R.d Sharma 2022 Mcqs Solutions for Class 9 Maths Chapter 19 - Surface Area And Volume Of A Right Circular Cylinder

Answer:

Let radius be 'r' and height be 'h'
∴ Original Curved Surface Area = $2\pi rh$
$\mathrm{Now}, \mathrm{new} \mathrm{radius}=2r, \mathrm{new} \mathrm{height}=\frac{h}{2}$
$$\begin{gathered} \mathrm{New} \mathrm{Curved} \mathrm{Surface} \mathrm{Area}=2\pi \left(2r\right)\left(\frac{h}{2}\right) \\ =2\pi rh \end{gathered}$$

Thus, if the radius is doubled and the height is halved in a cylinder, the curved surface area will be the same.

Hence, the correct answer is option (c).

Answer:

Let V₁ and V₂ be the volume of the two cylinders with radius r₁ and height h₁, and radius r₂ and height h₂, where $\frac{2r_1}{2r_2}=\frac{3}{1}, \frac{h_1}{h_2}=\frac{1}{3}$

So,

$V_1=\pi {r_1}^2{h}_{1 } .....\left(1\right)$

Now,

$V_2=\pi {r_2}^2{h}_{2 } .....\left(1\right)$ 

From equation (1) and (2), we have

$$\begin{gathered} \frac{V_1}{V_2}={\left(\frac{r_1}{r_2}\right)}^2 \left(\frac{h_1}{h_2}\right) \\ \Rightarrow \frac{V_1}{V_2}={\left(\frac{2r_1}{2r_2}\right)}^2 \left(\frac{h_1}{h_2}\right) \\ \Rightarrow \frac{V_1}{V_2}={\left(3\right)}^2 \left(\frac{1}{3}\right) \\ \Rightarrow \frac{V_1}{V_2}=\frac{3}{1} \end{gathered}$$

Thus, the ratio of their volumes is 3 : 1.

Hence, the correct answer is option (c).

Answer:

A right cylinder has one curved and two flat surfaces.



Thus, the number of surfaces in a right cylinder is 3.

Hence, the correct answer is option (c).

Answer:

From the following figure of a cylinder, it can be observed that the vertical cross section is the rectangle PQRS.

Thus, vertical cross-section of a right circular cylinder is always a rectangle.

Hence, the correct answer is option (b).

Answer:

Given: r is the radius and h is the height of the cylinder.

The volume of a cylinder is given by the formula

$V=\mathrm{\pi }{r}^{2}h$

Hence, the correct answer is option (b).

Answer:

In a hollow cylinder, there are two curved surface areas: inner and outer and one circular base with inner and outer surface area.

Thus, there are 4 surfaces in a hollow cylindrical object.

Hence, the correct answer is option (d).

Answer:

Let V₁ be the volume of the cylinder with radius r₁ and height h₁, then

$V_1=\pi {r_1}^2{h}_{1 }$

Now, let V₂ be the volume after changing the dimension, then

$r_2=2r_1 \mathrm{and} h_1=h_2$

So,

$\begin{array}{rcl}{V}_2& =& \pi {r_2}^2{h}_{2 } \\ & =& \pi {\times \left(2r_1\right)}^2\times h_1 \\ & =& 4\pi {r_1}^2{h}_1 \\ & =& 4V_1\end{array}$

Hence, the correct answer is option (d).

Answer:

Let V₁ be the volume of the cylinder with radius r₁ and height h₁, then

$V_1=\pi {r_1}^2{h}_{1 }$              .…. (1)

Now, let V₂ be the volume after changing the dimension, then

$r_2=r_1 \mathrm{and} h_1=2h_2$

So,

$\begin{array}{rcl}{V}_2& =& \pi {r_2}^2{h}_{2 } \\ & =& \pi {\times \left(r_1\right)}^2\times \left(2h_1 \right) \\ & =& 2\pi {r_1}^2{h}_1 \\ & =& 2V_1\end{array}$

Hence, the correct answer is option (a).

Answer:

Let V₁ be the volume of the cylinder with radius r₁ and height h₁, then

$V_1=\pi {r_1}^2{h}_{1 }$

Now, let V₂ be the volume after changing the dimension, then

$r_2=\frac{1}{2}{r}_1 \mathrm{and} h_1=2h_2$

So,

$\begin{array}{rcl}{V}_2& =& \pi {r_2}^2{h}_{2 } \\ & =& \pi {\times \left(\frac{1}{2}{r}_1\right)}^2\times 2h_1 \\ & =& \frac{1}{2}\pi {r_1}^2{h}_1 \\ & =& \frac{1}{2}{V}_1\end{array}$

Hence, the correct answer is option (c).

