Exercise Problems

Hint

The given vector is the resultant of two perpendicular vectors, one along the $X$-axis of magnitude 25 unit and the other along the $Y$-axis of magnitude 60 units. The resultant has a magnitude $A$ given by
$\begin{aligned} A &=\sqrt{(25)^2+(60)^2+2 \times 25 \times 60 \cos 90^{\circ}} \\ &=\sqrt{(25)^2+(60)^2}=65 . \end{aligned}$
The angle $\alpha$ between this vector and the $X$-axis is given by
$\tan \alpha=\frac{60}{25}$

Hint

The $x$-component of the $5.0 \mathrm{~m}$ vector $=5.0 \mathrm{~m} \cos 37^{\circ}$ $=4 \cdot 0 \mathrm{~m}$,
the $x$-component of the $3.0 \mathrm{~m}$ vector $=3.0 \mathrm{~m}$ and the $x$-component of the $2.0 \mathrm{~m}$ vector $=2.0 \mathrm{~m} \cos 90^{\circ}$ $=0$.
Hence, the $x$-component of the resultant
$=4.0 \mathrm{~m}+3.0 \mathrm{~m}+0=7.0 \mathrm{~m} \text {. }$
The $y$-component of the $5.0 \mathrm{~m}$ vector $=5.0 \mathrm{~m} \sin 37^{\circ}$
$=3.0 \mathrm{~m} \text {, }$
the $y$-component of the $3.0 \mathrm{~m}$ vector $=0$
and the $y$-component of the $2.0 \mathrm{~m}$ vector $=2.0 \mathrm{~m}$.
Hence, the $y$-component of the resultant
$=3.0 \mathrm{~m}+0+2.0 \mathrm{~m}=5.0 \mathrm{~m} \text {. }$
The magnitude of the resultant vector
$\begin{aligned} &=\sqrt{(7 \cdot 0 \mathrm{~m})^2+(5 \cdot 0 \mathrm{~m})^2} \\ &=8.6 \mathrm{~m} . \end{aligned}$
If the angle made by the resultant with the $X$-axis is $\theta$, then
$\tan \theta=\frac{y \text {-component }}{x \text {-component }}=\frac{5 \cdot 0}{7 \cdot 0} \text { or, } \theta=35 \cdot 5^{\circ} .$

Hint

The $x$-component of $\overrightarrow{O A}=(O A) \cos 90^{\circ}=0$.
The $x$-component of $\overrightarrow{O B}=(O B) \cos 0^{\circ}=O B$.
The $x$-component of $\overrightarrow{O C}=(O C) \cos 135^{\circ}=-\frac{1}{\sqrt{2}} O C$.
Hence, the $x$-component of the resultant
$=O B-\frac{1}{\sqrt{ } 2} O C \text {. } \dots(i)$
It is given that the resultant is zero and hence its $x$-component is also zero. From (i),
$O B=\frac{1}{\sqrt{2}} O C \text {. } \dots(ii)$
The $y$-component of $\overrightarrow{O A}=O A \cos 180^{\circ}=-O A$.
The $y$-component of $\overrightarrow{O B}=O B \cos 90^{\circ}=0$.
The $y$-component of $\overrightarrow{O C}=O C \cos 45^{\circ}=\frac{1}{\sqrt{2}} O C$.
Hence, the $y$-component of the resultant
$=\frac{1}{\sqrt{2}} O C-O A \dots(iii)$
As the resultant is zero, so is its $y$-component. From (iii),
$\frac{1}{\sqrt{2}} O C=O A, \text { or, } O C=\sqrt{ } 2 O A=5 \sqrt{ } 2 \mathrm{~m} .$
From (ii), $O B=\frac{1}{\sqrt{2}} O C=5 \mathrm{~m}$.

