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#### Page No 27:

Yes. A vector is defined by its magnitude and direction, so a vector can be changed by changing its magnitude and direction. If we rotate it through an angle, its direction changes and we can say that the vector has changed.

#### Page No 27:

No, it is not possible to obtain zero by adding two vectors of unequal magnitudes.
Example: Let us add two vectors $\stackrel{\to }{A}$ and $\stackrel{\to }{B}$ of unequal magnitudes acting in opposite directions. The resultant vector is given by

$R=\sqrt{{A}^{2}+{B}^{2}+2AB\mathrm{cos}\theta }\phantom{\rule{0ex}{0ex}}$

If two vectors are exactly opposite to each other, then

From the above equation, we can say that the resultant vector is zero (R = 0) when the magnitudes of the vectors $\stackrel{\to }{A}$ and $\stackrel{\to }{B}$ are equal (A = B) and both are acting in the opposite directions.

Yes, it is possible to add three vectors of equal magnitudes and get zero.
Lets take three vectors of equal magnitudes , given these three vectors make an angle of $120°$ with each other. Consider the figure below:

Lets examine the components of the three vectors.

Hence, proved.

#### Page No 27:

A zero vector has physical significance in physics, as the operations on the zero vector gives us a vector.
For any vector $\stackrel{\to }{A}$, assume that
$\stackrel{\to }{A}+\stackrel{\to }{0}=\stackrel{\to }{A}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{A}-\stackrel{\to }{0}=\stackrel{\to }{A}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{A}×\stackrel{\to }{0}=\stackrel{\to }{0}$
Again, for any real number $\lambda$, we have:
$\lambda \stackrel{\to }{0}=\stackrel{\to }{0}$
The
significance of a zero vector can be better understood through the following examples:
The displacement vector of a stationary body for a time interval is a zero vector.
Similarly, the velocity vector of the stationary body is a zero vector.
When a ball, thrown upward from the ground, falls to the ground, the displacement vector is a zero vector, which defines the displacement of the ball.

#### Page No 27:

Yes we can add three unit vectors to get a unit vector.
No, the answer does not change if two unit vectors are along the coordinate axes. Assume three unit vectors along the positive x-axis, negative x-axis and positive y-axis, respectively. Consider the figure given below:

The magnitudes of the three unit vectors ( ) are the same, but their directions are different.
So, the resultant of is a zero vector.
Now, $\stackrel{^}{j}+\stackrel{\to }{0}=\stackrel{^}{j}$          (Using the property of zero vector)
∴ The resultant of three unit vectors ()  is a unit vector ($\stackrel{^}{j}$).

#### Page No 28:

Yes, there are physical quantities like electric current and pressure which have magnitudes and directions, but are not considered as vectors because they do not follow vector laws of addition.

#### Page No 28:

Two forces are added using triangle rule, because force is a vector quantity. This statement is more appropriate, because we know that force is a vector quantity and only vectors are added using triangle rule.

#### Page No 28:

No, we cannot add two vectors representing physical quantities of different dimensions. However, we can multiply two vectors representing physical quantities with different dimensions.
Example: Torque, $\stackrel{\to }{\tau }=\stackrel{\to }{r}×\stackrel{\to }{F}$

#### Page No 28:

Yes, a vector can have zero components along a line and still have a nonzero magnitude.
Example: Consider a two dimensional vector $2\stackrel{^}{i}+0\stackrel{^}{j}$. This vector has zero components along a line lying along the Y-axis and a nonzero component along the X-axis. The magnitude of the vector is also nonzero.
Now, magnitude of $2\stackrel{^}{i}+0\stackrel{^}{j}$ = $\sqrt{{2}^{2}+{0}^{2}}=2$

#### Page No 28:

The direction of $-\stackrel{\to }{A}$ is opposite to $\stackrel{\to }{A}$. So, if vector $\stackrel{\to }{A}$ and $-\stackrel{\to }{A}$ make the angles ε1 and ε2 with the X-axis, respectively, then ε1 is equal to ε2 as shown in the figure:

Here, tan ε1 = tan ε2
Because these are alternate angles.
Thus, giving tan ε does not uniquely determine the direction of $\stackrel{\to }{A}$.