Answer:

Let r be the radius of the cylinder and h be its height

$$\begin{gathered} 2r=h \\ \Rightarrow r=\frac{h}{2} \end{gathered}$$

∴Total surface area (S) $=2\pi rh+\pi r^2$

$$\begin{gathered} \Rightarrow S=2\pi r\left(h+r\right) \\ \Rightarrow S=2\pi \times \frac{h}{2}\times \left(h+\frac{h}{2}\right) \\ \Rightarrow S=\pi \times h\times \left(\frac{3h}{2}\right) \\ \Rightarrow S=\frac{3\pi h^2}{2} \end{gathered}$$

Hence, the correct answer is option (b).

Answer:

Diameter of right circular cylindrical tunnel (d) = 2 m
Radius of right circular cylindrical tunnel (r) = $\frac{d}{2}=\frac{2}{2}=1$ m
Height of right circular cylindrical tunnel (h) = 40 m

∴ The area of the iron sheet required = Surface area of the cylinder 
                                                           $$\begin{gathered} =2\mathrm{\pi }rh \\ =2\mathrm{\pi }\times 1\times 40 \\ =80\mathrm{\pi } \end{gathered}$$

Hence, the correct answer is option (b).

Answer:

Let V₁ and V₂ be the volume of the two cylinders with h₁ and h₂ as their heights.

Let r₁ and r₂ be their base radius.

 $$\begin{gathered} V_1=V_2 \\ \mathrm{And}, \frac{h_1}{h_2}=\frac{1}{2} \end{gathered}$$

$$\begin{gathered} \because {\mathrm{V}}_1={\mathrm{V}}_2 \\ \Rightarrow \mathrm{\pi }{r_{\mathit{1}}}^2{h}_{\mathit{1}}\mathit{=}\mathrm{\pi }{r_{\mathit{2}}}^{\mathit{2}}{h}_2 \\ \mathit{\Rightarrow }\frac{{r_{\mathit{1}}}^{\mathit{2}}}{{r_{\mathit{2}}}^{\mathit{2}}}\mathit{=}\frac{h_{\mathit{2}}}{h_{\mathit{1}}} \\ \mathit{\Rightarrow }{\mathit{\left(}\frac{{\mathit{r}}_{\mathit{1}}}{{\mathit{r}}_{\mathit{2}}}\mathit{\right)}}^{\mathit{2}}=\frac{2}{1} \\ \mathit{\Rightarrow }\frac{r_{\mathit{1}}}{r_{\mathit{2}}}=\frac{\sqrt{2}}{1} \end{gathered}$$

Hence, the correct answer is option (b).

Answer:

Let V₁ and V₂ be the volume of the two cylinders with h₁ and h₂ as their heights.

Let r₁ and r₂ be their base radius.

$$\begin{gathered} V_1=V_2 \\ \mathrm{And}, r_2=\frac{1}{3}{r}_1 \end{gathered}$$

$$\begin{gathered} \because V_1=V_2 \\ \Rightarrow \mathrm{\pi }{r_1}^2{h}_1=\mathrm{\pi }{r_2}^2{h}_2 \\ \Rightarrow {r_1}^2{\mathrm{h}}_1={\left(\frac{1}{3}{\mathit{r}}_{\mathit{1}}\right)}^2{h}_2 \\ \Rightarrow {r_1}^2{h}_1=\frac{1}{9}{r_1}^2{h}_2 \\ \Rightarrow h_1=\frac{1}{9}{h}_2 \\ \Rightarrow 9h_1=h_2 \end{gathered}$$

Thus, the length will become 9 times.

Hence, the correct answer is option (c).

Answer:

Let V₁ be the volume of the cylinder with radius r₁ and height h₁, then

$V_1=\pi {r_1}^2{h}_{1 }$             

Now, let V₂ be the volume after changing the dimension, then

$h_1=2h_2$

$$\begin{gathered} \because V_1=V_2 \\ \Rightarrow \mathrm{\pi }{r_1}^2{h}_1=\mathrm{\pi }{r_2}^2{h}_2 \\ \Rightarrow {r_1}^2{h}_1={r_2}^2\times \left(2h_1\right) \\ \Rightarrow {r_1}^2={2r_2}^2 \\ \Rightarrow {r_2}^2=\frac{{r_1}^2}{2} \\ \Rightarrow r_2=\frac{r_1}{\sqrt{2}} \end{gathered}$$
Thus, the radius of the base should be multiplied by $\frac{1}{\sqrt{2}}$  so that the resulting cylinder has the same volume as the original cylinder.
Hence, the correct answer is option (b).