Hint

Let $O A=O B=O C=F$.
$x$-component of $\overrightarrow{O A}=F \cos 30^{\circ}=F \frac{\sqrt{3}}{2}$.
$x$-component of $\overrightarrow{O B}=F \cos 60^{\circ}=\frac{F}{2}$
$x$-component of $\overrightarrow{O C}=F \cos 135^{\circ}=-\frac{F}{\sqrt{2}}$.
$x$-component of $\overrightarrow{O A}+\overrightarrow{O B}-\overrightarrow{O C}$
$=\left(\frac{F \sqrt{ } 3}{2}\right)+\left(\frac{F}{2}\right)-\left(-\frac{F}{\sqrt{ } 2}\right)$
$=\frac{F}{2}(\sqrt{ } 3+1+\sqrt{ } 2) \text {. }$
$y$-component of $\overrightarrow{O A}=F \cos 60^{\circ}=\frac{F}{2}$.
$y$-component of $\overrightarrow{O B}=F \cos 150^{\circ}=-\frac{F \sqrt{3}}{2}$.
$y$-component of $\overrightarrow{O C}=F \cos 45^{\circ}=\frac{F}{\sqrt{2}}$.
$y$-component of $\overrightarrow{O A}+\overrightarrow{O B}-\overrightarrow{O C}$
$\begin{aligned} &=\left(\frac{F}{2}\right)+\left(-\frac{F \sqrt{ } 3}{2}\right)-\left(\frac{F}{\sqrt{ } 2}\right) \\ &=\frac{F}{2}(1-\sqrt{ } 3-\sqrt{ } 2) . \end{aligned}$
Angle of $\overrightarrow{O A}+\overrightarrow{O B}-\overrightarrow{O C}$ with the $X$-axis
$=\tan ^{-1} \frac{\frac{F}{2}(1-\sqrt{ } 3-\sqrt{ } 2)}{\frac{F}{2}(1+\sqrt{ } 3+\sqrt{ } 2)}=\tan ^{-1} \frac{(1-\sqrt{3}-\sqrt{ } 2)}{(1+\sqrt{ } 3+\sqrt{ } 2)} .$

Hint

$O A=O C$.
$\overrightarrow{O A}+\overrightarrow{O C}$ is along $\overrightarrow{O B}$ (bisector) and its magnitude is $2 R \cos 45^{\circ}=R \sqrt{ } 2$
$(\overrightarrow{O A}+\overrightarrow{O C})+\overrightarrow{O B}$ is along $\overrightarrow{O B}$ and its magnitude is $R \sqrt{ } 2+R=R(1+\sqrt{ } 2)$

Hint

Take the dotted lines as $X, Y$ axes. $x$-component of $\overrightarrow{O A}=4 \mathrm{~m}, x$-component of $\overrightarrow{O B}=6 \mathrm{~m} \cos \theta$
$x$-component of the resultant $=(4+6 \cos \theta) \mathrm{m}$.
But it is given that the resultant is along $Y$-axis. Thus, the $x$-component of the resultant $=0$
$4+6 \cos \theta=0$ or, $\cos \theta=-2 / 3$.

Hint

$\quad|\vec{A}|=\sqrt{5^2+1^2+(-2)^2}=\sqrt{30}$
The required unit vector is $\frac{\vec{A}}{|\vec{A}|}$
$=\frac{5}{\sqrt{30}} \vec{i}+\frac{1}{\sqrt{30}} \vec{j}-\frac{2}{\sqrt{30}} \vec{k}$

Hint

$\begin{aligned} &\text { We have }|\vec{a}+\vec{b}|^2=(\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})\\ &=\vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}\\ &=a^2+b^2+2 \vec{a} \cdot \vec{b} \text {. }\\ &|\vec{a}-\vec{b}|^2=(\vec{a}-\vec{b}) \cdot(\vec{a}-\vec{b})\\ &=a^2+b^2-2 \vec{a} \cdot \vec{b} \text {. }\\ &\text { If } \quad|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}| \text {, }\\ &a^2+b^2+2 \vec{a} \cdot \vec{b}=a^2+b^2-2 \vec{a} \cdot \vec{b}\\ &\text { or, } \quad \vec{a} \cdot \vec{b}=0\\ &\text { or, } \quad \vec{a} \perp \vec{b} \end{aligned}$

Hint

We have $\vec{a} \cdot \vec{b}=a b \cos \theta$ or, $\quad \cos \theta=\frac{\vec{a} \cdot \vec{b}}{a b}$ where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.
Now
$\begin{aligned} \vec{a} \cdot \vec{b} &=a_x b_x+a_y b_y+a_z b_z \\ &=2 \times 4+3 \times 3+4 \times 2=25 . \end{aligned}$
Also
$\begin{aligned} a &=\sqrt{a_x^2+a_y^2+a_z^2} \\ &=\sqrt{4+9+16}=\sqrt{29} \end{aligned}$
and
$b=\sqrt{b_x^2+b_y^2+b_z^2}=\sqrt{16+9+4}=\sqrt{29} .$
Thus,
$\cos \theta=\frac{25}{29}$
or,
$\theta=\cos ^{-1}\left(\frac{25}{29}\right) .$