#### Page No 28:

No, the vector sum of the unit vectors $\stackrel{\to }{i}$ and $\stackrel{\to }{j}$ is not a unit vector, because the magnitude of the resultant of $\stackrel{\to }{i}$ and $\stackrel{\to }{j}$ is not one.

Magnitude of the resultant vector is given by
R = $\sqrt{{1}^{2}+{1}^{2}+\mathrm{cos}90°}=\sqrt{2}$

Yes, we can multiply this resultant vector by a scalar number $\frac{1}{\sqrt{2}}$ to get a unit vector.

#### Page No 28:

A vector  $\stackrel{\to }{B}$ such that $\stackrel{\to }{A}\ne \stackrel{\to }{B}$, but A = B are as follows:

#### Page No 28:

No, we cannot have  $\stackrel{\to }{A}×\stackrel{\to }{B}=\stackrel{\to }{A}·\stackrel{\to }{B}$ with A ≠ 0 and B ≠ 0. This is because the left hand side of the given equation gives a vector quantity, while the right hand side gives a scalar quantity. However, if one of the two vectors is zero, then both the sides will be equal to zero and the relation will be valid.

#### Page No 28:

If $\stackrel{\to }{A}×\stackrel{\to }{B}=0$, then both the vectors are either parallel or antiparallel, i.e., the angle between the vectors is either .
($\stackrel{\to }{A}\stackrel{\to }{B}\mathrm{sin}\theta \stackrel{^}{n}=0$ $\because$ $\mathrm{sin}0°=\mathrm{sin}180°=0$)

Both the conditions can be satisfied:
(a) $\stackrel{\to }{A}=\stackrel{\to }{B},$ i.e., the two vectors are equal in magnitude and parallel to each other
(b) $\stackrel{\to }{A}\ne \stackrel{\to }{B}$, i.e., the two vectors are unequal in magnitude and parallel or anti parallel to each other

#### Page No 28:

If  , then we have $\stackrel{\to }{B}=k\stackrel{\to }{A}$ by putting the value of scalar k as $-1.5$.
However, we cannot say that $\frac{\stackrel{\to }{B}}{\stackrel{\to }{A}}$ = k, because a vector cannot be divided by other vectors, as vector division is not possible.

#### Page No 28:

(d) it is slid parallel to itself.

A vector is defined by its magnitude and direction. If we slide it to a parallel position to itself, then none of the given parameters, which define the vector, will change.

Let the magnitude of a displacement vector ($\stackrel{\to }{A}$) directed towards the north be 5 metres. If we slide it parallel to itself, then the direction and magnitude will not change.

#### Page No 28:

(c) 1, 2, 1

1,2 and 1 may represent the magnitudes of three vectors adding to zero.For example one of the vector of length 1 should make an angle of ${135}^{\circ }$ with x axis and the other vector of length 1 makes an angle of ${225}^{\circ }$ with x axis. The third vector of length 2 should lie along x axis.

#### Page No 28:

(c) α < β if A > B

The resultant of two vectors is closer to the vector with the greater magnitude.
Thus, α < β if A > B

#### Page No 28:

(d) None of these.

All the given options are incorrect. The component of a vector may be less than, greater than or equal to its magnitude, depending upon the vector and its components.

#### Page No 28:

(a) along the west

The vector product $\stackrel{\to }{A}×\stackrel{\to }{B}$ will point towards the west. We can determine this direction using the right hand thumb rule.

#### Page No 28:

(b) 14.1 cm2

Area of a circle, A = $\pi {r}^{2}$
On putting the values, we get:

The rules to determine the number of significant digits says that in the multiplication of two or more numbers, the number of significant digits in the answer should be equal to that of the number with the minimum number of significant digits. Here, 2.12 cm has a minimum of three significant digits. So, the answer must be written in three significant digits.

#### Page No 28:

(a) the value of a scalar
(c) a vector
(d) the magnitude of a vector

The value of a scalar, a vector and the magnitude of a vector do not depend on a given set of coordinate axes with different orientation. However, components of a vector depend on the orientation of the axes.