 

Answer:

Let V₁ be the volume of the cylinder with radius r and height h, then

$V_1=\pi r^{2}h$              .…. (1)

Now, let V₂ be the volume of the box, then

$V_2=x^{2}h$

∵ V₁ = $\frac{1}{4}$V₂

$$\begin{gathered} \Rightarrow \mathrm{\pi }{r}^{\mathit{2}}h\mathit{=}\frac{1}{4}{x}^{\mathit{2}}h \\ \Rightarrow \mathrm{\pi }{r}^{\mathit{2}}h\mathit{=}\frac{1}{4}{x}^{\mathit{2}}h \\ \mathit{\Rightarrow }{r}^{\mathit{2}}\mathit{=}\frac{1}{4\mathrm{\pi }}{x}^{\mathit{2}} \\ \mathit{\Rightarrow }r\mathit{=}\frac{x}{2\sqrt{\mathrm{\pi }}} \end{gathered}$$

Hence, the correct answer is option (b).

Answer:

Let h be the height of the cylinder with radius r.
The height h of a cylinder equals the circumference of the cylinder

$$\begin{gathered} \therefore 2\mathrm{\pi }r=h \\ \mathit{\Rightarrow }r\mathit{=}\frac{\mathit{h}}{\mathit{2}\mathit{\pi }} \end{gathered}$$

$\begin{array}{rcl}\mathrm{VolumeV} \mathrm{of} \mathrm{cylinder} \left(V\right)& =& \mathrm{\pi }{r}^{2}h\\ & =& \mathrm{\pi }{\left(\frac{h}{2\mathrm{\pi }}\right)}^{\mathit{2}}h\\ & \mathit{=}& \frac{h^{\mathit{3}}}{4\mathrm{\pi }}\end{array}$

Hence, the correct answer is option (a).

Answer:

Let S be the total surface area of the closed cylinder with radius r and height h, then

$$\begin{gathered} S=2\mathrm{\pi }{r}^{\mathit{2}}+2\mathrm{\pi }rh \\ S\mathit{=}2\pi r\mathit{(}\mathit{r}\mathit{+}\mathit{h}\mathit{)} \end{gathered}$$

Hence, the correct answer is option (a).

Answer:

Let r be the radius of the cylinder. It is given that the height drops 3 inches when 1 cubic foot of sand is poured out and 1 foot = 12 inch

$$\begin{gathered} \therefore 12\times 12\times 12=\mathrm{\pi }{r}^{2}h \\ \Rightarrow 12\times 12\times 12\mathit{=}\mathrm{\pi }{r}^{\mathit{2}}\times 3 \\ \Rightarrow 12\times 12\times 4\mathit{=}{\mathrm{\pi r}}^2\mathit{2} \\ \Rightarrow \frac{12\times 12\times 4}{\mathrm{\pi }}\mathit{=}{r}^{\mathit{2}} \\ \mathit{\Rightarrow }\frac{12\times 2}{\sqrt{\mathrm{\pi }}}\mathit{=}r \\ \mathit{\Rightarrow }r\mathit{=}\frac{24}{\sqrt{\pi }} \\ \mathit{\Rightarrow }2r=\frac{48}{\sqrt{\pi }} \end{gathered}$$

Thus, the diameter of the cylinder is $\frac{48}{\sqrt{\mathrm{\pi }}}$ inches.

Hence, the correct answer is option (b).

Answer:

Let the radius of the base of the cylinders be r and R.
Now, let the sheet with length a₁ be used to form a cylinder with volume v₁.

So,
 $$\begin{gathered} 2\mathrm{\pi }r={\mathrm{a}}_1 \\ \Rightarrow r=\frac{{\mathrm{a}}_1}{2\mathrm{\pi }} \end{gathered}$$
Volume $v_{\mathit{1}}\mathit{=}\mathrm{\pi }{r}^{\mathit{2}}h\mathit{=}\mathrm{\pi }{\left(\frac{a_{\mathrm{1}}}{\mathrm{2}\mathrm{\pi }}\right)}^{\mathit{2}}{a}_{\mathit{2}}\mathit{=}\frac{{a_1}^2{a}_2}{4\mathrm{\pi }}$
Similarly, let sheet with length a₂ be used to form a cylinder with volume v₂.