Hint

$\begin{aligned} \vec{B} \times \vec{C} &=(\vec{i}+2 \vec{k}) \times(\vec{j}-\vec{k}) \\ &=\vec{i} \times(\vec{j}-\vec{k})+2 \vec{k} \times(\vec{j}-\vec{k}) \\ &=\vec{i} \times \vec{j}-\vec{i} \times \vec{k}+2 \vec{k} \times \vec{j}-2 \vec{k} \times \vec{k} \\ &=\vec{k}+\vec{j}-2 \vec{i}-0=-2 \vec{i}+\vec{j}+\vec{k} \\ \vec{A} \cdot(\vec{B} \times \vec{C}) &=(2 \vec{i}-3 \vec{j}+7 \vec{k}) \cdot(-2 \vec{i}+\vec{j}+\vec{k}) \\ &=(2)(-2)+(-3)(1)+(7)(1) \\ &=0 \end{aligned}$

Hint

(a) $\quad V=\frac{4}{3} \pi R^3$
or, $\quad \frac{d V}{d R}=\frac{4}{3} \pi \frac{d}{d R}(R)^3=\frac{4}{3} \pi \cdot 3 R^2=4 \pi R^2$.
(b) At $R=20 \mathrm{~cm}$, the rate of change of volume with the radius is
$\begin{aligned} \frac{d V}{d R} &=4 \pi R^2=4 \pi\left(400 \mathrm{~cm}^2\right) \\ &=1600 \pi \mathrm{cm}^2 . \end{aligned}$
The change in volume as the radius changes from $20.0 \mathrm{~cm}$ to $20.1 \mathrm{~cm}$ is
$\begin{aligned} \Delta V &=\frac{d V}{d R} \Delta R \\ &=\left(1600 \pi \mathrm{cm}^2\right)(0.1 \mathrm{~cm}) \\ &=160 \pi \mathrm{cm}^3 . \end{aligned}$

Hint

(a)

$\begin{aligned} y &=x^2 \sin x \\ \frac{d y}{d x} &=x^2 \frac{d}{d x}(\sin x)+(\sin x) \frac{d}{d x}\left(x^2\right) \\ &=x^2 \cos x+(\sin x)(2 x) \\ &=x(2 \sin x+x \cos x) \end{aligned}$

(b)

$\begin{aligned} y &=\frac{\sin x}{x} \\ \frac{d y}{d x} &=\frac{x \frac{d}{d x}(\sin x)-\sin x\left(\frac{d x}{d x}\right)}{x^2} \\ &=\frac{x \cos x-\sin x}{x^2} . \end{aligned}$

(c)

$\begin{aligned} \frac{d y}{d x} &=\frac{d}{d x^2}\left(\sin x^2\right) \cdot \frac{d\left(x^2\right)}{d x} \\ &=\cos x^2(2 x) \\ &=2 x \cos x^2 \end{aligned}$

Hint

$\begin{aligned} y &=x+\frac{1}{x} \\ \frac{d y}{d x} &=\frac{d}{d x}(x)+\frac{d}{d x}\left(x^{-1}\right) \\ &=1+\left(-x^{-2}\right) \\ &=1-\frac{1}{x^2} . \end{aligned}$
For $y$ to be maximum or minimum,
$\frac{d y}{d x}=0$
or, $\quad 1-\frac{1}{x^2}=0$
Thus, $\quad x=1$ or $-1$.
For $x>0$ the only possible maximum or minimum is at $x=1$. At $x=1, y=x+\frac{1}{x}=2$.

Near $x=0, y=x+\frac{1}{x}$ is very large because of the term $\frac{1}{x}$. For very large $x$, again $y$ is very large because of the term $x$. Thus $x=1$ must correspond to a minimum. Thus, $y$ has only a minimum for $x>0$. This minimum occurs at $x=1$ and the minimum value of $y$ is $y=2$.

