt{5}\times 8+\frac{22}{7}\times 6\times 6+\frac{22}{7}\times 2\times 2\phantom{\rule{0ex}{0ex}}=224.88+113.14+12.57\phantom{\rule{0ex}{0ex}}=350.59{\mathrm{cm}}^{2}$

#### Page No 14.80:

#### Question 21:

The height of a cone is 10 cm. The cone is divided into two parts using a plane parallel to its base at the middle of its height. Find the ratio of the volumes of the two parts.

#### Answer:

$\mathrm{We}\mathrm{have},\phantom{\rule{0ex}{0ex}}\mathrm{Radius}\mathrm{of}\text{the}\mathrm{solid}\mathrm{cone},R=\mathrm{CP}\phantom{\rule{0ex}{0ex}}\text{H}\mathrm{eight}\mathrm{of}\mathrm{the}\mathrm{solid}\mathrm{cone},\mathrm{AP}=H\phantom{\rule{0ex}{0ex}}\text{R}\mathrm{adius}\mathrm{of}\mathrm{the}\mathrm{smaller}\mathrm{cone},\mathrm{QD}=r\phantom{\rule{0ex}{0ex}}\text{H}\mathrm{eight}\mathrm{of}\mathrm{the}\mathrm{smaller}\mathrm{cone},\mathrm{AQ}=h\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Also},\mathrm{AQ}=\frac{\mathrm{AP}}{2}\mathrm{i}.\mathrm{e}.h=\frac{H}{2}\mathrm{or}H=2h.....\left(i\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\mathrm{in}\u2206\mathrm{AQD}\mathrm{and}\u2206\mathrm{APC},\phantom{\rule{0ex}{0ex}}\angle \mathrm{QAD}=\angle \mathrm{PAC}\left(\mathrm{Common}\mathrm{angle}\right)\phantom{\rule{0ex}{0ex}}\angle \mathrm{AQD}=\angle \mathrm{APC}=90\xb0\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{by}\mathrm{AA}\mathrm{criteria}\phantom{\rule{0ex}{0ex}}\u2206\mathrm{AQD}~\mathrm{APC}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{AQ}}{\mathrm{AP}}=\frac{\mathrm{QD}}{\mathrm{PC}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{h}{H}=\frac{r}{R}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{h}{2h}=\frac{r}{R}\left[\mathrm{Using}\left(\mathrm{i}\right)\right]\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{2}=\frac{r}{R}\phantom{\rule{0ex}{0ex}}\Rightarrow R=2r.....\left(\mathrm{ii}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{As},\phantom{\rule{0ex}{0ex}}\mathrm{Volume}\mathrm{of}\mathrm{smaller}\mathrm{cone}=\frac{1}{3}\mathrm{\pi}{r}^{2}h\phantom{\rule{0ex}{0ex}}\mathrm{And},\phantom{\rule{0ex}{0ex}}\mathrm{Volume}\mathrm{of}\mathrm{solid}\mathrm{cone}=\frac{1}{3}\mathrm{\pi}{R}^{2}H\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\mathrm{\pi}{\left(2r\right)}^{2}\times \left(2h\right)\left[\mathrm{Using}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{8}{3}\mathrm{\pi}{r}^{2}h\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{So},\phantom{\rule{0ex}{0ex}}\mathrm{Volume}\mathrm{of}\mathrm{frustum}=\mathrm{Volume}\mathrm{of}\mathrm{solid}\mathrm{cone}-\mathrm{Volume}\mathrm{of}\mathrm{smaller}\mathrm{cone}\phantom{\rule{0ex}{0ex}}=\frac{8}{3}\mathrm{\pi}{r}^{2}h-\frac{1}{3}\mathrm{\pi}{r}^{2}h\phantom{\rule{0ex}{0ex}}=\frac{7}{3}\mathrm{\pi}{r}^{2}h\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\mathrm{the}\mathrm{ratio}\mathrm{of}\mathrm{the}\mathrm{volumes}\mathrm{of}\mathrm{the}\mathrm{two}\mathrm{parts}=\frac{\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{smaller}\mathrm{cone}}{\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{frustum}}\phantom{\rule{0ex}{0ex}}=\frac{\left({\displaystyle \frac{1}{3}}\mathrm{\pi}{r}^{2}h\right)}{\left({\displaystyle \frac{7}{3}}\mathrm{\pi}{r}^{2}h\right)}\phantom{\rule{0ex}{0ex}}=\frac{1}{7}\phantom{\rule{0ex}{0ex}}=1:7$

So, the ratio of the volume of the two parts of the cone is 1 : 7.

#### Page No 14.80:

#### Question 22:

A bucket, made of metal sheet, is in the form of a cone whose height is 35 cm and radii of circular ends are 30 cm and 12 cm. How many litres of milk it contains if it is full to the brim? If the milk is sold at Rs 40 per litre, find the amount received by the person.

#### Answer:

The given bucket is in the form of the frustum of a cone.

Height, *h* = 35 cm

*r*_{1} = 30 cm

*r*_{2} = 12 cm

$\mathrm{Volume}=\frac{\mathrm{\pi}}{3}h\left({{r}_{1}}^{2}+{{r}_{2}}^{2}+{r}_{1}{r}_{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi}}{3}\times 35\left({30}^{2}+{12}^{2}+30\times 12\right)\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi}}{3}\times 35\left(900+144+360\right)\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi}}{3}\times 35\times \left(1404\right)\phantom{\rule{0ex}{0ex}}=51433.2{\mathrm{cm}}^{3}$

= 51.4 litres

Selling price of the milk = Rs 40/litre

So, selling price of 51.4 litres of milk will be $51.4\times 40=\mathrm{Rs}2056$

#### Page No 14.80:

#### Question 23:

The diameters of the lower and upper ends of a bucket in the form of a frustum of a cone are 10 cm and 30 cm respectively. If its height is 24 cm,

(i) Find the area of the metal sheet used to make the bucket.

(ii) Why we should avoid the bucket made by ordinary plastic? (use π = 3.14)

#### Answer:

We have:

Radius of upper end of frustum, R = 15 cm; Radius of lower end of frustum, *r* = 5 cm; Height of frustum, *h *= 24 cm

We know,

Slant height, *l*^{2} = *h*^{2} + (*R* – *r*)^{2}

⇒ *l*^{2} = ((24)^{2} + (15 – 5)^{2}} = (576 + 100) = 676

⇒ *l *= 26 cm

(i)

$\begin{array}{rcl}\mathrm{Required}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{metal}\mathrm{sheet}& =& \mathrm{\pi}\left[{r}^{2}+l\left(R+r\right)\right]\mathrm{sq}.\mathrm{cm}\\ & =& 3.14\left[{5}^{2}+26\left(15+5\right)\right]{\mathrm{cm}}^{2}\\ & =& 3.14\times \left(25+520\right){\mathrm{cm}}^{2}\\ & =& 3.14\times 545{\mathrm{cm}}^{2}\\ & =& 1711.3{\mathrm{cm}}^{2}\end{array}$

(ii)Plastic is harmful to the environment and to protect the environment its use should be avoided.

#### Page No 14.80:

#### Question 24:

A reservoir in the form of the frustum of a right circular cone contains 44 × 10^{7} litres of water which fills it completely. The radii of the bottom and top of the reservoir are 50 metres and 100 metres respectively. Find the depth of water and the lateral surface area of the reservoir. (Take: π = 22/7)

#### Answer:

Let the depth of the frustum cone like reservoir is *h *m. The radii of the top and bottom circles of the frustum cone like reservoir are *r*_{1} =100m and *r*_{2} =50m respectively.

The volume of the reservoir is

Given that the volume of the reservoir is. Thus, we have

Hence, the depth of water in the reservoir is

The slant height of the reservoir is

The lateral surface area of the reservoir is

Hence, the lateral surface area is

#### Page No 14.80:

#### Question 1:

A metallic sphere 1 dm in diameter is beaten into a circular sheet of uniform thickness equal to 1 mm. Find the radius of the sheet.

#### Answer:

Radius of metallic sphere

Thickness of circular sheet

Let *r*_{1} be the radius of sheet.

Therefore,

Volume of circular sheet = volume of metallic sphere

Hence, the radius of circular sheet = 40.8 cm

#### Page No 14.80:

#### Question 2:

Three solid spheres of radii 3, 4 and 5 cm respectively are melted and converted into a single solid sphere. Find the radius of this sphere.

#### Answer:

Let R be the radius of single solid sphere.

Therefore,

Volume of single solid sphere = volume of all three spheres

Hence, the radius of single solid sphere = 6 cm.

#### Page No 14.81:

#### Question 3:

A spherical shell of lead, whose external diameter is 18 cm, is melted and recast into a right circular cylinder, whose height is 8 cm and diameter 12 cm. Determine the internal diameter of the shell.

#### Answer:

External radius of spherical shell

Let *r*_{2} be the internal radius of spherical shell.

Height of right circular cylinder *h* = 8 cm

& radius of right circular cylinder

Clearly, volume of spherical shell = volume of right circular cylinder

#### Page No 14.81:

#### Question 4:

A well with 10 m inside diameter is dug 8.4 m deep. Earth taken out of it is spread all around it to a width of 7.5 m to form an embankment. Find the height of the embankment.

#### Answer:

Radius of well

Depth of well h = 8.4 m

Clearly,

Volume of earth dugout

Let *h*' be the height of embankment

Clearly,

Embankment forms a cylindrical shell whose inner and outer radius are 5 m and 12.5 m respectively.

Volume of the embankment

But, volume of earth dugout = volume of the embankment

#### Page No 14.81:

#### Question 5:

In the middle of a rectangular field measuring 30 m × 20 m, a well of 7 m diameter and 10 m depth is dug. The earth so removed is evenly spread over the remaining part of the field. Find the height through which the level of the field is raised.

#### Answer:

Radius of dug

Depth of dug = 10 m

The volume of dug

The volume of dug = 385 m^{3}

Let the height through which the level of the field is raised will be *x*.

But, volume of dug = volume of earth spread over field.

#### Page No 14.81:

#### Question 6:

The inner and outer radii of a hollow cylinder are 15 cm and 20 cm, respectively. The cylinder is melted and recast into a solid cylinder of the same height. Find the radius of the base of new cylinder.

#### Answer:

Inner radius of hollow cylinder *r*_{1} = 15 cm

Outer radius of hollow cylinder *r*_{2} = 20 cm

The volume of hollow cylinder

Since,

The hollow cylinder is melted and recast into a solid cylinder of same height.

Let r be the radius of solid cylinder.

Therefore,

The volume of solid cylinder = volume of hollow cylinder.

Hence, the radius of solid cylinder is.

#### Page No 14.81:

#### Question 7:

Two cylindrical vessels are filled with oil. Their radii are 15 cm, 12 cm and heights 20 cm, 16 cm respectively. Find the radius of a cylindrical vessel 21 cm in height, which will just contain the oil of the two given vessels.

#### Answer:

Let *r* be the radius of vessel, which will just contain oil of the both vessels.

Therefore,

Volume of cylindrical vessel = sum of the volume of both vessels.

i.e.,

#### Page No 14.81:

#### Question 8:

A cylindrical bucket 28 cm in diameter and 72 cm high is full of water. The water is emptied into a rectangular tank 66 cm long and 28 cm wide. Find the height of the water level in the tank.

#### Answer:

Let h be the height of rectangular tank = volume of cylindrical bucket

Hence, the height of rectangular tank is 24 cm.

#### Page No 14.81:

#### Question 9:

A cubic cm of gold is drawn into a wire 0.1 mm in diameter, find the length of the wire.

#### Answer:

Let *x* be the length of wire.

Clearly, the volume of gold = volume of wire.

{Volume of gold = 1 cm^{3} = 1000 mm}

Hence, the length of wire is 127.3 m.

#### Page No 14.81:

#### Question 10:

A well of diameter 3 m is dug 14 m deep. The earth taken out of it is spread evenly all around it to a width of 4 m to form an embankment. Find the height of the embankment.

#### Answer:

We have,

Radius of well

Depth of well = 14 m

The volume of well

Therefore,

Volume of earth dugout = volume of well

Let *h* be the height of embankment.

Clearly, embankment form a cylindrical shell whose inner and outer radius are

and

Volume of embankment

Volume of embankment = volume of earth dugout

Hence, the height of embankment is 1.125 m

#### Page No 14.81:

#### Question 11:

A conical vessel whose internal radius is 10 cm and height 48 cm is full of water. Find the volume of water. If this water is poured into a cylindrical vessel with internal radius 20 cm, find the height to which the water level rises in it.

#### Answer:

Radius of conical vessel *r* = 10 cm

Height of conical vessel *h* = 48 cm

The volume of water = volume of conical vessel.

Let *h*' be the height of cylindrical vessel, which filled by the water of conical vessel,

Radius of cylindrical vessel = 20 cm

Clearly,

Volume of cylindrical vessel = volume of water

Thus, the volume of the cylindrical vessel and height of cylindrical vessel are and respectively.

#### Page No 14.81:

#### Question 12:

The vertical height of a conical tent is 42 dm and the diameter of its base is 5.4 m. Find the number of persons it can accommodate if each person is to be allowed 29.16 cubic dm.

#### Answer:

Height of conical tent *h* = 42 dm

The volume of conical tent

Since, each person is to be allowed 29.16 dm^{3},

Therefore,

#### Page No 14.81:

#### Question 13:

A right circular cylinder and a right circular cone have equal bases and equal heights. If their curved surfaces are in the ratio 8 : 5, determine the ratio of the radius of the base to the height of either of them.

#### Answer:

For right circular cylinder, let *r*_{1}* = r*, *h*_{1}* = h*.

For right circular cone, let *r*_{2} = *r*, *h*_{2} = *h*

Divide (*i*) and (*ii*),

#### Page No 14.81:

#### Question 14:

A sphere of diameter 5 cm is dropped into a cylindrical vessel partly filled with water. The diameter of the base of the vessel is 10 cm. If the sphere is completely submerged, by how much will the level of water rise?

#### Answer:

Radius of sphere radius of cylindrical vessel

When the sphere is completely submerged into the vessel. The level of water will be raised let *x* be height of level of raised water.

Therefore,

The volume of raised water in cylindrical vessel = volume of sphere

Hence, the level of water rise

#### Page No 14.81:

#### Question 15:

A spherical ball of iron has been melted and made into smaller balls. If the radius of each smaller ball is one-fourth of the radius of the original one, how many such balls can be made?

#### Answer:

Let radius of spherical ball = *r*

Then radius of smaller spherical ball

Let *n* be the no. of balls made by big spherical ball.

Clearly,

Volume of big spherical balls = *n* × volume of one smaller ball

Hence, the no. of balls = 64

#### Page No 14.81:

#### Question 16:

Find the depth of a cylindrical tank of radius 28 m, if its capacity is equal to that of a rectangular tank of size 28 m × 16 m × 11 m.