#### Page No 28:

(b) It is possible to have < and <

Statements (a), (c) and (d) are incorrect.
Given: $\stackrel{\to }{C}=\stackrel{\to }{A}+\stackrel{\to }{B}$
Here, the magnitude of the resultant vector may or may not be equal to or less than the magnitudes of $\stackrel{\to }{A}$ and $\stackrel{\to }{B}$ or the sum of the magnitudes of both the vectors if the two vectors are in opposite directions.

#### Page No 28:

(b) C must be less than

Here, we have three vector A, B and C.

Subtracting (i) from (ii), we get:
${\left|\stackrel{\to }{A}+\stackrel{\to }{B}\right|}^{2}-{\left|\stackrel{\to }{A}-\stackrel{\to }{B}\right|}^{2}=4\stackrel{\to }{A}.\stackrel{\to }{B}$

Using the resultant property $\stackrel{\to }{C}=\stackrel{\to }{A}+\stackrel{\to }{B}$, we get:
${\left|\stackrel{\to }{C}\right|}^{2}-{\left|\stackrel{\to }{A}-\stackrel{\to }{B}\right|}^{2}=4\stackrel{\to }{A}.\stackrel{\to }{B}\phantom{\rule{0ex}{0ex}}⇒{\left|\stackrel{\to }{C}\right|}^{2}={\left|\stackrel{\to }{A}-\stackrel{\to }{B}\right|}^{2}+4\stackrel{\to }{A}.\stackrel{\to }{B}\phantom{\rule{0ex}{0ex}}⇒{\left|\stackrel{\to }{C}\right|}^{2}={\left|\stackrel{\to }{A}-\stackrel{\to }{B}\right|}^{2}+4\left|\stackrel{\to }{A}\right|\left|\stackrel{\to }{B}\right|\mathrm{cos}120°$

Since cosine is negative in the second quadrant, must be less than $\left|A-B\right|$.

#### Page No 28:

(a) is equal to the sum of the x-components of the vectors
(b) may be smaller than the sum of the magnitudes of the vectors
(d) may be equal to the sum of the magnitudes of the vectors.

The x-component of the resultant of several vectors cannot be greater than the sum of the magnitudes of the vectors.

#### Page No 28:

(b) equal to AB
(c) less than AB
(d) equal to zero.

The magnitude of the vector product of two vectors and may be less than or equal to AB, or equal to zero, but cannot be greater than AB.

#### Page No 29:

From the above figure, we have:
Angle between $\stackrel{\to }{\mathrm{A}}$ and $\stackrel{\to }{\mathrm{B}}$ = 110° − 20° = 90°

Magnitude of the resultant vector is given by

Let β be the angle between .

Now, angle made by the resultant vector with the X-axis = 53° + 20° = 73°
∴ The resultant is 5 m and it makes an angle of 73° with the x-axis.

#### Page No 29:

Angle between , θ = 60° − 30° = 30°

Let β be the angle between .

Angle made by the resultant vector with the X-axis = 15° + 30° = 45°

∴ The magnitude of the resultant vector is 17.3 and it makes angle of 45° with the X-axis.

#### Page No 29:

First, we will find the components of the vector along the x-axis and y-axis. Then we will find the resultant x and y-components.
x
-component of
x-component of
x-component of $\stackrel{\to }{\mathrm{C}}$ = $\stackrel{\to }{\mathrm{C}}$cos315$°$ = 100 cos 315°

Resultant x-component $=\frac{100}{\sqrt{2}}-\frac{100}{\sqrt{2}}+\frac{100}{\sqrt{2}}=\frac{100}{\sqrt{2}}$

Now, y-component of
y-component of
y-component of
Resultant y-component$=\frac{100}{\sqrt{2}}+\frac{100}{\sqrt{2}}-\frac{100}{\sqrt{2}}=\frac{100}{\sqrt{2}}$

Magnitude of the resultant$=\sqrt{{\left(\frac{100}{\sqrt{2}}\right)}^{2}+{\left(\frac{100}{\sqrt{2}}\right)}^{2}}$
$=\sqrt{10000}=100$
Angle made by the resultant vector with the x-axis is given by

⇒ α = tan−1 (1) = 45°

∴ The magnitude of the resultant vector is 100 units and it makes an angle of 45° with the x-axis.