Hint

The area can be divided into strips by drawing ordinates between $x=0$ and $x=6$ at a regular interval of $d x$. Consider the strip between the ordinates at $x$ and $x+d x$. The height of this strip is $y=x^2$. The area of this strip is $d A=y d x=x^2 d x$.

The total area of the shaded part is obtained by summing up these strip-areas with $x$ varying from 0 to 6. Thus
$A=\int_0^6 x^2 d x$
$=\left[\frac{x^3}{3}\right]_0^6=\frac{216-0}{3}=72 .$

Hint

$\begin{aligned} & \int_0^t A \sin \omega t d t \\ =& A\left[\frac{-\cos \omega t}{\omega}\right]_0^t=\frac{A}{\omega}(1-\cos \omega t) . \end{aligned}$

Hint

We have
$\begin{aligned} v \frac{d v}{d x} &=-\omega^2 x \\ \text { or, } & v d v &=-\omega^2 x d x \\ \text { or, } & \int_{v_0}^v v d v &=\int_0^x-\omega^2 x d x \dots(i) \end{aligned}$
When summation is made on $-\omega^2 x d x$ the quantity to be varied is $x$. When summation is made on $v d v$ the quantity to be varied is $v$. As $x$ varies from 0 to $x$ the velocity varies from $v_0$ to $v$. Therefore, on the left the limits of integration are from $v_0$ to $v$ and on the right, they are from 0 to $x$. Simplifying (i),
$\begin{aligned} {\left[\frac{1}{2} v^2\right]_{v_0}^v } =-\omega^2\left[\frac{x^2}{2}\right]_0^x \\ \text { or, }  \frac{1}{2}\left(v^2-v_0^2\right) =-\omega^2 \frac{x^2}{2} \\ \text { or, }  v^2 =v_0^2-\omega^2 x^2 \\ \text { or, }  v =\sqrt{v_0^2-\omega^2 x^2} . \end{aligned}$

Hint

The total charge flown is the sum of all the $d q$ ‘s for $t$ varying from $t=0$ to $t=\tau$. Thus, the total charge flown is
$\begin{aligned} Q &=\int_0^\tau e^{-t / \tau} d t \\ &=\left[\frac{e^{-t / \tau}}{-1 / \tau}\right]_0^\tau=\tau\left(1-\frac{1}{e}\right) \end{aligned}$

Hint

$\begin{gathered} \quad \quad  21.6002 \\ 234 \\ 2732 \cdot 10 \\ \hline \end{gathered}$

$\begin{array}{r} 22 \\ \quad 234 \\ 2732 \\ \hline 2988 \end{array}$

The three numbers are arranged with their decimal points aligned (shown on the left part above). The column just left to the decimals has 4 as the doubtful digit. Thus, all the numbers are rounded to this column. The rounded numbers are shown on the right part above. The required expression is $2988 \times 13=38844$. As 13 has only two significant digits the product should be rounded off after two significant digits. Thus the result is 39000.

Hint

Angle between $\vec{A}$ and $\vec{B}=110^{\circ}-20^{\circ}=90^{\circ},|\vec{A}|=3 \mathrm{~m}$ and $|\vec{B}|=4 \mathrm{~m}$
Magnitude of the resultant vector is given by
$\begin{aligned} & R=\sqrt{A^2+B^2+2 A B \cos \theta} \\ & =\sqrt{3^2+4^2+2 \times 3 \times 4 \times \cos 90^{\circ}} \\ & =5 \mathrm{~m} \end{aligned}$
Let $\beta$ be the angle between $\vec{R}$ and $\vec{A}$.
$\begin{aligned} & \beta=\tan ^{-1}\left(\frac{A \sin 90^{\circ}}{A+B \cos 90^{\circ}}\right) \\ & =\tan ^{-1}\left(\frac{4 \sin 90^{\circ}}{3+4 \cos 90^{\circ}}\right) \\ & =\tan ^{-1} \frac{4}{3} \\ & =\tan ^{-1}(1.333) \\ & =53^{\circ} \end{aligned}$
Now, angle made by the resultant vector with the $X$-axis $=53^{\circ}+20^{\circ}=73^{\circ}$
$\therefore$ The resultant $\vec{R}$ is $5 \mathrm{~m}$ and it makes an angle of $73^{\circ}$ with the $\mathrm{x}$-axis.