#### Answer:

Let *x* be the depth of cylindrical tank.

The radius of tank r = 28 m.

Since,

The volume of cylindrical tank = volume of rectangular tank

${\mathrm{\pi r}}^{2}\mathrm{x}=28\times 16\times 11\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{22}{7}\times 28\times 28\times \mathrm{x}=28\times 16\times 11$

Thus, the depth of cylindrical tank = 2 m.

#### Page No 14.81:

#### Question 17:

A hemispherical bowl of internal radius 15 cm contains a liquid. The liquid is to be filled into cylindrical-shaped bottles of diameter 5 cm and height 6 cm. How many bottles are necessary to empty the bowl?

#### Answer:

Internal radius of hemispherical bowl *r *= 15 cm

The volume of bowl = volume of liquid

Volume of liquid = 2250 π cm^{3}

Since, the liquid filled into the cylindrical shaped bottles of radius and height 6 cm.

Let *n* be the no. of bottles.

Therefore,

The volume of liquid = *n* × volume of cylindrical shaped bottle

Hence, the no. of bottles = 60

#### Page No 14.81:

#### Question 18:

In a cylindrical vessel of diameter 24 cm, filled up with sufficient quantity of water, a solid spherical ball of radius 6 cm is completely immersed. Find the increase in height of water level.

#### Answer:

Radius of spherical ball *r* = 6 cm, radius of cylindrical vessel *r*_{1} = 12 cm

Since, the ball completely immersed into the vessel, the water level is increased.

Let the height of increased level.

Therefore,

The volume of increase water level = volume of ball

Hence, the level of water increased by 2 cm.

#### Page No 14.81:

#### Question 19:

A hemisphere of lead of radius 7 cm is cast into a right circular cone of height 49 cm. Find the radius of the base.

#### Answer:

Radius of hemisphere *r* = 7 cm

The volume of hemisphere

Since, the hemisphere cast into the right circular cone

The height of cone *h* = 49 cm

Let *x* be the radius of cone.

Clearly,

Volume of cone = volume of hemisphere

Thus, the radius of cone = 3.74 cm

#### Page No 14.82:

#### Question 20:

A solid metallic sphere of diameter 28 cm is melted and recast into a number of smaller cones, each of diameter 4$\frac{2}{3}$ cm and height 3 cm. Find the number of cones so formed.

#### Answer:

The radius of solid metallic sphere, = 14 cm

The volume of sphere

Given, the sphere is recast into smaller cones.

The radius of cone,

The height of cone *h* = 3 cm

Let *n* be the no. of smaller cones.

Clearly, the volume of solid sphere = *n* × volume of one smaller cone

$\frac{10976}{3}\mathrm{\pi}=n\times \frac{1}{3}\mathrm{\pi}\times {\left(\frac{7}{3}\right)}^{2}\times 3\phantom{\rule{0ex}{0ex}}n\times \frac{49}{3}=10976\phantom{\rule{0ex}{0ex}}n=\frac{10976\times 3}{49}\phantom{\rule{0ex}{0ex}}n=672$

Thus, the no. of smaller cones = 672

#### Page No 14.82:

#### Question 21:

The diameter of a copper sphere is 18 cm. The sphere is melted and is drawn into a long wire of uniform circular cross-section. If the length of the wire is 108 m, find its diameter.

#### Answer:

The radius of copper sphere, = 9 cm

The volume of sphere

Since,

The sphere is melted and drawn into a long circular wire of length 108 m = 10800 cm

Let *r* be the radius of wire,

Clearly,

The volume of wire = volume of sphere

And, *d* = 2*r** = *0.6 cm

Hence, the diameter of the wire will be equals to

#### Page No 14.82:

#### Question 22:

A hemisphere of lead of radius 7 cm is cast into a right circular cone of height 49 cm. Find the radius of the base.

#### Answer:

Radius of hemisphere *r* = 7 cm

The volume of hemisphere

Since,

The hemisphere cast into the right circular cone.

The height of cone *h* = 49 cm

Let *x* be the radius of cone.

Clearly,

Volume of cone = volume of hemisphere

Thus, the radius of cone = 3.74 cm

#### Page No 14.82:

#### Question 23:

A metallic sphere of radius 10.5 cm is melted and thus recast into small cones, each of radius 3.5 cm and height 3 cm. Find how many cones are obtained.

#### Answer:

The radius of sphere *r *= 10.5 cm

The volume of sphere

Let *n* be the number of cones obtained when the sphere is recast in to small cones, each of radius 3.5 and height 3 cm.

Then, volume of sphere = *n* × volume of cone

Hence, the no. of cones = 126

#### Page No 14.82:

#### Question 24:

A cone, a hemisphere and a cylinder stand on equal bases and have the same height. Show that their volumes are in the ratio 1 : 2 : 3.

#### Answer:

Let r be the radius of the base.

and h be the height.

Here, *h = r.*

Now,

The ratio of their volumes will be

Volume of cone : volume of hemisphere : volume of a cylinder

#### Page No 14.82:

#### Question 25:

A hollow sphere of internal and external diameters 4 and 8 cm respectively is melted into a cone of base diameter 8 cm. Find the height of the cone.

#### Answer:

Internal radius of hemisphere = 2 cm

External radius of hemisphere = 4 cm

Volume of hollow sphere

Since,

The hemisphere melted into a cone of radius *r* = 4 cm

Let *h* be height of the cone.

Clearly, the volume of cone = volume of hemisphere

Thus, the height is 14 cm.

#### Page No 14.82:

#### Question 26:

The largest sphere is carved out of a cube of side 10.5 cm. Find the volume of the sphere.

#### Answer:

The side of cube *a* = 10.5 cm.

Since, a largest sphere is curved out of that cube

*i.e.,* radius of sphere,

The volume of sphere

#### Page No 14.82:

#### Question 27:

Find the weight of a hollow sphere of metal having internal and external diameters as 20 cm and 22 cm, respectively if 1m^{3} of metal weighs 21*g*.

#### Answer:

External radius of hollow sphere, ${r}_{1}=\frac{22}{2}$ = 11 cm

Internal radius of hollow sphere, = 10 cm

The volume of hollow sphere = $\frac{4}{3}\mathrm{\pi}\left({{r}_{1}}^{3}-{{r}_{2}}^{2}\right)$

The volume of hollow sphere

The weight of hollow sphere

#### Page No 14.82:

#### Question 28:

A solid sphere of radius 'r' is melted and recast into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 4 cm, its height 24 cm and thickness 2 cm, find the value of 'r'.

#### Answer:

Volume of sphere …… (*i*)

Since,

The sphere is recast in to a hollow cylinder of uniform thickness 2 cm.

The external radius of hollow cylinder *r*_{1} = 4 cm

^{The internal radius of hollow cylinder }^{r}_{2}^{ = 4 − 2 = 2 cm}

and height, *h* = 24 cm

Clearly,

The volume of hollow cylinder = volume of sphere

#### Page No 14.82:

#### Question 29:

Lead spheres of diameter 6 cm are dropped into a cylindrical beaker containing some water and are fully submerged. If the diameter of the beaker is 18 cm and water rises by 40 cm. find the number of lead spheres dropped in the water.

#### Answer:

$\mathrm{Radius}\mathrm{of}\mathrm{sphere}=\frac{6}{2}=3\mathrm{cm}$

Let *n* be the no. of spheres are fully submerged.

and height of water raised = 40 cm

Clear,

The volume of raised water = n × volume of a sphere

Hence, no. of lead sphere = 90

#### Page No 14.82:

#### Question 30:

The height of a solid cylinder is 15 cm and the diameter of its base is 7 cm. Two equal conical holes each of radius 3 cm and height 4 cm are cut off. Find the volume of the remaining solid.

#### Answer:

The height of cylinder *h* = 15 cm

Radius of cylinder $r=\frac{7}{2}$

The volume of cylinder

The radius of conical holes = 3 cm

Height of conical holes = 4 cm.

The volume of conical holes

Clearly,

The volume of remaining solid

= vol. of cylinder − 2 × vol. of cone

$=183.75\mathrm{\pi}-24\mathrm{\pi}\phantom{\rule{0ex}{0ex}}=501.6{\mathrm{cm}}^{3}$

#### Page No 14.82:

#### Question 31:

A solid is composed of a cylinder with hemispherical ends. If the length of the whole solid is 108 cm and the diameter of the cylinder is 36 cm, find the cost of polishing the surface at the rate of 7 paise per cm^{2} .

#### Answer:

Height of the cylinder = height of entire solid - height of sphere 1 - height of sphere 2

= 108 - 18 - 18

= 108 − 36

= 72 cm

*r* = 18 cm

C.S.A. of cylinder

$=2\mathrm{\pi rh}\phantom{\rule{0ex}{0ex}}=2\mathrm{\pi}\times 18\times 72\phantom{\rule{0ex}{0ex}}=8138.88{\mathrm{cm}}^{2}$

C.S.A. of 2 hemispheres = surface area of a sphere

$=4{\mathrm{\pi r}}^{2}\phantom{\rule{0ex}{0ex}}=4\mathrm{\pi}{\left(18\right)}^{2}\phantom{\rule{0ex}{0ex}}=4069.44{\mathrm{cm}}^{2}$

Surface area of solid

= 8138.88 + 4069.44

= 12208.32 cm^{2}

Cost of polishing 1 cm^{2} = 7 paise = 0.07 Rs

Total cost

$=12208.32\times 0.07\phantom{\rule{0ex}{0ex}}=\mathrm{Rs}854.58$

#### Page No 14.82:

#### Question 32:

The surface area of a sphere is the same as the curved surface area of a cone having the radius of the base as 120 cm and height 160 cm. Find the radius of the sphere.

#### Answer:

Lateral height of cone

Surface area of sphere = surface area of cone

$4{{\mathrm{\pi r}}_{1}}^{2}=\mathrm{\pi rl}\phantom{\rule{0ex}{0ex}}{{\mathrm{r}}_{1}}^{2}=\frac{rl}{4}\phantom{\rule{0ex}{0ex}}{{\mathrm{r}}_{1}}^{2}=\frac{120\times 200}{4}\phantom{\rule{0ex}{0ex}}{{\mathrm{r}}_{1}}^{2}=6000\phantom{\rule{0ex}{0ex}}$

Radius of sphere

#### Page No 14.82:

#### Question 33:

A right circular cylinder and aright circular cone have equal bases and equal heights. If their curved surfaces are in the ratio 8 : 5, determine the ratio of the radius of the base to the height of either of them.

#### Answer:

In cylinder

radius = *r *

height = *h*

Surface area

In cone

radius = *r *

height = *h*

Slant height

Surface area

#### Page No 14.82:

#### Question 34:

A rectangular vessel of dimensions 20 cm × 16 cm × 11 cm is full of water. This water is poured into a conical vessel. The top of the conical vessel has its radius 10 cm. If the conical vessel is filled completely, determine its height.

#### Answer:

Volume of cone $=\frac{1}{3}{\mathrm{\pi r}}^{2}\mathrm{h}$

$\frac{1}{3}{\mathrm{\pi r}}^{2}\mathrm{h}=20\times 16\times 11\phantom{\rule{0ex}{0ex}}$

#### Page No 14.82:

#### Question 35:

If r_{1} and r_{2} be the radii of two solid metallic spheres and if they are melted into one solid sphere, prove that the radius of the new sphere is $\left({r}_{1}^{3}+{r}_{2}^{3}\right){.}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

#### Answer:

Volume of first sphere

Volume of second sphere

Total volume of new sphere

Say of radius of new sphere = *r*_{3}

Volume of new sphere

Hence,

So, radius of new sphere

#### Page No 14.82:

#### Question 36:

A solid metal sphere of 6 cm diameter is melted and a circular sheet of thickness 1 cm is prepared. Determine the diameter of the sheet.

#### Answer:

Diameter of sphere = 6 cm

Therefore,

Radius = 3 cm.

Therefore,

Surface area of sphere

Area of the circular sheet

Therefore,

Surface area of sphere = area of the circular sheet

Therefore,

Diameter of the sheet = 2 × 6 = 12 cm

#### Page No 14.82:

#### Question 37:

A hemispherical tank full of water is emptied by a pipe at the rate of $\frac{25}{7}$ litres per second. How much time will it take to half-empty the tank, If the tank is 3 metres in diameter?

#### Answer:

Volume of half of hemispherical tank

Amount of water emptied by pipe in 1 sec.

So, time taken

To half empty the tank line

#### Page No 14.83:

#### Question 38:

Find the number of coins, 1.5 cm is diameter and 0.2 cm thick, to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm.

#### Answer:

Volume of one coin

Volume of cylinder

So number of coins to be melted

#### Page No 14.83:

#### Question 39:

The radius of the base of a right circular cone of semi-vertical angle α is *r*. Show that its volume is $\frac{1}{3}{\mathrm{\pi r}}^{3}$ cot α and curved surface area is π*r*^{2} cosec α.

#### Answer:

$\mathrm{sin}\alpha =\frac{r}{l}\phantom{\rule{0ex}{0ex}}\Rightarrow r\mathrm{cos}ec\alpha =l\phantom{\rule{0ex}{0ex}}\mathrm{tan}\alpha =\frac{r}{h}\phantom{\rule{0ex}{0ex}}\Rightarrow rcot\alpha =h$

= π*r*^{2} cosec α.

#### Page No 14.83:

#### Question 40:

An iron pillar consists of a cylindrical portion 2.8 m high and 20 cm in diameter and a cone 42 cm high is surmounting it. Find the weight of the pillar, given that 1 cubic cm of iron weighs 7.5 gm.

#### Answer:

Volume of cylindrical portion

Volume of conical portion

Total number

So total height

#### Page No 14.83:

#### Question 41:

A circus tent is cylindrical to a height of 3 metres and conical above it. If its diameter is 105 m and the slant height of the conical portion is 53 m, calculate the length of the canvas 5 m wide to make the required tent.

#### Answer:

Total canvas used = curved area of cylinder + curved area of cone

So,

Area of canvas = 9735

Length × width = 9735 m^{2}

Length × 5 = 9735

So length

#### Page No 14.83:

#### Question 42:

Height of a solid cylinder is 10 cm and diameter 8 cm. Two equal conical hole have been made from its both ends. If the diameter oaf the holes is 6 cm and height 4 cm, find (i) volume of the cylinder, (ii) volume of one conical hole, (iii) volume of the remaining solid.

#### Answer:

Height = 10 cm.

Radius

(*i*) Volume of cylinder

= $160\mathrm{\pi}{\mathrm{cm}}^{3}$

(*ii*) Volume of conical hole diameter of

cone = 6 cm.

(*iii*) Volume of remaining solid

= $136\mathrm{\pi}{\mathrm{cm}}^{3}$

#### Page No 14.83:

#### Question 43:

The height of a solid cylinder is 15 cm and the diameter of its base is 7 cm. Two equal conical holes each of radius 3 cm, and height 4 cm are cut off. Find the volume of the remaining solid.