#### Page No 29:

Given:

(a) Magnitude of $\stackrel{\to }{a}$ is given by $\left|\stackrel{\to }{a}\right|=\sqrt{{4}^{2}+{3}^{2}}=\sqrt{16+9}=5$

(b)  Magnitude of $\stackrel{\to }{b}$ is  given by $\left|\stackrel{\to }{b}\right|=\sqrt{{3}^{2}+{4}^{2}}=\sqrt{9+16}=5$

(c) $\stackrel{\to }{a}+\stackrel{\to }{b}=\left(4\stackrel{^}{i}+3\stackrel{^}{j}\right)+\left(3\stackrel{^}{i}+4\stackrel{^}{j}\right)=\left(7\stackrel{^}{i}+7\stackrel{^}{j}\right)$

∴ Magnitude of vector $\stackrel{\to }{a}+\stackrel{\to }{b}$ is given by $\left|\stackrel{\to }{a}+\stackrel{\to }{b}\right|=\sqrt{49+49}=\sqrt{98}=7\sqrt{2}$

(d) $\stackrel{\to }{a}-\stackrel{\to }{b}=\left(4\stackrel{\to }{i}+3\stackrel{\to }{j}\right)-\left(3\stackrel{\to }{i}+4\stackrel{\to }{j}\right)=\stackrel{\to }{i}-\stackrel{\to }{j}$

∴ Magnitude of vector $\stackrel{\to }{a}-\stackrel{\to }{b}$ is given by $\left|\stackrel{\to }{a}-\stackrel{\to }{b}\right|=\sqrt{{\left(1\right)}^{2}+{\left(-1\right)}^{2}}=\sqrt{2}$

#### Page No 29:

First, let us find the components of the vectors along the x and y-axes. Then we will find the resultant x and y-components.

x
-component of m
x-component of $\stackrel{\to }{\mathrm{BC}}$ =1.5cos120°
$=-\frac{\left(1.5\right)}{2}=-7.5$ m
x-component of $\stackrel{\to }{\mathrm{DE}}$ = 1cos270°
= 1 × 0 = 0 m

y-component of $\stackrel{\to }{\mathrm{OA}}$ = 2 sin 30° = 1
y-component of $\stackrel{\to }{\mathrm{BC}}$ = 1.5 sin 120°
$=\frac{\left(\sqrt{3}×1.5\right)}{2}=1.3$
y-component of $\stackrel{\to }{\mathrm{DE}}$ = 1 sin 270° = −1
x-component of resultant

y-component of resultant Ry = 1 + 1.3 − 1 = 1.3 m

If it makes an angle α with the positive x-axis, then

∴ α = tan−1 (1.32)

#### Page No 29:

Let the two vectors be $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$.
Now,

(a) If the resultant vector is 1 unit, then

Squaring both sides, we get:

Hence, the angle between them is 180°.

(b) If the resultant vector is 5 units, then

Squaring both sides, we get:

25 + 24 cos θ = 25
⇒ 24 cos θ = 0
⇒ cos θ = 90°

Hence, the angle between them is 90°.

(c) If the resultant vector is 7 units, then

Squaring both sides, we get:

25 + 24 cos θ = 49,
⇒ 24 cos θ = 24
⇒ cos θ = 1
⇒ θ = cos−1 1 = 0°

Hence, the angle between them is 0°.

#### Page No 29:

The displacement of the car is represented by $\stackrel{\to }{\mathrm{AD}}$.

Magnitude of $\stackrel{\to }{\mathrm{AD}}$ is given by

Now,

Hence, the displacement of the car is 6.02 km along the direction with positive the x-axis.