Hint

Angle between $\vec{A}$ and $\vec{B}$ is $\theta=60^{\circ}-30^{\circ}=30^{\circ}$
$|\vec{A}|$ and $|\vec{B}|=10$ unit
$R=\sqrt{10^2+10^2+2 \cdot 10 \cdot 10 \cdot \cos 30^{\circ}}=19.3$
$\beta$ be the angle between $\vec{R}$ and $\vec{A}$
$\beta=\tan ^{-1}\left(\frac{10 \sin 30^{\circ}}{10+10 \cos 30^{\circ}}\right)=\tan ^{-1}\left(\frac{1}{2+\sqrt{3}}\right)=\tan ^{-1}(0.26795)=15^{\circ}$
Resultant makes $15^{\circ}+30^{\circ}=45^{\circ}$ angle with $x$-axis.

Hint

$x$ component of $\vec{A}=100 \cos 45^{\circ}=100 / \sqrt{2}$ unit
$x$ component of $\vec{B}=100 \cos 135^{\circ}=100 / \sqrt{2}$
$x$ component of $\dot{C}=100 \cos 315^{\circ}=100 / \sqrt{2}$
Resultant $x$ component $=100 / \sqrt{2}-100 / \sqrt{2}+100 / \sqrt{2}=100 / \sqrt{2}$
$y$ component of $\vec{A}=100 \sin 45^{\circ}=100 / \sqrt{2}$ unit
$y$ component of $\vec{B}=100 \sin 135^{\circ}=100 / \sqrt{2}$
$y$ component of $\vec{C}=100 \sin 315^{\circ}=-100 / \sqrt{2}$
Resultant $y$ component $=100 / \sqrt{2}+100 / \sqrt{2}-100 / \sqrt{2}=100 / \sqrt{2}$
Resultant $=100$
Tan $\alpha=\frac{y \text { component }}{x \text { component }}=1$
$\Rightarrow \alpha=\tan ^{-1}(1)=45^{\circ}$
The resultant is 100 unit at $45^{\circ}$ with $x$-axis.

Hint

Given: $\vec{a}=4 \vec{i}+3 \vec{j}$ and $\vec{b}=3 \vec{i}+4 \vec{j}$
(a) Magnitude of $\vec{a}$ is given by $|\vec{a}|=\sqrt{4^2+3^2}=\sqrt{16+9}=5$
Magnitude of $\vec{b}$ is given by $|\vec{b}|=\sqrt{3^2+4^2}=\sqrt{9+16}=5$
(c) $\vec{a}+\vec{b}=(4 \hat{i}+3 \hat{j})+(3 \hat{i}+4 \hat{j})=(7 \hat{i}+7 \hat{j})$
$\therefore$ Magnitude of vector $\vec{a}+\vec{b}$ is given by $|\vec{a}+\vec{b}|=\sqrt{49+49}=\sqrt{98}=7 \sqrt{2}$
(d) $\vec{a}-\vec{b}=(4 \vec{i}+3 \vec{j})-(3 \vec{i}+4 \vec{j})=\vec{i}-\vec{j}$
$\therefore$ Magnitude of vector $\vec{a}-\vec{b}$ is given by $|\vec{a}-\vec{b}|=\sqrt{(1)^2+(-1)^2}=\sqrt{2}$

Hint

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 $\overrightarrow{O A}=2 \cos 30^{\circ}=\sqrt{3}$
x-component of $\overrightarrow{B C}=1.5 \cos 120^{\circ}$
$=-\frac{(1.5)}{2}=-7.5$
$\begin{aligned} & \text { x-component of } \overrightarrow{D E}=1 \cos 270^{\circ} \\ & \quad=1 \times 0=0 \mathrm{~m} \end{aligned}$
$\mathrm{y}$-component of $\overrightarrow{O A}=2 \sin 30^{\circ}=1$
$\mathrm{y}$-component of $\overrightarrow{B C}=1.5 \sin 120^{\circ}$
$=\frac{(\sqrt{3} \times 1.5)}{2}=1.3$
$\begin{aligned} & \text { y-component of } \overrightarrow{D E}=1 \sin 270^{\circ}=-1 \\ & \text { x-component of resultant } R_x=\sqrt{3}-0.75+0=0.98 \mathrm{~m} \\ & y \text {-component of resultant } R_y=1+1.3-1=1.3 \mathrm{~m} \\ & \therefore \text { Resultant, } R=\sqrt{\left(R_x\right)^2+\left(R_y\right)^2} \\ & =\sqrt{(0.98)^2+(1.3)^2} \\ & =\sqrt{0.96+1.69} \\ & =\sqrt{2.65} \\ & =1.6 \mathrm{~m} \end{aligned}$
If it makes an angle $\alpha$ with the positive $x$-axis, then
$\begin{aligned} & \tan \alpha=\frac{y \text {-component }}{\mathrm{x} \text { – component }} \\ & =\frac{1.3}{0.98}=1.332 \\ & \therefore \alpha=\tan ^{-1}(1.32) \end{aligned}$