#### Answer:

Volume of cylinder

Volume of cones

Volume of remaining solid

#### Page No 14.83:

#### Question 44:

A solid is composed of a cylinder with hemispherical ends. If the length of the whole solid is 108 cm and the diameter of the cylinder is 36 cm, find the cost of polishing the surface at the rate of 7 paise per cm^{2}. (Use π = 3.1416)

#### Answer:

Height of the cylinder

C.S.A. of cylinder

C.S.A. of 2 hemisphere = surface of a sphere

Total surface

Total cost

#### Page No 14.83:

#### Question 45:

The largest sphere is to be curved out of a right circular cylinder of radius 7 cm and height 14 cm. Find the volume of the sphere.

#### Answer:

The radius of largest sphere

Volume of sphere

#### Page No 14.83:

#### Question 46:

A tent is in the form of a right circular cylinder surmounted by a cone. The diameter of the base of the cylinder or the cone is 24 m. The height of the cylinder is 11 m. If the vertex of the cone is 16 m above the ground, find the area of the canvas required for making the tent.

(Use π = 22/7)

#### Answer:

Surface area of cylindrical part

Height of cone

Surface area of conical part

S.A. of conical part

Total area

So canvas required will be 1320 m^{2}

#### Page No 14.83:

#### Question 47:

A toy is in the form of a cone mounted on a hemisphere of radius 3.5 cm. The total height of the toy is 15.5 cm find the total surface area and volume of the toy.

#### Answer:

Radius of hemisphere = 3.5 cm

Total height of the toy = 15.5 cm.

Surface area of cone

Therefore,

Surface area of cone

Surface area of hemisphere

Therefore,

Total surface area of the toy

Volume of cone

Volume of hemisphere

Therefore,

Total volume of the toy

#### Page No 14.83:

#### Question 48:

A cylindrical container is filled with ice-cream, whose diameter is 12 cm and height is 15 cm. the whole ice-cream is distributed to 10 children in equal cones having hemispherical tops. If the height of the conical portion is twice the diameter of its base, find the diameter of the ice-cream.

#### Answer:

Volume of cylindrical container

Amount of ice-cream distributed to 10 children

Therefore,

Height of conical portion = 2 × diameter of its bars

Let the diameter of bare = *r*

Height = 2*r*

Therefore,

Volume of the cones

Therefore,

Volume of the cones = amount distributed

#### Page No 14.83:

#### Question 49:

Find the volume of a solid in the form of a right circular cylinder with hemi-spherical ends whose total length is 2.7 m and the diameter of each hemi-spherical end is 0.7 m.

#### Answer:

Radius of hemispherical ends = radius of cylinder

Total length = 2.7 m.

Height of cylinder

Volume of two hemispheres

Volume of cylinders

Hence,

Volume of solid = 0.1797+0.77 = 0.95 m^{3}

#### Page No 14.83:

#### Question 50:

A tent of height 8.25 m is in the form of a right circular cylinder with diameter of base 30 m and height 5.5 m, surmounted by a right circular cone of the same base. Find the cost of the canvas of the tent at the rate of Rs 45 per m^{2}.

#### Answer:

Total surface area of the tent

Therefore,

Cost of the canvas of the tent

#### Page No 14.83:

#### Question 51:

An iron pole consisting of a cylindrical portion 110 cm high and of base diameter 12 cm is surmounted by a cone 9 cm high. Find the mass of the pole, given that 1 cm^{3} of iron has 8 gram mass approximately.

(Use : π = 355/115)

#### Answer:

Volume of cylinder

Volume of cone

Therefore,

Total volume

Therefore,

Mass of the pole

#### Page No 14.83:

#### Question 52:

The interior of a building is in the form of a cylinder of base radius 12 m and height 3.5 m, surmounted by a cone of equal base and slant height 12.5 m. Find the internal curved surface area and the capacity of the building.

#### Answer:

Height of the cone

Capacity (volume) of cone

Capacity (volume) of cylinder

Therefore,

Total capacity of building

Internal curved surface area of the building

#### Page No 14.84:

#### Question 53:

A right angled triangle with sides 3 cm and 4 cm is revolved around its hypotenuse. Find the volume of the double cone thus generated.

#### Answer:

The double cone so formed is as in figure.

Hypotenuse AC

Area of$=\frac{1}{3}\times \frac{22}{7}\times \frac{12}{5}\times \frac{12}{5}\times 5\phantom{\rule{0ex}{0ex}}=\frac{1056}{35}\phantom{\rule{0ex}{0ex}}=30\frac{6}{35}$

$\frac{1}{2}\times 3\times 4=\frac{1}{2}\times 5\times OB\phantom{\rule{0ex}{0ex}}OB=\frac{12}{5}$

Volume of double cone = volume of cone 1 + cone 2

$=\frac{1}{3}\times \frac{22}{7}\times \frac{12}{5}\times \frac{12}{5}\times 5\phantom{\rule{0ex}{0ex}}=\frac{1056}{35}\phantom{\rule{0ex}{0ex}}=30\frac{6}{35}{\mathrm{cm}}^{2}$

#### Page No 14.84:

#### Question 54:

A toy is in the form of a cone mounted on a hemisphere with the same radius. The diameter of the base of the conical portion is 6 cm and its height is 4 cm. Determine the surface area of the toy. (Use π = 3.14)

#### Answer:

Radius of hemisphere and the cone are the same.

So, r = 3 cm

Surface area of the cone

$=\mathrm{\pi rl}\phantom{\rule{0ex}{0ex}}=3.14\times 3\times \sqrt{{3}^{2}+{4}^{2}}\phantom{\rule{0ex}{0ex}}=47.1{\mathrm{cm}}^{2}$

Surface area of the hemisphere

$=2{\mathrm{\pi r}}^{2}\phantom{\rule{0ex}{0ex}}=2\times 3.14\times 9\phantom{\rule{0ex}{0ex}}=56.52{\mathrm{cm}}^{2}$

Total surface area of the toy = Surface area of the cone + surface area of the hemisphere

=47.1 + 56.52 cm^{2}

=103.62 cm^{2}

#### Page No 14.84:

#### Question 55:

Find the mass of a 3.5 m long lead pipe, if the external diameter of the pipe is 2.4 cm, thickness of the metal is 2 mm and the mass of 1 cm^{3} of lead is 11.4 grams.

#### Answer:

Length of the pipe (h) = 3.5 cm = 300 cm

External radius of the pipe (R) = $\frac{2.4}{2}=1.2cm$

Mass of 1 cm^{3} pipe is 11.4 gm

Total mass = 4184 × 11.4 = 5517.6 gm = 5.518 kg.

#### Page No 14.84:

#### Question 56:

A solid is in the form of a cylinder with hemispherical ends. Total height of the solid is 19 cm and the diameter of the cylinder is 7 cm. Find the volume and total surface area of the solid.

#### Answer:

Therefore,

Total surface area of the solid

#### Page No 14.84:

#### Question 57:

A golf ball has diameter equal to 4.2 cm. Its surface has 200 dimples each of radius 2 mm. Calculate the total surface area which is exposed to the surroundings assuming that the dimples are hemispherical.

#### Answer:

Surface area of ball

Total surface area exposed

#### Page No 14.84:

#### Question 58:

The radii of the ends of a bucket of height 24 cm are 15 cm and 5 cm. Find its capacity. (Take π = 22/7)

#### Answer:

Therefore,

Capacity of the bucket

#### Page No 14.84:

#### Question 59:

The radii of the ends of a bucket 30 cm high are 21 cm and 7 cm. Find its capacity in litres and the amount of sheet required to make this bucket.

#### Answer:

Height of the bucket = 30 cm.

Therefore,

Capacity of the bucket

The slant height of the bucket

$l=\sqrt{{h}^{2}+{\left({r}_{1}-{r}_{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{900+{\left(21-7\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{900+196}\phantom{\rule{0ex}{0ex}}=\sqrt{1096}=33.105\mathrm{cm}$

Total C.S.A. of the bucket

$=\mathrm{\pi}\left({\mathit{r}}_{\mathit{1}}\mathit{+}{\mathit{r}}_{\mathit{2}}\right)\times l\phantom{\rule{0ex}{0ex}}\mathit{=}\mathrm{\pi}\left(21+7\right)\times 33.1$

$=88\times 33.1\phantom{\rule{0ex}{0ex}}\approx 2913{\mathrm{cm}}^{2}$

Area of the base

Total sheet required to make this bucket

$=2913+154\phantom{\rule{0ex}{0ex}}=3067{\mathrm{cm}}^{2}$

#### Page No 14.84:

#### Question 60:

The radii of the ends of a frustum of a right circular cone are 5 metres and 8 metres and its lateral height is 5 metres. Find the lateral surface and volume of the frustum.

#### Answer:

Lateral surface area of frustum

$=\mathrm{\pi}\left(\mathrm{r}+\mathrm{R}\right)\mathrm{l}\phantom{\rule{0ex}{0ex}}=\mathrm{\pi}\left(5+8\right)\times 5\phantom{\rule{0ex}{0ex}}=204.28{\mathrm{m}}^{2}$

Height of cone

#### Page No 14.84:

#### Question 61:

A frustum of a cone is 9 cm thick and the diameters of its circular ends are 28 cm and 4 cm. Find the volume and lateral surface area of the frustum.

(Take π = 22/7).

#### Answer:

Volume

#### Page No 14.84:

#### Question 62:

A bucket is in the form of a frustum of a cone and holds 15.25 litres of water. The diameters of the top and bottom are 25 cm and 20 cm respectively. Find its height and area of tin used in its construction.

#### Answer:

Since volume of frustum

Area of the required

#### Page No 14.84:

#### Question 63:

If a cone of radius 10 cm is divided into two parts by drawing a plane through the mid-point of its axis, parallel to its base. Compare the volumes of the two parts.

#### Answer:

Consider a cone of radius R and height H.

Let a cone of height is cutout from this cone whose base is parallel to the original cone. Now is

Volume of cone ABC

Volume of cone ADE

Now volume of remaining frustum of cons EDCB

= volume of cone ABC − volume of cone AED

$=\frac{7}{8}(volumeofconeABC)$

Comparing the volume of cone AED and frustum EDBC we get, the ratio 1 : 7.

#### Page No 14.84:

#### Question 64:

A tent is of the shape of a right circular cylinder upto a height of 3 metres and then becomes a right circular cone with a maximum height of 13.5 metres above the ground. Calculate the cost of painting the inner side of the tent at the rate of Rs 2 per square metre, if the radius of the base is 14 metres.

#### Answer:

Let *r* m be the radius of cylindrical base of cylinder of height by *m*

*r* = 14 m and *h*_{1} = 3m

Curved surface area of cylinder

The radius of cylindrical box of cylinder is also equal to the radius of right circular cons.

Let h_{2} be the height of cone and *l* be the slant height of cone

Curved surface area of the cone

Curved surface of area of cone

Therefore,

Total area of tent which is to be painted

= curved surface area of cylinder + curved surface area of cone

Now cost of painting 1 m^{2} of inner side of tent = Rs. 2

Cost of painting 1034 m^{2} inner side of tent

#### Page No 14.84:

#### Question 65:

An oil funnel of tin sheet consists of a cylindrical portion 10 cm long attached to a frustum of a cone. If the total height be 22 cm, the diameter of the cylindrical portion 8 cm and the diameter of the top of the funnel 18 cm, find the area of the tin required.

(Use π = 22/7).

#### Answer:

Let *r*_{1} and *r*_{2} (*r*_{1} > *r*_{2}) be the radii of the ends of the frustum of a cone. Suppose *l* and *h* be the slant height and height of frustum of cone.

Let h_{1}_{ }be the height of the cylindrical portion.

Now, h_{1} = 10 cm

Total height of the funnel = 22 cm

height of frustum of cone, h = 22−10 = 12 cm

$l=\sqrt{{\left({r}_{1}-{r}_{2}\right)}^{2}+{h}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow l=\sqrt{{\left(9-4\right)}^{2}+{\left(12\right)}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow l=\sqrt{25+144}\phantom{\rule{0ex}{0ex}}\Rightarrow l=\sqrt{169}\phantom{\rule{0ex}{0ex}}\Rightarrow l=13\mathrm{cm}\phantom{\rule{0ex}{0ex}}$

Curved surface area of funnel = CSA of frustum of cone + CSA of cylinder

Curved surface area of funnel = $\mathrm{\pi}\left({r}_{1}+{r}_{2}\right)l+2\mathrm{\pi}{r}_{2}{h}_{1}$

Therefore,

Area of the tin required

$=\mathrm{\pi}\left({r}_{1}+{r}_{2}\right)l+2\mathrm{\pi}{r}_{2}{h}_{1}\phantom{\rule{0ex}{0ex}}=\mathrm{\pi}\left[\left({r}_{\mathit{1}}+{r}_{2}\right)l+2{r}_{2}{h}_{1}\right]\phantom{\rule{0ex}{0ex}}=\mathrm{\pi}\left[\left(9+4\right)\times 13+2\times 4\times 10\right]\phantom{\rule{0ex}{0ex}}=\mathrm{\pi}\left[169+80\right]\phantom{\rule{0ex}{0ex}}=\mathrm{\pi}\left[249\right]\phantom{\rule{0ex}{0ex}}=249\mathrm{\pi}{\mathrm{cm}}^{2}$

#### Page No 14.84:

#### Question 66:

A solid cylinder of diameter 12 cm and height 15 cm is melted and recast into toys with the shape of a right circular cone mounted on a hemisphere of radius 3 cm.If the height of the toy is 12 cm, find the number of toys so formed.

#### Answer:

Diameter of cylinder = 12 cm

Therefore

*r* = 6 cm

Height = 15 cm

Therefore,

Volume of cylinder

Therefore,

Volume of toy = volume of cone + volume of hemisphere

Therefore,

No. of toys

#### Page No 14.84:

#### Question 67:

A container open at the top, is in the form of a frustum of a cone of height 24 cm with radii of its lower and upper circular ends as 8 cm and 20 cm respectively. Find the cost of milk which can completely fill the container at the rate of Rs 21 per litre.

(Use $\mathrm{\pi}=\frac{22}{7}$)

#### Answer:

Volume of container = $\frac{1}{3}\mathrm{\pi}h\left[{R}^{2}+Rr+{r}^{2}\right]$

$=\frac{1}{3}\times \frac{22}{7}\times 24\left[{\left(20\right)}^{2}+20\times 8+{8}^{2}\right]\phantom{\rule{0ex}{0ex}}=\frac{22\times 8}{7}\times \left(400+160+64\right)\phantom{\rule{0ex}{0ex}}=\frac{22\times 8}{7}\times 624\phantom{\rule{0ex}{0ex}}=15689.14{\mathrm{cm}}^{3}\phantom{\rule{0ex}{0ex}}=\frac{15689.14}{1000}\mathit{}l\left[\because 1{\mathrm{cm}}^{3}=\frac{1}{1000}l\right]\phantom{\rule{0ex}{0ex}}\mathit{=}15.69\mathit{}l$

Cost of milk which can completely fill the container is given by

15.69 ⨯ 21

= Rs 329.49

#### Page No 14.85:

#### Question 68:

A cone of maximum size is carved out from a cube of edge 14 cm . Find the surface area of the cone and of the remaining solid left out after the cone carved out .