Volume $v_{\mathit{2}}\mathit{=}\mathrm{\pi }{R}^{\mathit{2}}h\mathit{=}\mathrm{\pi }{\mathit{\left(}\frac{{\mathit{a}}_{\mathit{2}}}{\mathit{2}\mathit{\pi }}\mathit{\right)}}^{\mathit{2}}{a}_1\mathit{=}\frac{{a_2}^2{a}_1}{4\mathrm{\pi }}$
Now,
 $$\begin{gathered} \frac{v_1}{v_2}=\frac{{a_1}^2{a_2}^2}{{a_2}^2{a}_1} \\ \Rightarrow \frac{v_1}{v_2}=\frac{a_1}{a_2} \\ \Rightarrow v_1{a}_2=v_2{a}_1 \end{gathered}$$

Hence, the correct answer is option (c).
 

Answer:

Let the radius and height of the original circular cylinder be r and h and the radius and height of the new circular cylinder be R and H.

H = 6h        (Given)              .....(1)

Base area of the original circular cylinder (a) = $\mathrm{\pi }{r}^2$                  .....(2)
Base area of the new circular cylinder (A) = $\mathrm{\pi }{R}^2$                    .....(3)
∵ Base area of the new circular cylinder (A) =  $\frac{1}{9}\times$Base area of the original circular cylinder (a)
$$\begin{gathered} \Rightarrow \mathrm{\pi }{R}^2=\frac{1}{9}\mathrm{\pi }{r}^2 \left[\mathrm{From} \left(2\right) \mathrm{and} \left(3\right)\right] \\ \Rightarrow {\mathrm{R}}^2=\frac{1}{9}{r}^2 \\ \mathit{\Rightarrow }\mathrm{R}\mathit{=}\sqrt{\frac{\mathit{1}}{\mathit{9}}{r}^{\mathit{2}}} \\ \Rightarrow \mathrm{R}\mathit{=}\frac{1}{3}r .....\left(4\right) \end{gathered}$$

Lateral surface area of the original circular cylinder (s) = $2\mathrm{\pi }rh$                  .....(5)
Lateral surface area of the new circular cylinder (S) = $2\mathrm{\pi }RH$                    
$$\begin{gathered} =2\mathrm{\pi }RH \\ =2\mathrm{\pi }\left(\frac{1}{3}r\right)\left(6\mathrm{h}\right) \left[\mathrm{From} \left(1\right) \mathrm{and} \left(4\right)\right] \\ =2\times \left(2\mathrm{\pi }r\mathrm{h}\right) \\ =2\mathrm{s} \left[\mathrm{From} \left(5\right)\right] \end{gathered}$$

Thus, the factor by which the lateral surface of the cylinder increases, is 2.

Hence, the correct answer is option (d).

Answer:

Given: The volume of cuboid = 64 cm³            .....(1)
Let l, b and h be the length, breadth and height of the cuboid.

The volume of cuboid = l × b × h             .....(2)    
From (1) and (2), we get
l × b × h = 64 cm³
The possible 3 numbers whose multiplication will result in 64 are
Case I: 1 × 8 × 8
Case II: 2 × 4 × 8
Case III: 4 × 4 × 4
Case IV: 1 × 4 ×16
Case V: 1 × 2× 32

Lateral surface area of cuboid = 2 × (l + b)× h 
All the possible lateral surface area of the cuboid are

Case I: l = 8, b = 8, h = 1 
⇒ Lateral surface area of cuboid = 2 × (8 + 8) × 1 
= 2 ×16
= 32 cm²

Case II: l = 8, b = 4, h = 2
⇒ Lateral surface area of cuboid = 2 × (8 + 4) × 2 
= 2 × 12 × 2
= 48 cm²

Case III: l = 4, b = 4, h = 4
⇒ Lateral surface area of cuboid = 2 × (4 + 4) × 4 
= 2 × 8 × 4
= 64 cm²

Case IV: l = 16, b = 4, h = 1
⇒ Lateral surface area of cuboid = 2 × (16 + 4) × 1 
= 2 × 22 × 1
= 44 cm²

Case V: l = 32, b = 2, h = 1
⇒ Lateral surface area of cuboid = 2 × (32 + 2) × 1 
= 2 × 34 × 1
= 68 cm²

Thus, the minimum possible lateral surface area of the cuboid is 32 cm².

Hence, the required answer is option (a).