#### Page No 29:

Consider that the queen is initially at point A as shown in the figure.
Let AB be x ft.
So, DE = (2 $-$ x) ft
In ∆ABC, we have:
...(i)
Also, in ∆DCE, we have:
...(ii)
From (i) and (ii), we get:
$\frac{x}{2}=\frac{\left(2-x\right)}{4}\phantom{\rule{0ex}{0ex}}⇒2\left(2-x\right)=4x\phantom{\rule{0ex}{0ex}}⇒4-2x=4x\phantom{\rule{0ex}{0ex}}⇒6x=4\phantom{\rule{0ex}{0ex}}⇒x=\frac{2}{3}\mathrm{ft}$

(a) In ∆ABC, we have:
$\mathrm{AC}=\sqrt{{\mathrm{AB}}^{2}+{\mathrm{BC}}^{2}}$

(b) In ∆CDE, we have:
DE
CD = 4 ft

(c) In ∆AGE, we have:
$\mathrm{AE}=\sqrt{{\mathrm{AG}}^{2}+{\mathrm{GE}}^{2}}$

#### Page No 29:

Displacement vector of the mosquito, $\stackrel{\to }{r}=7\stackrel{^}{i}+4\stackrel{^}{i}+3\stackrel{^}{k}$

(a) Magnitude of displacement$=\sqrt{{7}^{2}+{4}^{2}+{3}^{2}}$

(b) The components of the displacement vector are 7 ft, 4 ft and 3 ft along the X, Y and Z-axes, respectively.

#### Page No 29:

Given: $\stackrel{\to }{a}$ is a vector of magnitude 4.5 units due north.

Case (a):
$3\left|\stackrel{\to }{a}\right|=3×4.5=13.5$
$3\stackrel{\to }{a}$ is a vector of magnitude 13.5 units due north.

Case (b):

$-4\stackrel{\to }{a}$ is a vector of magnitude 6 units due south.

#### Page No 29:

Let the two vectors be .
Angle between the vectors, θ = 60°

(a) The scalar product of two vectors is given by .

∴

(b) The vector product of two vectors is given by .
∴

#### Page No 29:

According to the polygon law of vector addition, the resultant of these six vectors is zero.
Here, a = b = c = d = e = f (magnitudes), as it is a regular hexagon. A regular polygon has all sides equal to each other.
So,
[As the resultant is zero, the x-component of resultant Rx is zero]

Note: Similarly, it can be proven that .

#### Page No 29:

We have:
$\stackrel{\to }{a}=2\stackrel{\mathit{\to }}{\mathit{i}}+3\stackrel{\mathit{\to }}{\mathit{j}}+4\stackrel{\to }{k}\phantom{\rule{0ex}{0ex}}\stackrel{\to }{b}=3\stackrel{\mathit{\to }}{\mathit{i}}+4\stackrel{\mathit{\to }}{\mathit{j}}+5\stackrel{\to }{k}\phantom{\rule{0ex}{0ex}}$

Using scalar product, we can find the angle between vectors $\stackrel{\to }{a}$ and $\stackrel{\to }{b}$.
i.e.,
So, $\theta ={\mathrm{cos}}^{-1}\left(\frac{\stackrel{\to }{a}.\stackrel{\to }{b}}{\left|\stackrel{\to }{a}\right|\left|\stackrel{\to }{b}\right|}\right)$

∴ The required angle is ${\mathrm{cos}}^{-1}\frac{38}{\sqrt{1450}}.$

#### Page No 29:

To prove: $\stackrel{\to }{\mathrm{A}}.\left(\stackrel{\to }{\mathrm{A}}×\stackrel{\to }{\mathrm{B}}\right)=0\phantom{\rule{0ex}{0ex}}$

Proof: Vector product is given by .

is a vector which is perpendicular to the plane containing . This implies that it is also perpendicular to $\stackrel{\to }{\mathrm{A}}$. We know that the dot product of two perpendicular vectors is zero.
$\stackrel{\to }{\mathrm{A}}.\left(\stackrel{\to }{\mathrm{A}}×\stackrel{\to }{\mathrm{B}}\right)=0$

Hence, proved.