#### Answer:

The base of the largest right circular cone will be the circle inscribed in a face of the cube and its height will be equal to an edge of the cube.

Given, Edge of the cube = 14 cm

Radius of base of the cone, $r=\frac{14}{2}=7cm$

Height of the cone, *h* = 14 cm

Slant height of the cone, $l=\sqrt{{h}^{2}+{r}^{2}}$

$\Rightarrow l=\sqrt{{\left(7\right)}^{2}+{\left(14\right)}^{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow l=\sqrt{49+196}=\sqrt{245}=7\sqrt{5}\mathrm{cm}$

Surface area of the cone

$=\mathrm{\pi r}\left(\mathrm{r}+\mathrm{l}\right)\phantom{\rule{0ex}{0ex}}=\mathrm{\pi}\left(7\right)\left(7+7\sqrt{5}\right)\phantom{\rule{0ex}{0ex}}=154\left(1+\sqrt{5}\right){\mathrm{cm}}^{2}$

Surface area of the remaining solid = Surface area of the cube − surface area of the cone

$=6{a}^{2}-\frac{1}{3}{\mathrm{\pi r}}^{2}\mathrm{h}\phantom{\rule{0ex}{0ex}}=6\times {\left(14\right)}^{2}-\left[154\left(1+\sqrt{5}\right)\right]\phantom{\rule{0ex}{0ex}}=1022+154\sqrt{5}{\mathrm{cm}}^{2}$

#### Page No 14.85:

#### Question 69:

A cone of radius 4 cm is divided into two parts by drawing a plane through the mid point of its axis and parallel to its base . Compare the volumes of two parts.

#### Answer:

Radius of the cone, R = 4 cm

Let the height be H.

Since the plane divides the cone into two parts through the mid point so, a small cone and a frustum will be formed.

$OC=CA=\frac{H}{2}$

Let the radius of the smaller cone be *r* cm.

In ∆OCD and ∆OAB,

∠OCD = ∠OAB (90°)

∠COD = ∠AOB (Common)

∴∆OCD ∼ ∆OAB (AA Similarly criterion)

$\Rightarrow \frac{OA}{OC}=\frac{AB}{CD}=\frac{OB}{OD}\left(correspondingsidesareproportional\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{H}{{\displaystyle \frac{H}{2}}}=\frac{4}{r}\phantom{\rule{0ex}{0ex}}\Rightarrow r=2cm$

Now volume of the smaller cone = $\frac{1}{3}\mathrm{\pi}{\left(\mathrm{CD}\right)}^{2}\times \mathrm{OC}=\frac{{\displaystyle 1}}{{\displaystyle 3}}\mathrm{\pi}{\left(2\right)}^{2}\times \frac{\mathrm{H}}{2}=\frac{2\mathrm{\pi H}}{3}\mathrm{cm}\phantom{\rule{0ex}{0ex}}$

Height of the frustum of the cone = $\frac{H}{2}$

Volume of frustum of cone = $\frac{1}{3}\mathrm{\pi h}\left[{\mathrm{r}}_{1}^{2}+{\mathrm{r}}_{1}{\mathrm{r}}_{2}+{\mathrm{r}}_{2}^{2}\right]$

$=\frac{1}{3}\mathrm{\pi}\left(\frac{\mathrm{H}}{2}\right)\left[{\left(4\right)}^{2}+{\left(2\right)}^{2}+4\times 2\right]\phantom{\rule{0ex}{0ex}}=\frac{14\mathrm{\pi H}}{3}{\mathrm{cm}}^{3}$

$\frac{\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{smaller}\mathrm{cone}}{\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{frustum}\mathrm{of}\mathrm{the}\mathrm{cone}}=\frac{\frac{2\mathrm{\pi H}}{3}}{{\displaystyle \frac{14\mathrm{\pi H}}{3}}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{\displaystyle \mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{smaller}\mathrm{cone}}}{{\displaystyle \mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{frustum}\mathrm{of}\mathrm{the}\mathrm{cone}}}=\frac{1}{7}$

Hence, the ratio of the volumes of the two parts will be 1 : 7.

#### Page No 14.85:

#### Question 70:

A wall 24 m , 0.4 m thick and 6 m high is constructed with the bricks each of dimensions 25 cm $\times $ 16 cm $\times $ 10 cm . If the mortar occupies $\frac{1}{10}th$ of the volume of the wall, then find the number of bricks used in constructing the wall.

#### Answer:

Dimensions of the wall are 24 m × 0.4 m × 6 m

Volume of the wall = 24 m × 0.4 m × 6 m = 57.6 m^{3}

Dimensions of the bricks are 25 m × 16 m × 10 m

Volume of each brick = 4000 cm^{3} = 0.004 m^{3 }

Volume of mortar = $\frac{1}{10}\times Volumeofthewall=\frac{1}{10}\times 57.6=5.76{m}^{3}$

Volume of all the bricks = Volume of the wall − Volume of mortar

$=57.6-5.76\phantom{\rule{0ex}{0ex}}=51.84{m}^{3}$

Let the number of bricks used in making the wall be n.

$\frac{Volumeofallthebricks}{Volumeofeachbrick}=n\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{51.84}{0.004}=n\phantom{\rule{0ex}{0ex}}\Rightarrow n=12960$

Hence, 12960 bricks are used to make the wall.

#### Page No 14.85:

#### Question 71:

A bucket is in the form of a frustum of a cone and holds 28.490 litres of water . The radii of the top and bottom are 28 cm and 21 cm respectively . Find the height of the bucket .

#### Answer:

Radii r_{1} = 21 cm and r_{2} = 28 cm

Let the height of the bucket be h.

Volume of the bucket which is in the form of the frustum

$V=\frac{1}{3}\mathrm{\pi h}\left({\mathrm{r}}_{1}^{2}+{\mathrm{r}}_{1}{\mathrm{r}}_{2}+{\mathrm{r}}_{2}^{2}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{3}\mathrm{\pi h}\left({28}^{2}+28\times 21+{21}^{2}\right)=28490\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{h}=14.9\approx 15\mathrm{cm}$

Hence, the height of the bucket is 15 cm.

#### Page No 14.85:

#### Question 72:

Marbles of diameter 1.4 cm are dropped into a cylindrical beaker of diameter 7 cm containing some water . Find the number of marbles that should be dropped into the beaker so that the water level rises by 5.6 cm .

#### Answer:

Diameter of the marbles = 1.4 cm

Radius of the marbles, r = $\frac{1.4}{2}=0.7cm$

Diameter of the cylinderical beaker = 7 cm

Radius of the beaker = $\frac{7}{2}=3.5cm$

Rise in the level of water = 5.6 cm

let the number of marbles be n.

$n\times volumeofthemarbles=volumeofthewaterrisen\phantom{\rule{0ex}{0ex}}\Rightarrow n=\frac{volumeofthewaterrisen}{volumeofthemarbles}\phantom{\rule{0ex}{0ex}}\Rightarrow n=\frac{\mathrm{\pi}{\left({\displaystyle \frac{7}{2}}\right)}^{2}\left(5.6\right)}{{\displaystyle \frac{4}{3}}\mathrm{\pi}{\left(0.7\right)}^{3}}=150$

Hence, the number marbles will be 150.

#### Page No 14.85:

#### Question 73:

Two cones with same base radius 8 cm and height 15 cm are joined together along their bases. Find the surface area of the shape formed.

#### Answer:

Radius, r = 8 cm

Height, h = 15 cm

$l=\sqrt{{h}^{2}+{r}^{2}}\phantom{\rule{0ex}{0ex}}l=\sqrt{{15}^{2}+{8}^{2}}\phantom{\rule{0ex}{0ex}}l=\sqrt{225+64}\phantom{\rule{0ex}{0ex}}l=\sqrt{289}=17cm$

When the two cones are joined through their bases so, the total surface area of the shape formed will be sum of the curved surface areas

of the two cones.

Curved surface area of the cone

$CSA=\mathrm{\pi rl}\phantom{\rule{0ex}{0ex}}=\mathrm{\pi}\left(8\right)\left(17\right)\phantom{\rule{0ex}{0ex}}=427.4{\mathrm{cm}}^{2}$

similarly curved surface area of the other cone will also be the same.

Hence, the total surface area of the solid formed = 427.4 + 427.4 = 854.8 cm^{2}

#### Page No 14.85:

#### Question 74:

From a solid cube of side 7 cm , a conical cavity of height 7 cm and radius 3 cm is hollowed out . Find the volume of the remaining solid.

#### Answer:

Side length of the cube, a = 7 cm

Height of the cone, h = 7 cm

radius, r = 3 cm

Volume of the remaining solid = Volume of the cube − volume of the cone

$V={a}^{3}-\frac{1}{3}{\mathrm{\pi r}}^{2}\mathrm{h}\phantom{\rule{0ex}{0ex}}\mathrm{V}={7}^{3}-\frac{1}{3}\mathrm{\pi}{\left(3\right)}^{2}\times 7\phantom{\rule{0ex}{0ex}}\mathrm{V}=343-66=277{\mathrm{cm}}^{3}$

#### Page No 14.85:

#### Question 75:

Two solid cones *A *and* B* are placed in a cylindrical tube as shown in fig .16.76. The ratio of their capacities are 2: 1 . Find the heights and capacities of the cones . Also, find the volume of the remaining portion of the cylinder.

figure

#### Answer:

V1 : V2 = 2 : 1

Diameter of the cylinder = 6 cm

Radius, r = 3 cm

Height of the cylinder = 21 cm

Let the height of one cone be H.

So, the height of the other cone will be 21 − H.

$\frac{{V}_{1}}{{V}_{2}}=\frac{\mathrm{\pi}{\left(3\right)}^{2}\mathrm{H}}{\mathrm{\pi}{\left(3\right)}^{2}\left(21-\mathrm{H}\right)}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2}{1}=\frac{H}{21-H}\phantom{\rule{0ex}{0ex}}\Rightarrow 42-2H=H\phantom{\rule{0ex}{0ex}}\Rightarrow H=14cm$

Height of one of the cones will be 14 cm and of the other will be 21 − H = 21 − 14 = 7 cm

Volume of cone with height 14 cm = ${V}_{1}=\mathrm{\pi}{\left(3\right)}^{2}\times 14=396{\mathrm{cm}}^{3}$

Volume of cone with height 7 cm = ${V}_{2}=\frac{1}{3}\mathrm{\pi}{\left(3\right)}^{2}\times 7=66{\mathrm{cm}}^{3}$

Volume of the remaining portion of the cylinder = $Volumeofthecylinder-volumeofcone1-volumeofcone2$

$\Rightarrow V=\mathrm{\pi}{\left(3\right)}^{2}\times 21-396-66\phantom{\rule{0ex}{0ex}}=594-396-66\phantom{\rule{0ex}{0ex}}=132{\mathrm{cm}}^{3}$

#### Page No 14.85:

#### Question 76:

An icecream cone full of icecream having radius 5 cm and height 10 cm as shown in fig. 16.77. Calculate the volume of icecream , provided that its 1/ 6 part is left unfilled with icecream .

figure

#### Answer:

Ice cream above the cup is in the form of a hemisphere

So, volume of the ice above the cup = $\frac{2}{3}{\mathrm{\pi r}}^{3}=\frac{2}{3}\mathrm{\pi}{\left(5\right)}^{3}{\mathrm{cm}}^{3}$

Volume of the cup = $\frac{1}{3}\mathrm{\pi}{\left(\mathrm{r}\right)}^{2}\mathrm{h}=\frac{{\displaystyle 1}}{{\displaystyle 3}}\mathrm{\pi}{\left(5\right)}^{2}\left(5\right)=\frac{{\displaystyle 1}}{{\displaystyle 3}}\mathrm{\pi}\times 125$

Now, 1/6 part of the total is left unfilled. So, 5/6 is filled.

So, the volume of ice cream

$=\frac{5}{6}\left[Volumeofhemisphericalcup+volumeofcone\right]\phantom{\rule{0ex}{0ex}}=\frac{5}{6}\left[\frac{2\times 125\mathrm{\pi}}{3}+\frac{125\mathrm{\pi}}{3}\right]\phantom{\rule{0ex}{0ex}}=\frac{5}{6}\times \frac{125\mathrm{\pi}}{3}\left[2+1\right]\phantom{\rule{0ex}{0ex}}=327.38c{m}^{3}$

#### Page No 14.86:

#### Question 1:

The diameter of a sphere is 6 cm. It is melted and drawn in to a wire of diameter 2 mm. The length of the wire is

(a) 12 m

(b) 18 m

(c) 36 m

(d) 66 m

#### Answer:

The diameter of a sphere = 6 cm

Then radius of a sphere

The diameter of a wire = 2 mm

Then radius of wire

Now,

Volume of sphere = volume of wire

Here,

*r* = radius

*l* = length of wire

To remove the decimal from base we should multiply both numerator and denumerator by 100,

We get,

Hence, the correct answer is choice (*c*).

#### Page No 14.86:

#### Question 2:

A metallic sphere of radius 10.5 cm is melted and then recast into small cones, each of radius 3.5 cm and height 3 cm. The number of such cones is

(a) 63

(b) 126

(c) 21

(d) 130

#### Answer:

Radius of metallic sphere = 10.5 cm

Therefore,

Volume of the sphere

Now,

Radius of the cone = 3.5 cm

and Height of the cone = 3 cm

Therefore,

Volume of the cone

Number of cone

Dividing eq. (*i*) and (*ii*) we get

Number of cone

Number of cone = 126

Hence, the correct answer is choice (*b*).

#### Page No 14.86:

#### Question 3:

A solid is hemispherical at the bottom and conical above. If the surface areas of the two parts are equal, then the ratio of its radius and the height of its conical part is

(a) 1 : 3

(b) 1 : $\sqrt{3}$

(c) 1 : 1

(d) $\sqrt{3}$ : 1

#### Answer:

Let *r* be the radius of the base and h be the height of conical part.

Since,

Surface area of both part of solid is equal.

i.e.,

But,

Squaring on both side,

Then we get,

From equation (*i*) putting the value of *l* in above equation

Hence, the correct answer is choice (*b*).

#### Page No 14.87:

#### Question 4:

A solid sphere of radius r is melted and cast into the shape of a solid cone of height r, the radius of the base of the cone is

(a) 2r

(b) 3r

(c) r

(d) 4r

#### Answer:

Volume of sphere = volume of the cone

Hence, the correct answer is choice (*a*).