Hint

Let the two vectors be $\vec{a}$ and $\vec{b}$
Now,
$|\vec{a}|=3 \text { and }|\vec{b}|=4$
(a) If the resultant vector is 1 unit, then
$\begin{aligned} & \sqrt{\vec{a}^2+\vec{b}^2+2 \cdot \vec{a} \cdot \vec{b} \cos \theta}=1 \\ & \Rightarrow \sqrt{3^2+4^2+2 \cdot 3 \cdot 4 \cos \theta}=1 \end{aligned}$
Squaring both sides, we get:
$\begin{aligned} & 25+24 \cos \theta=1 \\ & \Rightarrow \theta=180^{\circ} \end{aligned}$
Hence, the angle between them is $180^{\circ}$.
(b) If the resultant vector is 5 units, then
$\begin{aligned} & \sqrt{\vec{a}^2+\vec{b}^2+2 \cdot \vec{a} \cdot \vec{b} \cos \theta}=5 \\ & \Rightarrow \sqrt{3^2+4^2+2 \cdot 3 \cdot 4 \cos \theta}=5 \end{aligned}$
Squaring both sides, we get:
$\begin{aligned} & 25+24 \cos \theta=25 \\ \Rightarrow & \cos \theta=90^{\circ} \end{aligned}$
Hence, the angle between them is $90^{\circ}$.
(c) If the resultant vector is 7 units, then
$\begin{aligned} & \sqrt{\vec{a}^2+\vec{b}^2+2 \cdot \vec{a} \cdot \vec{b} \cos \theta}=1 \\ & \Rightarrow \sqrt{3^2+4^2+2 \cdot 3 \cdot 4 \cos \theta}=7 \end{aligned}$
Squaring both sides, we get:
$\begin{aligned} & 25+24 \cos \theta=49, \\ & \Rightarrow \theta=\cos ^{-1} 1=0^{\circ} \end{aligned}$
Hence, the angle between them is $0^{\circ}$.

Hint

The displacement of the car is represented by $\overrightarrow{A D}$.
$\begin{aligned} & \overrightarrow{A D}=2 \hat{i}+0.5 \hat{j}+4 \hat{i} \\ & =6 \hat{i}+0.5 \hat{j} \end{aligned}$
Magnitude of $\overrightarrow{A D}$ is given by
$\begin{aligned} & A D=\sqrt{A E^2+D E^2} \\ & =\sqrt{6^2+(0.5)^2} \\ & =\sqrt{36+0.25}=6.02 \mathrm{~km} \end{aligned}$
Now,
$\begin{aligned} & \tan \theta=\frac{D E}{A E}=\frac{1}{12} \\ & \Rightarrow \theta=\tan ^{-1}\left(\frac{1}{12}\right) \end{aligned}$
Hence, the displacement of the car is $6.02 \mathrm{~km}$ along the direction $\tan ^{-1}\left(\frac{1}{12}\right)$ with positive the $x$-axis.