#### Page No 29:

Given: and $\stackrel{\mathit{\to }}{\mathit{B}}\mathit{=}4\stackrel{\mathit{^}}{\mathit{i}}\mathit{+}3\stackrel{\mathit{^}}{\mathit{j}}\mathit{+}2\stackrel{\mathit{^}}{\mathit{k}}\phantom{\rule{0ex}{0ex}}$
The vector product of $\stackrel{\mathit{\to }}{\mathit{A}}\mathit{×}\stackrel{\mathit{\to }}{\mathit{B}}$ can be obtained as follows:

#### Page No 29:

Given: are mutually perpendicular. $\stackrel{\mathit{\to }}{\mathit{A}}\mathit{×}\stackrel{\mathit{\to }}{\mathit{B}}$ is a vector with its direction perpendicular to the plane containing .

∴ The angle between is either 0° or 180°.

i.e., $\stackrel{\to }{\mathrm{C}}×\left(\stackrel{\to }{\mathrm{A}}×\stackrel{\to }{\mathrm{B}}\right)=0$

However, the converse is not true. For example, if two of the vectors are parallel, then also, $\stackrel{\to }{\mathrm{C}}×\left(\stackrel{\to }{\mathrm{A}}×\stackrel{\to }{\mathrm{B}}\right)=0$

So, they need not be mutually perpendicular.

#### Page No 29:

The particle moves on the straight line XX' at a uniform speed ν.

In $∆\mathrm{POQ}$, we have:
OQ = OP sin θ

This product is always equal to the perpendicular distance from point O. Also, the direction of this product remains constant.
So, irrespective of the the position of the particle, the magnitude and direction of $\stackrel{\to }{\mathrm{OP}}×\stackrel{\to }{\mathrm{\nu }}$ remain constant.
is independent of the position P.

#### Page No 29:

According to the problem, the net electric and magnetic forces on the particle should be zero.

i.e., $\stackrel{\to }{F}=q\stackrel{\to }{E}+q\left(\stackrel{\to }{\nu }×\stackrel{\to }{B}\right)=0$

So, the direction of should be opposite to the direction of $\stackrel{\to }{E}$. Hence, $\stackrel{\to }{\nu }$ should be along the positive z-direction.

Again, E = νB sin θ

For ν to be minimum,
So, the particle must be projected at a minimum speed of $\frac{E}{B}$ along the z-axis.

#### Page No 29:

To prove:
Suppose that $\stackrel{\to }{A}$ is perpendicular to is along the west direction.
Also, $\stackrel{\mathit{\to }}{\mathit{B}}$ is perpendicular to and $\stackrel{\to }{C}$ are along the south and north directions, respectively.

$\stackrel{\to }{A}$ is perpendicular to $\stackrel{\mathit{\to }}{\mathit{B}}$, so there dot or scalar product is zero.
i.e., $\stackrel{\to }{\mathrm{A}}·\stackrel{\to }{\mathrm{B}}=\left|\stackrel{\to }{\mathrm{A}}\right|\left|\stackrel{\to }{\mathrm{B}}\right|\mathrm{cos}\theta =\left|\stackrel{\to }{\mathrm{A}}\right|\left|\stackrel{\to }{\mathrm{B}}\right|\mathrm{cos}90°=0\phantom{\rule{0ex}{0ex}}$
$\stackrel{\to }{B}$ is perpendicular to $\stackrel{\to }{C}$, so there dot or scalar product is zero.
i.e., $\stackrel{\mathit{\to }}{C}\mathit{·}\stackrel{\mathit{\to }}{B}\mathit{=}\left|\stackrel{\mathit{\to }}{C}\right|\left|\stackrel{\mathit{\to }}{B}\right|cos\theta =\left|\stackrel{\mathit{\to }}{C}\right|\left|\stackrel{\mathit{\to }}{B}\right|cos90\mathit{°}=0$

Hence, proved.

#### Page No 29:

Note: Students should draw the graph y = 2x2 on a graph paper for results.

To find a slope at any point, draw a tangent at the point and extend the line to meet the x-axis. Then find tan θ as shown in the figure.

The above can be checked as follows:

Here, x = x-coordinate of the point where the slope is to be measured

#### Page No 29:

y = sin x   ...(i)

Now, consider a small increment ∆x in x.
Then y + ∆y = sin (x + ∆x)   ...(ii)
Here, ∆y is the small change in y.