#### Page No 14.87:

#### Question 5:

The material of a cone is converted into the shape of a cylinder of equal radius. If height of the cylinder is 5 cm, then height of the cone is

(a) 10 cm

(b) 15 cm

(c) 18 cm

(d) 24 cm

#### Answer:

A cone is converted into a cone.

So,

Volume of cone = Volume of cylinder

Hence, the correct answer is choice (*b*).

#### Page No 14.87:

#### Question 6:

A circus tent is cylindrical to a height of 4 m and conical above it. If its diameter is 105 m and its slant height is 40 m, the total area of the canvas required in m^{2} is

(a) 1760

(b) 2640

(c) 3960

(d) 7920

#### Answer:

For conical portion

Curved surface area of the conical portion

For cylindrical portion we have

Then,

Curved surface area of cylindrical portion

Area of canvas used for making the tent

Hence, the correct answer is choice (*d*).

#### Page No 14.87:

#### Question 7:

The number of solid spheres, each of diameter 6 cm that could be moulded to form a solid metal cylinder of height 45 cm and diameter 4 cm, is

(a) 3

(b) 4

(c) 5

(d) 6

#### Answer:

Here,

Diameter of sphere = 6 cm

Radius of sphere

Volume of the sphere

Now,

Diameter of cylinder = 4 cm

^{Radius of cylinder }

Height of the cylinder = 45 cm

Then,

Volume of the cylinder

The number of solid sphere

The number of solid sphere is 5.

Hence, the correct answer is choice (*c*).

#### Page No 14.87:

#### Question 8:

A sphere of radius 6 cm is dropped into a cylindrical vessel partly filled with water. The radius of the vessel is 8 cm. If the sphere is submerged completely, then the surface of the water rises by

(a) 4.5 cm

(b) 3

(c) 4 cm

(d) 2 cm

#### Answer:

Radius of the sphere = 6 cm.

Volume of the sphere

and

Radius of the cylinder = 8 cm

Volume of the cylinder

Therefore,

Volume of the sphere = volume of the cylinder

or

Hence, the correct answer is choice (*a*).

#### Page No 14.87:

#### Question 9:

If the radii of the circular ends of a bucket of height 40 cm are of lengths 35 cm and 14 cm, then the volume of the bucket in cubic centimeters, is

(a) 60060

(b) 80080

(c) 70040

(d) 80160

#### Answer:

Height of the bucket = 40 cm

Radius of the upper part of bucket = 35 cm

R_{1} = 35 cm and

R_{2} = 14 cm

The volume of the bucket

Hence, the correct answer is choice (*b*).

#### Page No 14.87:

#### Question 10:

If a cone is cut into two parts by a horizontal plane passing through the mid-point of its axis, the ratio of the volumes of the upper part and the cone is

(a) 1 : 2

(b) 1: 4

(c) 1 : 6

(d) 1 : 8

#### Answer:

Since,

Therefore,

In and

$\frac{O\text{'}V}{OV}=\frac{O\text{'}C}{OA}$

The ratio of the volume of upper part and the cone,

From eq. (i) and (ii),

We get,

Hence, the correct answer is choice (*d*).

#### Page No 14.87:

#### Question 11:

The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to the base. If its volume be $\frac{1}{27}$ of the volume of the given cone, then the height above the base at which the section has been made, is

(a) 10 cm

(b) 15 cm

(c) 20 cm

(d) 25 cm

#### Answer:

Let VAB be cone of height 30 cm and base radius r_{1} cm.

Suppose it is cut off by a plane parallel to the base at a height h_{2} from the base of the cone.

Clearly $\u2206VOD~\u2206VO\text{'}B$

Therefore,

$\frac{OV}{O\text{'}V}=\frac{OD}{O\text{'}B}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{h}_{1}}{30}=\frac{{r}_{2}}{{r}_{1}}\phantom{\rule{0ex}{0ex}}$

But,

$VolumeofconeVCD=\frac{1}{27}VolumeofconeVAB\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{3}\mathrm{\pi}{\left({\mathrm{r}}_{2}\right)}^{2}{\mathrm{h}}_{1}=\frac{1}{27}\left(\frac{1}{3}\mathrm{\pi}{\left({\mathrm{r}}_{1}\right)}^{2}30\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(\frac{{\mathrm{r}}_{2}}{{\mathrm{r}}_{1}}\right)}^{2}{\mathrm{h}}_{1}=\frac{10}{9}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(\frac{{\mathrm{h}}_{1}}{30}\right)}^{2}{\mathrm{h}}_{1}=\frac{10}{9}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{h}}_{1}=10$

Hence,

Required height

Hence, the correct answer is choice (*c*).

#### Page No 14.87:

#### Question 12:

A solid consists of a circular cylinder with an exact fitting right circular cone placed at the top. The height of the cone is *h*. If the total volume of the solid is 3 times the volume of the cone, then the height of the circular is

(a) 2*h*

(b) $\frac{2h}{3}$

(c) $\frac{3h}{2}$

(d) 4*h*

#### Answer:

Let *r* be the radius of the base of solid.

Clearly,

The volume of solid = 3 × volume of cone

Vol. of cone + Vol. of cylinder = 3 Volume of cone

Vol. of cylinder = 2 Vol. of cone

Thus,

The height of cylinder

Hence, the correct answer is choice (*b*).

#### Page No 14.87:

#### Question 13:

A reservoir is in the shape of a frustum of a right circular cone. It is 8 m across at the top and 4 m across at the bottom. If it is 6 m deep, then its capacity is

(a) 176 m^{3}

(b) 196 m^{3}

(c) 200 m^{3}

(d) 110 m^{3}

#### Answer:

The volume of reservoir

The volume of reservoir = 176 m^{2}

Hence, the correct answer is choice (*a*).

#### Page No 14.87:

#### Question 14:

Water flows at the rate of 10 metre per minute from a cylindrical pipe 5 mm in diameter. How long will it take to fill up a conical vessel whose diameter at the base is 40 cm and depth 24 cm?

(a) 48 minutes 15 sec

(b) 51 minutes 12 sec

(c) 52 minutes 1 sec

(d) 55 minutes

#### Answer:

The radius of cylindrical pipe

The volume per minute of water flow from the pipe

The radius of cone

Depth of cone = 24 cm

The volume of cone

The time it will take to fill up a conical vessel

Hence, the correct answer is choice (*b*).

#### Page No 14.87:

#### Question 15:

A cylindrical vessel 32 cm high and 18 cm as the radius of the base, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, the radius of its base is

(a) 12 cm

(b) 24 cm

(c) 36 cm

(c) 48 cm

#### Answer:

Volume of sand filled in cylindrical vessel

Clearly,

The volume of conical heap = volume of sand

Hence, the correct answer is choice (*c*).

#### Page No 14.88:

#### Question 16:

The curved surface area of a right circular cone of height 15 cm and base diameter 16 cm is

(a) 60π cm^{2}

(b) 68π cm^{2}

(c) 120π cm^{2}

(d) 136π cm^{2}

#### Answer:

Height,

The C.S.A. of cone

Hence, the correct answer is choice (*d*).

#### Page No 14.88:

#### Question 17:

A right triangle with sides 3 cm, 4 cm and 5 cm is rotated about the side of 3 cm to form a cone. The volume of the cone so formed is

(a) 12π cm^{3}

(b) 15π cm^{3}

(c) 16π cm^{3}

(d) 20π cm^{3}

#### Answer:

Radius of cone *VAOB*

*r* = 4 cm

Height of cone *VAOB*

*h* = 3 cm

The volume of cone *VAOB*

Hence, the correct answer is choice (*a*).

#### Page No 14.88:

#### Question 18:

The curved surface area of a cylinder is 264 m^{2} and its volume is 924 m^{3}. The ratio of its diameter to its height is

(a) 3 : 7

(b) 7 : 3

(c) 6 : 7

(d) 7 : 6

#### Answer:

The C.S.A. of cylinder

S = 264 m^{2}

The volume of cylinder

V = 924 m^{3}

From eq. (*i*) and (*ii*),

We get

Putting the value in (*i*)

Hence, the correct answer is choice (*b*).

#### Page No 14.88:

#### Question 19:

A cylinder with base radius of 8 cm and height of 2 cm is melted to form a cone of height 6 cm. The radius of the cone is

(a) 4 cm

(b) 5 cm

(c) 6 cm

(d) 8 cm

#### Answer:

Volume of cylinder

Let *r* be the radius of cone

But,

The volume of cone = volume of cylinder

Hence, Radius of cone = 8 cm.

Hence, the correct answer is choice (*d*).

#### Page No 14.88:

#### Question 20:

The volumes of two spheres are in the ratio 64 : 27. The ratio of their surface areas is

(a) 1 : 2

(b) 2 : 3

(c) 9 : 16

(d) 16 : 9

#### Answer:

Ist sphere

…… (*i*)

IInd sphere

…… (*ii*)

Divide (*i*) by (*ii*) we get,

Now, the ratio of their C.S.A

Hence,

Hence, the correct answer is choice (*d*).

#### Page No 14.88:

#### Question 21:

If three metallic spheres of radii 6 cm, 8 cm and 10 cm are melted to form a single sphere, the diameter of the sphere is

(a) 12 cm

(b) 24 cm

(c) 30 cm

(d) 36 cm

#### Answer:

Let r be the radius of single sphere.

Now,

The volume of single sphere = sum of volume of three spheres

Hence, the diameter = 20 × *r* = 24 cm

Hence, the correct answer is choice (*b*).

#### Page No 14.88:

#### Question 22:

The surface area of a sphere is same as the curved surface area of a right circular cylinder whose height and diameter are 12 cm each. The radius of the sphere is

(a) 3 cm

(b) 4 cm

(c) 6 cm

(d) 12 cm

#### Answer:

Let r be the radius of sphere

But,

Surface area of sphere = C.S.A. of cylinder

Hence, the correct answer is choice (*c*).

#### Page No 14.88:

#### Question 23:

The volume of the greatest sphere that can be cut off from a cylindrical log of wood of base radius 1 cm and height 5 cm is

(a) $\frac{4}{3}\mathrm{\pi}$

(b) $\frac{10}{3}\mathrm{\pi}$

(c) 5$\mathrm{\pi}$

(d) $\frac{20}{3}\mathrm{\pi}$

#### Answer:

The radius of greatest sphere cut off from cylindrical log of wood should be radius of cylindrical log.

*i.e.,* *r* = 1 cm

The volume of sphere

Hence, the correct answer is choice (*a*).

#### Page No 14.88:

#### Question 24:

A cylindrical vessel of radius 4 cm contains water. A solid sphere of radius 3 cm is lowered into the water until it is completely immersed. The water level in the vessel will rise by

(a) $\frac{2}{9}$ cm

(b) $\frac{4}{9}$ cm

(c) $\frac{9}{4}$ cm

(d) $\frac{9}{2}$ cm

#### Answer:

The radius of sphere, *r* = 3 cm

The volume of sphere

$=\frac{4}{3}{\mathrm{\pi r}}^{3}\phantom{\rule{0ex}{0ex}}=\frac{4}{3}\mathrm{\pi}{\left(3\right)}^{3}\phantom{\rule{0ex}{0ex}}=36\mathrm{\pi}{\mathrm{cm}}^{3}$

Since,

The sphere fully immersed into the vessel, the level of water be raised by *x cm*.

Then,

The volume of raised water = volume of sphere

Hence, the correct answer is choice (*c*).

#### Page No 14.88:

#### Question 25:

12 spheres of the same size are made from melting a solid cylinder of 16 cm diameter and 2 cm height. The diameter of each sphere is

(a) $\sqrt{3}$ cm

(b) 2 cm

(c) 3 cm

(d) 4 cm

#### Answer:

The volume of solid cylinder = 12 × volume of one sphere

The required diameter *d* = 2 × 2 = 4 cm

Hence, the correct answer is choice (*d*).

#### Page No 14.88:

#### Question 26:

A solid metallic spherical ball of diameter 6 cm is melted and recast into a cone with diameter of the base as 12 cm. The height of the cone is

(a) 2 cm

(b) 3 cm

(c) 4 cm

(d) 6 cm

#### Answer:

Clearly,

The volume of recasted cone = volume of sphere

Hence, the correct answer is choice (*b*).

#### Page No 14.88:

#### Question 27:

A hollow sphere of internal and external diameters 4 cm and 8 cm respectively is melted into a cone of base diameter 8 cm. The height of the cone is

(a) 12 cm

(b) 14 cm

(c) 15 cm

(d) 18 cm

#### Answer:

External radius

Internal radius

The volume of hollow sphere

$V=\frac{4}{3}\mathrm{\pi}\left({\mathrm{R}}^{3}-{\mathrm{r}}^{3}\right)\phantom{\rule{0ex}{0ex}}=\frac{4}{3}\mathrm{\pi}\left({4}^{3}-{2}^{3}\right)$

Let *h* be the height of cone.

Clearly,

The volume of recasted cone = volume of hollow sphere

$\frac{1}{3}{\mathrm{\pi r}}^{2}\mathrm{h}=\frac{4}{3}\mathrm{\pi}\left({4}^{3}-{2}^{3}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow {4}^{2}\mathrm{h}=4\left({4}^{3}-{2}^{3}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{h}=14\mathrm{cm}$

Hence, the height of cone = 14 cm

Hence, the correct answer is choice (*b*).

#### Page No 14.88:

#### Question 28:

A solid piece of iron of dimensions 49 × 33 × 24 cm is moulded into a sphere. The radius of the sphere is

(a) 21 cm

(b) 28 cm

(c) 35 cm

(d) none of these

#### Answer:

The volume of iron piece = 49 × 33 × 24 cm^{3}

Let, *r* is the radius sphere.

Clearly,

The volume of sphere = volume of iron piece

Hence, the correct answer is choice (*a*).

#### Page No 14.89:

#### Question 29:

The ratio of lateral surface area to the total surface area of a cylinder with base diameter 1.6 m and height 20 cm is

(a) 1 : 7

(b) 1 : 5

(c) 7 : 1

(d) 8 : 1

#### Answer:

The ratio of lateral surface to the total surface area of cylinder

Hence, the correct answer is choice (*b*).

#### Page No 14.89:

#### Question 30:

A solid consists of a circular cylinder surmounted by a right circular cone. The height of the cone is *h*. If the total height of the solid is 3 times the volume of the cone, then the height of the cylinder is

(a) 2h

(b) $\frac{3h}{2}$

(c) $\frac{h}{2}$

(d) $\frac{2h}{2}$

#### Answer:

**
Disclaimer: **In the the question, the statement given is incorrect. Instead of total height of solid being equal to 3 times the volume

of cone, the volume of the total solid should be equal to 3 times the volume of the cone.

Let *x* be the height of cylinder.