Hint

$\begin{aligned} & \text { In } \triangle A B C, \tan \theta=\frac{x}{2} \\ & \text { and in } \triangle D C E, \tan \theta=\frac{2-x}{4} \\ & \tan \theta=\frac{x}{2}=\frac{(2-x)}{4} \\ & \Rightarrow 6 x=4 \\ & \Rightarrow x=\frac{2}{3} \mathrm{ft} \\ & \text { (a) } \ln \triangle A B C, A C=\sqrt{A^2+B C^2} \\ & =\sqrt{\left(\frac{2^2}{3}\right)}+2^2 \\ & =\sqrt{\frac{4}{9}+4}=\sqrt{\frac{40}{9}} \\ & =\frac{2}{3} \sqrt{10} \mathrm{ft} \end{aligned}$
(b) In $\triangle C D E, D E=2-\frac{2}{3}=\frac{6-2}{3}=\frac{4}{3} \mathrm{ft}$ since $C D=4 \mathrm{ft}$, therefore $C E$
$\begin{aligned} & =\sqrt{C D^2+D E^2} \\ & =\sqrt{4^2+\left(\frac{4^2}{3}\right)}=\frac{4}{3} \sqrt{10} \mathrm{ft} \end{aligned}$
$\begin{aligned} & \text { (c) In } \triangle A G E, A E=\sqrt{\mathrm{AG}^2+G E^2} \\ & =\sqrt{2^2+2^2} \\ & =\sqrt{8}=2 \sqrt{2} \mathrm{ft} \end{aligned}$

Hint

Displacement vector
$\vec{r}=7 \hat{i}+4 \hat{i}+3 \hat{k}$
(a) Magnitude of displacement
$\begin{aligned} & =\sqrt{7^2+4^2+3^2} \\ & =\sqrt{74} \mathrm{ft} \end{aligned}$
(b) Components of the displacement vector are $7 \mathrm{ft}, 4 \mathrm{ft}$ and $3 \mathrm{ft}$.

Hint

Given: $\vec{a}$ is a vector of magnitude $4.5$ units due north.
Case (a):
$3|\vec{a}|=3 \times 4.5=13.5$
$\therefore 3 \vec{a}$ is a vector of magnitude $13.5$ units due north.
Case (b):
$|-4 \vec{a}|=4 \times 4.5=18$ units
$\therefore-4 \vec{a}$ is a vector of magnitude 18 units due south.

Hint

Let the two vectors be $|\vec{a}|=2 m$ and $|\vec{b}|=3 m$.
Angle between the vectors, $\theta=60^{\circ}$
(a) The scalar product of two vectors is given by $\vec{a} \cdot \vec{b}=|\vec{a}| \cdot|\vec{b}| \cos \theta^{\circ}$
$\begin{aligned} & \therefore \vec{a} \cdot \vec{b}=|\vec{a}| \cdot|\vec{b}| \cos 60^{\circ} \\ & =2 \times 3 \times \frac{1}{2}=3 m^2 \end{aligned}$
(b) The vector product of two vectors is given by $|\vec{a} \times \vec{b}|=|\vec{a}||\vec{a}| \sin \theta^{\circ}$.
$\begin{aligned} & \therefore|\vec{a} \times \vec{b}|=|\vec{a}||\vec{a}| \sin 60^{\circ} \\ & =2 \times 3 \times \frac{\sqrt{3}}{2} \\ & =3 \sqrt{3} m^2 \end{aligned}$

Hint

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, $R_x=A \cos 0+A \cos \frac{\pi}{3}+A \cos \frac{2 \pi}{3}+A \cos \frac{3 \pi}{3}+A \cos \frac{4 \pi}{3}+A \cos \frac{5 \pi}{3}=0$
[As the resultant is zero, the $x$-component of resultant $R_X$ is zero]
$\Rightarrow \cos 0+\cos \frac{\pi}{3}+\cos \frac{2 \pi}{3}+\cos \frac{3 \pi}{3}+\cos \frac{4 \pi}{3}+\cos \frac{5 \pi}{5}=0$

Hint

We have:
$\begin{aligned} & \vec{a}=2 \vec{i}+3 \vec{j}+4 \vec{k} \\ & \vec{b}=3 \vec{i}+4 \vec{j}+5 \vec{k} \end{aligned}$
Using scalar product, we can find the angle between vectors $\vec{a}$ and $\vec{b}$. i.e.,
$\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta$
So, $\theta=\cos ^{-1}\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\right)$
$=\cos ^{-1}\left(\frac{2 \times 3+3 \times 4+4 \times 5}{\sqrt{\left(2^2+3^2+4^2\right)} \sqrt{\left(3^2+4^2+5^2\right)}}\right)$
$=\cos ^{-1}\left(\frac{38}{\sqrt{29} \sqrt{50}}=\cos ^{-1} \frac{38}{\sqrt{1450}}\right)$
$\therefore$ The required angle is $\cos ^{-1} \frac{38}{\sqrt{1450}}$.