Subtracting (ii) from (i), we get:
y = sin (x + ∆x) − sin x

= 0.0157

#### Page No 29:

Given: i = i0 et/RC
∴ Rate of change of current $=\frac{di}{dt}$

$=\frac{-{i}_{0}}{RC}×{e}^{-t/RC}$
On applying the conditions given in the questions, we get:
(a) At
(b) At
(c) At $=\frac{-{i}_{0}}{RC{e}^{10}}$

#### Page No 29:

Electric current in a discharging R-C circuit is given by the below equation:
i
= i0et/RC   ...(i)
Here, i0 = 2.00 A
R = 6 × 105 Ω
C = 0.0500 × 10−6 F
= 5 × 10−7 F

On substituting the values of R, C and i0 in equation (i), we get:
i = 2.0 et/0.3   ...(ii)

According to the question, we have:

(a) current at t = 0.3 s

(b) rate of change of current at t = 0.3 s
$\frac{di}{dt}=\frac{-{i}_{0}}{RC}·{e}^{-t/RC}$
When t = 0.3 s, we have:

(c) approximate current at t = 0.31 s

#### Page No 30:

The given equation of the curve is y = 3x2 + 6x + 7.

The area bounded by the curve and the X-axis with coordinates x1 = 5 and x2 = 10 is given by

#### Page No 30:

The given equation of the curve is y = sin x.

The required area can found by integrating y w.r.t x within the proper limits.

#### Page No 30:

The given function is y = ex.
When x = 0, y = e−0 = 1

When x increases, the value of y decrease. Also, only when x = ∞, y = 0

So, the required area can be determined by integrating the function from 0 to ∞.

#### Page No 30:

ρ = mass/length = a + bx

So, the SI unit of ρ is kg/m.

(a)
SI unit of a = kg/m
SI unit of b = kg/m2
(From the principle of homogeneity of dimensions)

(b) Let us consider a small element of length dx at a distance x from the origin as shown in the figure given below:

dm = mass of the element
= ρdx
= (a + dx) dx
∴ Mass of the rod = ∫ dm

#### Page No 30:

According to the question, we have:

Momentum is zero at time, t = 0
Now, dp = [(10 N) + (2 Ns−1)t]dt

On integrating the above equation, we get:

#### Page No 30:

Changes in a function of y and the independent variable x are related as follows:

Integrating of both sides, we get:
dy = ∫x2 dx
$⇒y=\frac{{x}^{3}}{3}+c$, where c is a constant
y as a function of x is represented by $y=\frac{{x}^{3}}{3}+c$.

#### Page No 30:

(a) 1001
Number of significant digits = 4
(b) 100.1
Number of significant digits = 4
(c) 100.10
Number of significant digits = 5
(d) 0.001001
Number of significant digits = 4

#### Page No 30:

The metre scale is graduated at every millimetre.
i.e., 1 m = 1000 mm
The minimum number of significant digits may be one (e.g., for measurements like 4 mm and 6 mm) and the maximum number of significant digits may be 4 (e.g., for measurements like 1000 mm). Hence, the number of significant digits may be 1, 2, 3 or 4.

#### Page No 30:

(a) In 3472, 7 comes after the digit 4. Its value is greater than 5. So, the next two digits are neglected and 4 is increased by one.
∴ The value becomes 3500.
(b) 84
(c) 2.6
(d) 29

#### Page No 30:

Length of the cylinder, l = 4.54 cm
Radius of the cylinder, r = 1.75 cm
Volume of the cylinder, V = $\pi$r2l
= ($\pi$) × (4.54) × (1.75)2

The minimum number of significant digits in a particular term is three. Therefore, the result should have three significant digits, while the other digits should be rounded off.

∴ Volume, V = $\pi$r2l
= (3.14) × (1.75) × (1.75) × (4.54)
= 43.6577 cm3
Since the volume is to be rounded off to 3 significant digits, we have:
V = 43.7 cm3

#### Page No 30:

∴ Rounding off to three significant digits, the average thickness becomes 2.17 mm.