Since, volume of the total solid should be equal to 3 times the volume of the cone,

So,

$\frac{1}{3}{\mathrm{\pi r}}^{2}\mathrm{h}+{\mathrm{\pi r}}^{2}\mathrm{x}=3\left(\frac{1}{3}{\mathrm{\pi r}}^{2}\mathrm{h}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{1}{3}{\mathrm{\pi r}}^{2}\mathrm{h}-{\mathrm{\pi r}}^{2}\mathrm{h}+{\mathrm{\pi r}}^{2}\mathrm{x}=0\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{\pi r}}^{2}\mathrm{x}=\frac{2}{3}{\mathrm{\pi r}}^{2}\mathrm{h}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{x}=\frac{2}{3}\mathrm{h}$

Hence, the height of cylindrical part

Hence, the correct answer is choice (*d*).

#### Page No 14.89:

#### Question 31:

The maximum volume of a cone that can be carved out of a solid hemisphere of radius r is

(a) $3{\mathrm{\pi r}}^{2}$

(b) $\frac{{\mathrm{\pi r}}^{3}}{3}$

(c) $\frac{{\mathrm{\pi r}}^{2}}{3}$

(d) $3{\mathrm{\pi r}}^{3}$

#### Answer:

Radius of hemisphere = *r*

Therefore,

The radius of cone = *r*

and height *h = r*

Then,

Volume of cone

Hence, the correct answer is choice (*b*).

#### Page No 14.89:

#### Question 32:

The radii of two cylinders are in the ratio 3 : 5. If their heights are in the ratio 2 : 3, then the ratio of their curved surface areas is

(a) 2 : 5

(b) 5 : 2

(c) 2 : 3

(d) 3 : 5

#### Answer:

Given that

and

Then,

The ratio of C.S.A. of cylinders

Hence, the correct answer is choice (*a*).

#### Page No 14.89:

#### Question 33:

A right circular cylinder of radius *r* and height *h* (*h* = 2*r*) just encloses a sphere of diameter

(a) *h*

(b) *r*

(c) 2*r*

(d) 2*h*

#### Answer:

Radius of cylinder = *r*

Height = *h*

= 2*r*

Since, the sphere fitted the cylinder.

*i.e.,* diameter of sphere = height of cylinder.

Hence, the correct answer is choice (*c*).

#### Page No 14.89:

#### Question 34:

The radii of the circular ends of a frustum are 6 cm and 14 cm. If its slant height is 10 cm, then its vertical height is

(a) 6 cm

(b) 8 cm

(c) 4 cm

(d) 7 cm

#### Answer:

Radii of circular ends of frustum

Slant height

{squaring on both sides}

Hence, the correct answer is choice (*a*).

#### Page No 14.89:

#### Question 35:

The height and radius of the cone of which the frustum is a part are h_{1} and r_{1} respectively. If h_{2} and r_{2} are the heights and radius of the smaller base of the frustum respectively and h_{2} : h_{1} = 1 : 2, then r_{2} : r_{1} is equal to

(a) 1 : 3

(b) 1 : 2

(c) 2 : 1

(d) 3 : 1

#### Answer:

Since,

are similar triangles,

*i.e.,* In

$\frac{OA}{O\text{'}L}=\frac{OV}{O\text{'}V}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{r}_{1}}{{r}_{2}}=\frac{{h}_{1}}{{h}_{1}-{h}_{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \left({h}_{1}-{h}_{2}\right){r}_{1}={h}_{1}{r}_{2}$

$\Rightarrow {r}_{1}{h}_{1}-{r}_{1}{h}_{2}={h}_{1}{r}_{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {r}_{1}{h}_{1}-{h}_{1}{r}_{2}={r}_{1}{h}_{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {h}_{1}\left({r}_{1}-{r}_{2}\right)={r}_{1}{h}_{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\left({r}_{1}-{r}_{2}\right)}{{r}_{1}}=\frac{{h}_{2}}{{h}_{1}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\left({r}_{1}-{r}_{2}\right)}{{r}_{1}}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 1-\frac{{r}_{2}}{{r}_{1}}=\frac{1}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{{r}_{2}}{{r}_{1}}=1-\frac{1}{2}=\frac{1}{2}$

Thus, ${r}_{2}:{r}_{1}=1:2$

Hence, the correct answer is choice (*b*).

#### Page No 14.89:

#### Question 36:

The diameters of the ends of a frustum of a cone are 32 cm and 20 cm. If its slant height is 10 cm, then its lateral surface area is

(a) 321 π cm^{2}

(b) 300 π cm^{2}

(c) 260 π cm^{2}

(d) 250 π cm^{2}

#### Answer:

Slant height = 10 cm

Total lateral surface area

= 260 $\mathrm{\pi}{\mathrm{cm}}^{2}$

Hence, the correct answer is choice (*c*).

#### Page No 14.89:

#### Question 37:

A solid frustum is of height 8 cm. If the radii of its lower and upper ends are 3 cm and 9 cm respectively, then its slant height is

(a) 15 cm

(b) 12 cm

(c) 10 cm

(d) 17 cm

#### Answer:

Hence, the correct answer is choice (*c*).

#### Page No 14.89:

#### Question 38:

The radii of the ends of a bucket 16 cm height are 20 cm and 8 cm. The curved surface area of the bucket is

(a) 1760 cm^{2}

(b) 2240 cm^{2}

(c) 880 cm^{2}

(d) 3120 cm^{2}

#### Answer:

Radius of top of bucket *r*_{1} = 20 cm

Radius of bottom of bucket *r*_{2} = 8 cm

Height of bucket = 16 cm

The curved surface area of bucket

C.S.A. of bucket

= $1760c{m}^{2}$

Hence, the correct answer is choice (*a*).

#### Page No 14.89:

#### Question 39:

The diameters of the top and the bottom portions of a bucket are 42 cm and 28 cm respectively. If the height of the bucket is 24 cm, then the cost of painting its outer surface at the rate of 50 paise / cm^{2} is

(a) Rs. 1582.50

(b) Rs. 1724.50

(c) Rs. 1683

(d) Rs. 1642

#### Answer:

Radius of top of bucket

Radius of bottom of bucket

Height of bucket, *h* = 24 cm.

C.S.A. of the bucket

= 2750 cm^{2}

Area of bottom

The cost of painting its C.S. ,

Hence, the correct answer is choice (*c*).

#### Page No 14.89:

#### Question 40:

If four times the sum of the areas of two circular faces of a cylinder of height 8 cm is equal to twice the curve surface area, then diameter of the cylinder is

(a) 4 cm

(b) 8 cm

(c) 2 cm

(d) 6 cm

#### Answer:

Let *r* be the radius of cylinder.

Area of circular base of cylinder

The height of cylinder *h* = 8 cm

The C.S.A. of cylinder

Clearly,

The diameter of cylinder

Hence, the correct answer is choice (*b*).

#### Page No 14.89:

#### Question 41:

If the radius of the base of a right circular cylinder is halved, keeping the height the same, then the ratio of the volume of the cylinder thus obtained to the volume of original cylinder is

(a) 1 : 2

(b) 2 : 1

(c) 1 : 4

(d) 4 : 1 [CBSE 2012]

#### Answer:

Let the radius and height of the original cylinder be *R* and *h*, respectively.

Now, the radius of the new cylinder = $\frac{R}{2}$

Then, the ratio of the volume of the cylinder thus obtained to the volume of original cylinder is given by

$\mathrm{\pi}{\left(\frac{R}{2}\right)}^{2}h:\mathrm{\pi}{R}^{2}h\phantom{\rule{0ex}{0ex}}=\frac{1}{4}:1\phantom{\rule{0ex}{0ex}}=1:4$

Hence, the correct answer is option C.

#### Page No 14.90:

#### Question 42:

A metalic solid cone is melted to form a solid cylinder of equal radius. If the height of the cylinder is 6 cm, then the height of the cone was

(a) 10 cm

(b) 12 cm

(c) 18 cm

(d) 24 cm [CBSE 2014]

#### Answer:

Let the height of the cone be *h*.

Volume of cylinder = Volume of cone

$\Rightarrow \mathrm{\pi}{r}^{2}\left(6\right)=\frac{1}{3}\mathrm{\pi}{r}^{2}h\phantom{\rule{0ex}{0ex}}\Rightarrow h=18\mathrm{cm}$

Hence, the correct answer is option C.

#### Page No 14.90:

#### Question 43:

A rectangular sheet of paper 40 cm ⨯ 22 cm, is rolled to form a hollow cylinder of height 40 cm. The radius of the cylinder (in cm) is [CBSE 2014]

(a) 3.5

(b) 7

(c) $\frac{80}{7}$

(d) 5

#### Answer:

Let the radius of the cylinder be *r* cm.

Curved surface area of cylinder = Area of rectangular sheet

$\Rightarrow 2\mathrm{\pi}r\left(40\right)=40\times 22\phantom{\rule{0ex}{0ex}}\Rightarrow 2\times \frac{22}{7}\times r\times 40=40\times 22\phantom{\rule{0ex}{0ex}}\Rightarrow r=3.5\mathrm{cm}$

Hence, the correct answer is option A.

#### Page No 14.90:

#### Question 44:

The number of solid spheres, each of diameter 6 cm that can be made by melting a solid metal cylinder of height 45 cm and diameter 4 cm is [CBSE 2014]

(a) 3

(b) 5

(c) 4

(d) 6

#### Answer:

Let the number of solid spheres be *n*.

Now, Volume of *n* solid sphere = Volume of cylinder

$\Rightarrow n\times \frac{4}{3}\times \frac{22}{7}\times {\left(\frac{6}{2}\right)}^{3}=\frac{22}{7}\times {\left(\frac{4}{2}\right)}^{2}\times 45\phantom{\rule{0ex}{0ex}}\Rightarrow n\times \frac{4}{3}\times 27=4\times 45\phantom{\rule{0ex}{0ex}}\Rightarrow n=5$

Hence, the correct answer is option B.

#### Page No 14.90:

#### Question 45:

Volumes of two spheres are in the ratio 64 ; 27 . The ratio of their surface areas is

(a) 3 : 4 (b) 4 : 3 (c) 9 : 16 (d) 16 : 9

#### Answer:

Volume of the spheres are in the ratio 64 : 27.

$\Rightarrow {V}_{1}:{V}_{2}=64:27\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{4}{3}{\mathrm{\pi r}}_{1}^{3}:\frac{4}{3}{\mathrm{\pi r}}_{2}^{3}=64:27\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{r}}_{1}^{3}:{\mathrm{r}}_{2}^{3}=64:27\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(\frac{{\mathrm{r}}_{1}}{{\mathrm{r}}_{2}}\right)}^{3}=\frac{64}{27}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(\frac{{\mathrm{r}}_{1}}{{\mathrm{r}}_{2}}\right)}^{3}={\left(\frac{4}{3}\right)}^{3}$

Thus, ratio of the radii = 4 : 3

Ratio of the surface area of the spheres will be

$\frac{4{\mathrm{\pi r}}_{1}^{2}}{4{\mathrm{\pi r}}_{2}^{2}}\phantom{\rule{0ex}{0ex}}={\left(\frac{{\displaystyle {r}_{1}}}{{\displaystyle {r}_{2}}}\right)}^{2}={\left(\frac{4}{3}\right)}^{2}=\frac{16}{9}$

So, the ratio is 16 : 9.

Hence the correct answer is option (d).

#### Page No 14.90:

#### Question 46:

A right circular cylinder of radius *r* and height* h* (*h > *2*r*) just encloses a sphere of diameter

(a) *r * (b) 2*r *(c) *h * (d) 2*h*

#### Answer:

Since *h* > 2*r* where h is the height of the cylinder and *r* is the radius

So, when a sphere is enclosed in it, the radius of the sphere will be *r*.

Thus, the diameter of the sphere will be 2*r*.

Hence, the correct answer is option (b)

#### Page No 14.90:

#### Question 47:

In a right circular cone , the cross-section made by a plane parallel to the base is a

(a) circle (b) frustyum of a cone (c) sphere (d) hemisphere

#### Answer:

When a plane parallel to the base of a cone cuts it, then a frustum and a smaller cone is formed.

The cross-section thus formed will be a circle.

Hence, the correct answer is option (a).

#### Page No 14.90:

#### Question 48:

If two solid-hemisphere s of same base radius *r* are joined together along their bases , then curved surface area of this new solid is

(a) $4{\mathrm{\pi r}}^{2}$ (b) $6{\mathrm{\pi r}}^{2}$ (c) $3{\mathrm{\pi r}}^{2}$ (d) $8{\mathrm{\pi r}}^{2}$

#### Answer:

Base radius of the hemisphere = r

Since the two hemispheres are joined end to end, it becomes a complete sphere.

Curved surface area of the new solid = total surface area of the sphere.

$\mathrm{Curved}\mathrm{surface}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{new}\mathrm{solid}=4{\mathrm{\pi r}}^{2}$

Hence, the correct answer is option (a).

#### Page No 14.90:

#### Question 49:

The diameters of two circular ends of the bucket are 44 cm and 24 cm . The height of the bucket is 35 cm . The capacity of the bucket is

(a) 32.7 litres (b) 33.7 litres (c) 34.7 litres (d) 31.7 litres

#### Answer:

The bucket is in the form of a frustum.

The diameters are respectively, ${d}_{1}=44cmand{d}_{2}=24cm$

Radii of the circular ends = ${r}_{1}=22cmand{r}_{2}=12cm$

$Volume,V=\frac{1}{3}\mathrm{\pi h}\left({\mathrm{r}}_{1}^{2}+{\mathrm{r}}_{1}{\mathrm{r}}_{2}+{\mathrm{r}}_{2}^{2}\right)\phantom{\rule{0ex}{0ex}}\mathrm{V}=\frac{1}{3}\mathrm{\pi}\times 35\left({22}^{2}+22\times 12+{12}^{2}\right)$

$V=32706.6c{m}^{3}\phantom{\rule{0ex}{0ex}}=32.7liters$

Hence, the correct answer is option (a)

#### Page No 14.90:

#### Question 50:

Q

#### Answer:

Radius of the bigger sphere = r cm

Radius of smaller spheres = r_{1 }cm

$\frac{Volumeofbiggersphere}{Volumeofsmallspheres}=\frac{{\displaystyle \frac{4}{3}{\mathrm{\pi r}}^{3}}}{\frac{4}{3}\pi {r}_{1}^{3}}=\frac{{\displaystyle {r}^{3}}}{{r}_{1}^{3}}=8\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(\frac{r}{{r}_{1}}\right)}^{3}={\left(\frac{2}{1}\right)}^{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \left(\frac{{\displaystyle r}}{{\displaystyle {r}_{1}}}\right)=\left(\frac{{\displaystyle 2}}{{\displaystyle 1}}\right)$

Hence, r : r_{1 }= 2 : 1.

Hence, the correct answer is option (a).

#### Page No 14.90:

#### Question 1:

A funnel is the combination of ____________ of a cone and __________.

#### Answer:

A funnel is the combination of _____frustum_____ of a cone and ____cylinder____.

#### Page No 14.90:

#### Question 2:

A cylindrical pencil sharpened at one edge is the combination of a _________and a _________.

#### Answer:

A cylindrical pencil sharpened at one edge is the combination of a ____cone____ and a ____cylinder____.

#### Page No 14.90:

#### Question 3:

A surahi is the combination of a _______ and a ________.

#### Answer:

A surahi is the combination of a ____cylinder____ and a ____sphere____.

#### Page No 14.90:

#### Question 4:

A plumb line (sahul) is the combination of a _________ and a _________.

#### Answer:

A plumb line (sahul) is the combination of a ____cone____ and a _____hemisphere_____.

#### Page No 14.91:

#### Question 5:

The shape of a glass (tumbler) is usually in the form of __________.

#### Answer:

The shape of a glass (tumbler) is usually in the form of __frustum of a cone__.

#### Page No 14.91:

#### Question 6:

The shape of a gilli, in the gilli-danda game, is a combination of _______and a ________.

#### Answer:

The shape of a gilli, in the gilli-danda game, is a combination of __two cones__ and a __cylinder__.

#### Page No 14.91:

#### Question 7:

A shuttle cock used for playing badminton has the shape of the combination of ________ and _________.

#### Answer:

A shuttle cock used for playing badminton has the shape of the combination of __hemisphere__ and __frustum of a cone__.

#### Page No 14.91:

#### Question 8:

A cone is cut through a plane parallel to its base and then the cone that is formed on one side of that plane is removed. The new part that is left over on the other side of the plane is called a _________.

#### Answer:

A cone is cut through a plane parallel to its base and then the cone that is formed on one side of that plane is removed. The new part that is left over on the other side of the plane is called a __frustum of cone__.

#### Page No 14.91:

#### Question 9:

Two identical solid cubes of side *a* are joined end to end. The total surface area of the resulting cuboid is _________.

#### Answer:

Length of the resulting cuboid, *l* = *a* + *a* = 2*a*

Breadth of the resulting cuboid, *b* = *a*

Height of the resulting cuboid, *h* = *a*

∴ Total surface area of the resulting cuboid

= 2(*lb* + *bh* + *h**l*)

= 2(2*a* × *a* + *a* × *a + a *× 2*a*)

= 2(2*a*^{2} + *a*^{2} + 2*a*^{2})

= 2 × 5*a*^{2}

= 10*a*^{2}

Two identical solid cubes of side *a* are joined end to end. The total surface area of the resulting cuboid is ______10 a^{2}_______.

#### Page No 14.91:

#### Question 10:

A spherical steel ball of radius *r* is melted to make eight new identical balls. The radius of each new ball is _________.

#### Answer:

Let the radius of each new ball be *R*.

∴ Volume of spherical steel ball = 8 × Volume of each new ball

$\Rightarrow \frac{4}{3}\mathrm{\pi}{r}^{3}=8\times \frac{4}{3}\mathrm{\pi}{R}^{3}\phantom{\rule{0ex}{0ex}}\Rightarrow {R}^{3}=\frac{{r}^{3}}{8}={\left(\frac{r}{2}\right)}^{3}\phantom{\rule{0ex}{0ex}}\Rightarrow R=\frac{r}{2}$

A spherical steel ball of radius *r* is melted to make eight new identical balls. The radius of each new ball is $\overline{)\frac{r}{2}}$.

#### Page No 14.91:

#### Question 11:

If a solid cone of base radius *r *and height *h *is placed over a solid cylinder having same base radius and height as that of the cone, then the curved surface area of the solid so formed is __________.

#### Answer:

Slant height of the cone, *l* = $\sqrt{{r}^{2}+{h}^{2}}$

Curved surface area of the solid

= Curved surface area of cone + Curved surface area of cylinder

$=\mathrm{\pi}rl+2\mathrm{\pi}rh\phantom{\rule{0ex}{0ex}}=\mathrm{\pi}r\left(l+2h\right)\phantom{\rule{0ex}{0ex}}=\mathrm{\pi}r\left(\sqrt{{r}^{2}+{h}^{2}}+2h\right)$

If a solid cone of base radius *r *and height *h *is placed over a solid cylinder having same base radius and height as that of the cone, then the curved surface area of the solid so formed is $\overline{)\mathrm{\pi}r\left(\sqrt{{r}^{2}+{h}^{2}}+2h\right)}$.

#### Page No 14.91:

#### Question 12:

A solid ball is exactly fitted inside the cuboidal box of side *a*. The volume of the ball is _________.

#### Answer:

If a solid ball is exactly fitted inside the cuboidal box, then the diameter of the ball is equal to the side of the cuboidal box.

Let the radius of the ball be *r* units.

$\therefore 2r=a\phantom{\rule{0ex}{0ex}}\Rightarrow r=\frac{a}{2}$

Volume of the ball = $=\frac{4}{3}\mathrm{\pi}{r}^{3}=\frac{4}{3}\mathrm{\pi}{\left(\frac{a}{2}\right)}^{3}=\frac{1}{6}\mathrm{\pi}{a}^{3}$

A solid ball is exactly fitted inside the cuboidal box of side *a*. The volume of the ball is $\overline{)\frac{1}{6}\mathrm{\pi}{a}^{3}}$.

#### Page No 14.91:

#### Question 13:

A solid cylinder of radius *r *and height *h* is placed over other cylinder of same height and radius. The total surface area of the shape so formed is _______.

#### Answer:

Total height of the new cylinder, *H* = *h* + *h* = 2*h*

Radius of the new cylinder = *r*

∴ Total surface area of the shape (or new cylinder)

$=2\mathrm{\pi}rH+2\mathrm{\pi}{r}^{2}\phantom{\rule{0ex}{0ex}}=2\mathrm{\pi}r\left(H+r\right)\phantom{\rule{0ex}{0ex}}=2\mathrm{\pi}r\left(2h+r\right)$

A solid cylinder of radius *r *and height *h* is placed over other cylinder of same height and radius. The total surface area of the shape so formed is $\overline{)2\mathrm{\pi}r\left(2h+r\right)}$.

#### Page No 14.91:

#### Question 14:

Two identical solid hemispheres of equal base radius *r* cm are stuck together along their bases. The total surface area of the combination is _________.

#### Answer:

When two identical solid hemispheres of equal base radius are stuck together along their bases, then the new solid so formed is a sphere. The radius of the sphere is same as the base radius of each solid hemisphere.

Radius of the sphere = *r* cm

∴ Total surface area of the combination = Total surface area of sphere = $4\mathrm{\pi}{r}^{2}$ cm^{2}

Two identical solid hemispheres of equal base radius *r* cm are stuck together along their bases. The total surface area of the combination is $\overline{)4\mathrm{\pi}{r}^{2}{\mathrm{cm}}^{2}}$.

#### Page No 14.91:

#### Question 15:

The radii of the ends of the frustum of a cone are *r*_{1} and *r*_{2 }cm and If *h *cm is the height of the frustum, then the volume of the frustum of the cone in cm^{3 }is _________.

#### Answer:

The radii of the ends of the frustum of a cone are *r*_{1} and *r*_{2 }cm and if *h *cm is the height of the frustum, then the volume of the frustum of the cone in cm^{3 }is $\overline{)\frac{1}{3}\mathrm{\pi}h\left({r}_{1}^{2}+{r}_{2}^{2}+{r}_{1}{r}_{2}\right)}$.

#### Page No 14.91:

#### Question 16:

If *A*_{1}, and *A*_{2}, denote areas of circular bases of the frustum of a cone and *h *is its height, then the volume of the frustum is _________.

#### Answer:

Let *r*_{1} and *r*_{2} be the radius of the top and bottom faces of the frustum of cone.

∴ ${A}_{1}=\mathrm{\pi}{r}_{1}^{2}\Rightarrow {r}_{1}=\sqrt{\frac{{A}_{1}}{\mathrm{\pi}}}$ .....(1)

and

${A}_{2}=\mathrm{\pi}{r}_{2}^{2}\Rightarrow {r}_{2}=\sqrt{\frac{{A}_{2}}{\mathrm{\pi}}}$ .....(2)

Volume of frustum of cone

$=\frac{1}{3}\mathrm{\pi}h\left({r}_{1}^{2}+{r}_{2}^{2}+{r}_{1}{r}_{2}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\mathrm{\pi}h\left(\frac{{A}_{1}}{\mathrm{\pi}}+\frac{{A}_{2}}{\mathrm{\pi}}+\sqrt{\frac{{A}_{1}}{\mathrm{\pi}}}\sqrt{\frac{{A}_{2}}{\mathrm{\pi}}}\right)\left[\mathrm{Using}\left(1\right)\mathrm{and}\left(2\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{3}h\left({A}_{1}+{A}_{2}+\sqrt{{A}_{1}{A}_{2}}\right)$

If *A*_{1}, and *A*_{2}, denote areas of circular bases of the frustum of a cone and *h *is its height, then the volume of the frustum is $\overline{)\frac{1}{3}h\left({A}_{1}+{A}_{2}+\sqrt{{A}_{1}{A}_{2}}\right)}$.

#### Page No 14.91:

#### Question 17:

The radii of the ends of a frustum of a cone are *r*_{1} and *r*_{2} cm. If *h *cm is the height of the frustum then the height of the cone of which the frustum is a part, is ________.

#### Answer:

Let the height of the cone be *H* cm.

Here, ∆OAB ~ ∆OCD

$\therefore \frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{AB}}{\mathrm{CD}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{H-h}{H}=\frac{{r}_{1}}{{r}_{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{h}{H}=1-\frac{{r}_{1}}{{r}_{2}}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow \frac{h}{H}=\frac{{r}_{2}-{r}_{1}}{{r}_{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow H=\left(\frac{{r}_{2}}{{r}_{2}-{r}_{1}}\right)h$

The radii of the ends of a frustum of a cone are *r*_{1} and *r*_{2} cm. If *h *cm is the height of the frustum then the height of the cone of which the frustum is a part, is $\overline{)\left(\frac{{r}_{2}}{{r}_{2}-{r}_{1}}\right)h}$.

#### Page No 14.91:

#### Question 18:

The radii of the ends of a frustum of a cone are *r*_{1} and *r*_{2}. If *l* is the slant height of the frustum, then the slant height of the cone of which the frustum is apart, is ________.

#### Answer:

Let the slant height of the cone be *L *units.

Here, ∆OAB $~$ ∆OCD

$\therefore \frac{\mathrm{OB}}{\mathrm{OD}}=\frac{\mathrm{AB}}{\mathrm{CD}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{L-l}{L}=\frac{{r}_{1}}{{r}_{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow 1-\frac{l}{L}=\frac{{r}_{1}}{{r}_{2}}$

$\Rightarrow \frac{l}{L}=1-\frac{{r}_{1}}{{r}_{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{l}{L}=\frac{{r}_{2}-{r}_{1}}{{r}_{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow L=\left(\frac{{r}_{2}}{{r}_{2}-{r}_{1}}\right)l$

The radii of the ends of a frustum of a cone are *r*_{1} and *r*_{2}. If *l* is the slant height of the frustum, then the slant height of the cone of which the frustum is apart, is $\overline{)\left(\frac{{r}_{2}}{{r}_{2}-{r}_{1}}\right)l}$.

#### Page No 14.91:

#### Question 19:

The capacity of a cylindrical vessel with a hemispherical portion raised upward at the bottom as shown in the following figure is ________.

#### Answer:

From the given figure, the radius of the hemispherical portion is same as the radius of the cylinder.

Capacity of the cylindrical vessel with a hemispherical portion raised upward at the bottom

= Volume of the cylinder − Volume of the hemisphere

$=\mathrm{\pi}{r}^{2}h-\frac{2}{3}\mathrm{\pi}{r}^{3}\phantom{\rule{0ex}{0ex}}=\mathrm{\pi}{r}^{2}\left(h-\frac{2}{3}r\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\mathrm{\pi}{r}^{2}\left(3h-2r\right)$

The capacity of a cylindrical vessel with a hemispherical portion raised upward at the bottom as shown in the following figure is $\overline{)\frac{1}{3}\mathrm{\pi}{r}^{2}\left(3h-2r\right)}$.

#### Page No 14.91:

#### Question 20:

If a marble of radius 2.1 cm is put into a cylindrical cup, full of water, of radius 5 cm and height 6 cm, then the volume of water that flows out of the cup is ________.

#### Answer:

Radius of the marble, *r* = 2.1 cm

Now,

Volume of water that flows out of the cylindrical cup

= Volume of the marble

$=\frac{4}{3}\mathrm{\pi}{r}^{3}\phantom{\rule{0ex}{0ex}}=\frac{4}{3}\times \frac{22}{7}\times {\left(2.1\right)}^{3}\phantom{\rule{0ex}{0ex}}=38.808{\mathrm{cm}}^{3}$

If a marble of radius 2.1 cm is put into a cylindrical cup, full of water, of radius 5 cm and height 6 cm, then the volume of water that flows out of the cup is ____38.808 cm ^{3}____.

#### Page No 14.92:

#### Question 21:

Three solid cubes have a face diagonal of $4\sqrt{2}$ cm each. Three other solid cubes have a face diagonal of $8\sqrt{2}$ cm each. All the cubes are melted together to form a cube. The side of the cube so formed is of length ___________.

#### Answer:

We know that each face of a cube is a square.

Let the side of solid cube having face diagonal of $4\sqrt{2}$ cm each be *x* cm.

Diagonal of a square = $\sqrt{2}\times \mathrm{Side}=\sqrt{2}x$

$\therefore \sqrt{2}x=4\sqrt{2}$

⇒* x *= 4 cm

Now, let the side of solid cube having face diagonal of $8\sqrt{2}$ cm each be *y* cm.

$\therefore \sqrt{2}y=8\sqrt{2}$

⇒* y *= 8 cm

Suppose the side of the bigger cube obtained on melting the given cubes be *a* cm.

∴ Volume of the bigger cube = 3 × Volume of cube having side 4 cm + 3 × Volume of cube having side 8 cm

$\Rightarrow {a}^{3}=3\times {4}^{3}+3\times {8}^{3}\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{3}=192+1536=1728\phantom{\rule{0ex}{0ex}}\Rightarrow {a}^{3}={\left(12\right)}^{3}\phantom{\rule{0ex}{0ex}}\Rightarrow a=12\mathrm{cm}$

Three solid cubes have a face diagonal of $4\sqrt{2}$ cm each. Three other solid cubes have a face diagonal of $8\sqrt{2}$ cm each. All the cubes are melted together to form a cube. The side of the cube so formed is of length ____12 cm____.

#### Page No 14.92:

#### Question 22:

In Fig. 14.78, a circle is inscribed in a square *ABCD *and the square is circumscribed by a circle. If the radius of the smaller circle is *r* сm, then the area of the shaded region in cm^{2} is _________.

#### Answer:

Side of the square = Diameter of the smaller circle = 2*r*

Length of diagonal of square = $\sqrt{2}$ × Side of the square = $2\sqrt{2}r$

∴ Radius of the bigger circle = $\frac{\mathrm{Length}\mathrm{of}\mathrm{diagonal}\mathrm{of}\mathrm{square}}{2}=\frac{2\sqrt{2}r}{2}=\sqrt{2}r$

Area of the shaded region

= $\frac{1}{4}$(Area of the bigger circle − Area of the square)

$=\frac{1}{4}\left[\mathrm{\pi}{\left(\sqrt{2}r\right)}^{2}-{\left(2r\right)}^{2}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\left(2\mathrm{\pi}{r}^{2}-4{r}^{2}\right)\phantom{\rule{0ex}{0ex}}=\left(\frac{\mathrm{\pi}-2}{2}\right){r}^{2}{\mathrm{cm}}^{2}$

In Fig. 14.78, a circle is inscribed in a square *ABCD *and the square is circumscribed by a circle. If the radius of the smaller circle is *r* сm, then the area of the shaded region in cm^{2} is $\overline{)\left(\frac{\mathrm{\pi}-2}{2}\right){r}^{2}}$.

#### Page No 14.92:

#### Question 23:

The volume of the greatest right circular cone, which can be cut from a cube of side 4 cm is __________.

#### Answer:

The greatest right circular cone will be such that the diameter of the base of the cone will be equal to the side of the cube and the height of the cone will be equal to the side of the cube.

Radius of the greatest right circular cone which can be cut from the given cube, *r *= $\frac{4}{2}$ = 2 cm

Height of the greatest right circular cone which can be cut from the given cube, *h* = 4 cm

∴ Volume of the greatest right circular cone which can be cut from the given cube

$=\frac{1}{3}\mathrm{\pi}{r}^{2}h\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\mathrm{\pi}\times {2}^{2}\times 4\phantom{\rule{0ex}{0ex}}=\frac{16}{3}\mathrm{\pi}{\mathrm{cm}}^{3}$

The volume of the greatest right circular cone, which can be cut from a cube of side 4 cm is $\overline{)\frac{16}{3}\mathrm{\pi}{\mathrm{cm}}^{3}}$.

#### Page No 14.92:

#### Question 24:

The volume of a cuboid is $24\sqrt{42}{\mathrm{cm}}^{3}$. Its length is $5\sqrt{2}$ cm, breadth and height are in the ratio $\sqrt{3}:\sqrt{7}$. The height of the cuboid is ________.

#### Answer:

Let the breadth and height of the cuboid be $\sqrt{3}x$ and $\sqrt{7}x$, respectively.

Volume of the cuboid = Length × Breadth × Height

$\therefore 5\sqrt{2}\times \sqrt{3}x\times \sqrt{7}x=24\sqrt{42}{\mathrm{cm}}^{3}\phantom{\rule{0ex}{0ex}}\Rightarrow 5\sqrt{42}{x}^{2}=24\sqrt{42}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}=\frac{24}{5}\phantom{\rule{0ex}{0ex}}\Rightarrow x=\sqrt{\frac{24}{5}}\mathrm{cm}$

∴ Height of the cuboid = $\sqrt{7}x=\sqrt{7}\times \sqrt{\frac{24}{5}}=\frac{2}{5}\sqrt{210}\mathrm{cm}$

The volume of a cuboid is $24\sqrt{42}{\mathrm{cm}}^{3}$. Its length is $5\sqrt{2}$ cm, breadth and height are in the ratio $\sqrt{3}:\sqrt{7}$. The height of the cuboid is $\overline{)\frac{2}{5}\sqrt{210}\mathrm{cm}}$.

#### Page No 14.92:

#### Question 25:

The sum of the length, breadth and height of a cuboid is $5\sqrt{3}$ cm and length of its diagonal is $3\sqrt{5}$ cm, then its total surface area is ___________.

#### Answer:

Let the length, breadth and height of the cuboid be *l*, *b* and *h*, respectively.

Given:

$l+b+h=5\sqrt{3}\mathrm{cm}$ .....(1)

Length of diagonal of cuboid = $\sqrt{{l}^{2}+{b}^{2}+{h}^{2}}=3\sqrt{5}\mathrm{cm}$

$\Rightarrow {l}^{2}+{b}^{2}+{h}^{2}=45{\mathrm{cm}}^{2}$ .....(2)

Now,

${\left(l+b+h\right)}^{2}={l}^{2}+{b}^{2}+{h}^{2}+2lb+2bh+2hl\phantom{\rule{0ex}{0ex}}\Rightarrow 2\left(lb+bh+hl\right)={\left(5\sqrt{3}\mathrm{cm}\right)}^{2}-45{\mathrm{cm}}^{2}\left[\mathrm{Using}\left(1\right)\mathrm{and}\left(2\right)\right]\phantom{\rule{0ex}{0ex}}\Rightarrow 2\left(lb+bh+hl\right)=75-45=30{\mathrm{cm}}^{2}$

Total surface area of the cuboid = $2\left(lb+bh+hl\right)=30{\mathrm{cm}}^{2}$

The sum of the length, breadth and height of a cuboid is $5\sqrt{3}$ cm and length of its diagonal is $3\sqrt{5}$ cm, then its total surface area is ____30 cm ^{3}___.

#### Page No 14.92:

#### Question 1:

The radii of the base of a cylinder and a cone are in the ratio 3 : 4 and their heights are in the ratio 2 : 3. What is the ratio of their volumes?

#### Answer:

Let *r*_{1} and *r*_{2} be the radii of the base of a cylinder and a cone.

The volume of cylinder …… (*i*)

The volume of cone …… (*ii*)

Dividing (*i*) by (*ii*), the, we get

#### Page No 14.92:

#### Question 2:

If the heights of two right circular cones are in the ratio 1 : 2 and the perimeters of their bases are in the ratio 3 : 4, what is the ratio of their volumes?

#### Answer:

Given that,

Therefore,

The ratios of volume of their cones will be

#### Page No 14.92:

#### Question 3:

If a cone and a sphere have equal radii and equal volumes. What is the ratio of the diameter of the sphere to the height of the cone?

#### Answer:

Given that,

A cone and a sphere have equal radii and equal volume

*i.e.,* volume of cone = volume of sphere

#### Page No 14.93:

#### Question 4:

A cone, a hemisphere and a cylinder stand on equal bases and have the same height. What is the ratio of their volumes?

#### Answer:

Let *r* be the radius of the base.

And *h* is the height.

Here, *h = r*.

Now,

The ratio of their volumes will be

Volume of cone: volume of hemisphere: volume of a cylinder

#### Page No 14.93:

#### Question 5:

The radii of two cylinders are in the ratio 3 : 5 and their heights are in the ratio 2 : 3. What is the ratio of their curved surface areas?

#### Answer:

Given that,

Now, the ratio of their curved surface area

#### Page No 14.93:

#### Question 6:

Two cubes have their volumes in the ratio 1 : 27. What is the ratio of their surface areas?

#### Answer:

The rate of the value of cubes = 1:27

…… (*i*)

Now,

The ratio of their surface area

Hence,

#### Page No 14.93:

#### Question 7:

Two right circular cylinders of equal volumes have their heights in the ratio 1 : 2. What is the ratio of their radii?

#### Answer:

Let *r*_{1} and *r*_{2} be the radii of two right circular cylinders and *h*_{1} and *h*_{2} be the heights.

Since,

Both the cylinder has the same volume.

Therefore,

#### Page No 14.93:

#### Question 8:

If the volumes of two cones are in the ratio 1 : 4 and their diameters are in the ratio 4 : 5, then write the ratio of their weights.

#### Answer:

The ratio of the volume of cones

And

*i.e., *

Hence,

#### Page No 14.93:

#### Question 9:

A sphere and a cube have equal surface areas. What is the ratio of the volume of the sphere to that of the cube?

#### Answer:

Surface area of sphere = sphere area of cube

*i.e.,* 4π*r*^{2} = 6*a*^{2}

…… (*i*)

#### Page No 14.93:

#### Question 10:

What is the ratio of the volume of a cube to that of a sphere which will fit inside it?

#### Answer:

Ratio of sphere

Now,

Volume of cube

Volume of sphere

The ratio of their volumes

#### Page No 14.93:

#### Question 11:

What is the ratio of the volumes of a cylinder, a cone and a sphere, if each has the same diameter and same height?

#### Answer:

Given that the diameter and the height of the cylinder, cone and sphere are the same.

The volume of cylinder, =$\mathrm{\pi}{\left(\frac{\mathrm{d}}{2}\right)}^{2}\mathrm{d}$

The volume of cone, =$\frac{1}{3}\mathrm{\pi}{\left(\frac{\mathrm{d}}{2}\right)}^{2}\mathrm{d}$

And the volume of sphere, =$\frac{4}{3}\mathrm{\pi}{\left(\frac{\mathrm{d}}{2}\right)}^{3}$

Therefore,

The ratio of their volumes,

${v}_{1}={v}_{2}={v}_{3}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{\pi}{\left(\frac{\mathrm{d}}{2}\right)}^{2}\mathrm{d}=\frac{1}{3}\mathrm{\pi}{\left(\frac{\mathrm{d}}{2}\right)}^{2}\mathrm{d}=\frac{4}{3}\mathrm{\pi}{\left(\frac{\mathrm{d}}{2}\right)}^{3}\phantom{\rule{0ex}{0ex}}\Rightarrow 3:1:2$

Hence, the ratio is 3 : 1 : 2

#### Page No 14.93:

#### Question 12:

A sphere of maximum volume is cut-out from a solid hemisphere of radius r, what is the ratio of the volume of the hemisphere to that of the cut-out sphere?

#### Answer:

Since, a sphere of maximum volume is cut out from a solid hemisphere of radius.

*i.e.,* radius of sphere

Therefore,

The volume of sphere

…… (*i*)

The volume of hemisphere …… (*ii*)

Divide (*i*) by (*ii*).

#### Page No 14.93:

#### Question 13:

A metallic hemisphere is melted and recast in the shape of a cone with the same base radius *R* as that of the hemisphere. If *H *is the height of the cone, then write the values of $\frac{H}{R}.$

#### Answer:

Given,

Radius of the hemisphere = Radius of the cone.

Now,

Volume of the hemisphere

and

Volume of the cone

Volume of the hemisphere = volume of the cone

#### Page No 14.93:

#### Question 14:

A right circular cone and a right circular cylinder have equal base and equal height. If the radius of the base and height are in the ratio 5 : 12, write the ratio of the total surface area of the cylinder to that of the cone.

#### Answer:

Given that

*i.e.*

Since,

Right, circular cone and right circular cylinder have equal base and equal right.

Therefore,

The total surface area of cylinder

The total surface area of cone

Now,

Hence,

#### Page No 14.93:

#### Question 15:

A cylinder, a cone and a hemisphere are of equal base and have the same height. What is the ratio of their volumes?

#### Answer:

Let the diameter of the base for all three be *x* cm and height be *y* cm.

For hemisphere radius

Height

(As height of the hemisphere is equal to the radius of hemisphere)

For cone

Radius

Height

(As height is same for all)

For cylinder

Radius

Height

The ratio of their volume is

= cylinder volume : conic volume : hemispherical volume

#### Page No 14.93:

#### Question 16:

The radii of two cones are in the ratio 2 : 1 and their volumes are equal. What is the ratio of their heights?

#### Answer:

Let the radius of the first cone = 2*x*

And height of the first cone = *h*_{1}

Then,

The radius of the second cone = *x*

Height of the second cone = *h*_{2}

Then,

Since,

The volumes of the two cones are equal.

Or

#### Page No 14.93:

#### Question 17:

Two cones have their heights in the ratio 1 : 3 and radii 3 : 1. What is the ratio of their volumes?

#### Answer:

Let the radius of the cone is 3*x* and *x*,

And the height of the cone is *y* and 3*y*.

Then,

Volume of the first cone

Volume of the second cone

Then the radius of their volume

Or

#### Page No 14.93:

#### Question 18:

A hemisphere and a cone have equal bases. If their heights are also equal, then what is the ratio of their curved surfaces?

#### Answer:

The base of the cone and hemisphere are equal. So radius of the two is also equal.

and

Height of the hemisphere = height of the cone

Then the slant height of the cone

Now, the curved surface area of

Hemisphere

and

The curved surface area of cone

Putting the value of *l* from eq. (*i*)

We get

Now,

#### Page No 14.93:

#### Question 19:

If r_{1} and r_{2} denote the radii of the circular bases of the frustum of a cone such that r_{1} > r_{2}, then write the ratio of the height of the cone of which the frustum is a part to the height fo the frustum.

#### Answer:

Since,

Therefore,

In

Hence,

The ratio of the height of cone of which the frustum is a part to the height fo the frustum.

$\frac{OV}{OO\text{'}}=\frac{h}{{h}_{1}}=\frac{{r}_{1}}{{r}_{1}-{r}_{2}}$

Hence,

#### Page No 14.93:

#### Question 20:

If the slant height of the frustum of a cone is 6 cm and the perimeters of its circular bases are 24 cm and 12 cm respectively. What is the curved surface area of the frustum?

#### Answer:

The parameter of upper base

The parameter of lower base

The surface area of frustum

#### Page No 14.93:

#### Question 21:

If the areas of circular bases of a frustum of a cone are 4 cm^{2} and 9 cm^{2} respectively and the height of the frustum is 12 cm. What is the volume of the frustum?

#### Answer:

Area of circular bases of frustum is

The height of frustum *h* = 12 cm

Now, the volume of frustum

#### Page No 14.93:

#### Question 22:

The surface area of a sphere is 616 cm^{2} . Find its radius.

#### Answer:

The surface area of sphere = 616k cm^{2}

We know that

Taking squire root both the side

#### Page No 14.93:

#### Question 23:

A cylinder and a cone are of the same base radius and of same height. Find the ratio of the value of the cylinder to that of the cone

#### Answer:

Since, cylinder and a cone both are have same radius and height.

Therefore,

#### Page No 14.93:

#### Question 24:

The slant height of the frustum of a cone is 5 cm. If the difference between the radii of its two circular ends is 4 cm, write the height of the frustum.

#### Answer:

Slant height of the Frustum = 5 cm

Squaring both sides we get

Height of the Frustum = 3 cm

#### Page No 14.93:

#### Question 25:

Volume and surface area of a solid hemisphere are numerically equal. What is the diameter of hemisphere?

#### Answer:

Let the radius of the hemisphere be *r *units.

Volume of a hemisphere = Surface area of the hemisphere

$\Rightarrow \frac{2}{3}\mathrm{\pi}{r}^{3}=3\mathrm{\pi}{r}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{2}{3}r=3\phantom{\rule{0ex}{0ex}}\Rightarrow r=\frac{9}{2}\phantom{\rule{0ex}{0ex}}\Rightarrow d=9\mathrm{units}$

Hence, diameter of the hemisphere is equal to 9 units.

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