Hint

Vector product is given by $\vec{A} \times \vec{B}=|\vec{A}||\vec{B}| \sin \hat{n}$
$|\vec{A}||\vec{B}| \sin \hat{n}$
is a vector which is perpendicular to the plane containing
$\vec{A}$ and $\vec{B}$. This implies that it is also perpendicular to $\vec{A}$. We know that the dot product of two perpendicular vectors is zero.
$\therefore \vec{A} \cdot(\vec{A} \times \vec{B})=0$

Hint

Given:
$\begin{aligned} & \vec{A}=2 \hat{i}+3 \hat{j}+4 \hat{k} \text { and } \\ & \vec{B}=4 \hat{i}+3 \hat{j}+2 \hat{k} \end{aligned}$
The vector product of $\vec{A} \times \vec{B}$ can be obtained as follows:
$\begin{aligned} & \vec{A} \times \vec{B}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 4 & 3 & 2 \end{array}\right| \\ & =\hat{i}(6-12)-\hat{j}(4-16)+\hat{k}(6-12) \\ & =-6 \hat{i}+12 \hat{j}-6 \hat{k} \end{aligned}$

Hint

Given that $\vec{A}, \vec{B}$ and $\vec{C}$ are mutually perpendicular
$\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}$ is a vector which direction is perpendicular to the plane containing $\overrightarrow{\mathrm{A}}$ and $\overrightarrow{\mathrm{B}}$
Also $\vec{C}$ is perpendicular to $\vec{A}$ and $\vec{B}$
Angle between $\overrightarrow{\mathrm{C}}$ and $\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}$ is $0^{\circ}$ or $180^{\circ}$ (figure below)
So, $\overrightarrow{\mathrm{C}} \times(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{B}})=0$
The converse is not true.
For example, if two of the vector are parallel, (figure below), then also $\overrightarrow{\mathrm{C}} \times(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}})=0$
So, they need not be mutually perpendicular.

Hint

It is independent of the position $\mathrm{P}$ because
The particle moves on the straight line PP’ at speed v. From the figure,
$\overrightarrow{\mathrm{OP}} \times v=(O P) v \sin \theta \hat{n}=v(O P) \sin \theta \hat{n}=v(O Q) \hat{n}$
It can be seen from the figure, $O Q=O P \sin \theta=O P^{\prime} \sin \theta$
So, whatever may be the position of the particle, the magnitude and direction of $\overrightarrow{\mathrm{OP}} \times \vec{v}$ remain constant.
$\overrightarrow{\mathrm{OP}} \times \overrightarrow{\mathrm{V}}$ is independent of the position $\mathrm{P}$.

Hint

According to the problem, the net electric and magnetic forces on the particle should be zero.
i.e.,
$\begin{aligned} & \vec{F}=q \vec{E}+q(\vec{\nu} \times \vec{B})=0 \\ & \Rightarrow E=-(\vec{\nu} \times \vec{B}) \end{aligned}$
So, the direction of $\vec{\nu} \times \vec{B}$ should be opposite to the direction of $\vec{E}$ Hence, $\vec{\nu}$ should be along the positive $z$-direction.
Again, $E=v B \sin \theta$
$\Rightarrow \nu=\frac{E}{B} \sin \theta$
For $v$ to be minimum, $\theta=90$ and, thus, $\nu_{\min }=\frac{E}{B}$
So, the particle must be projected at a minimum speed of $\frac{E}{B}$ along the $z$-axis.

Hint

$y=\sin x \quad \ldots(i)$
Now, consider a small increment $\Delta \mathrm{x}$ in $\mathrm{x}$. Then $\mathrm{y}+\Delta \mathrm{y}=\sin (\mathrm{x}+\Delta \mathrm{x}) \quad$…(ii) Here, $\Delta y$ is the small change in $y$.
Subtracting (ii) from (i), we get:
$\begin{aligned} & \Delta y=\sin (x+\Delta x)-\sin x \\ & =\sin \left(\frac{\pi}{3}+\frac{\pi}{100}\right)-\sin \frac{\pi}{3} \\ & =0.0157 \end{aligned}$

Hint

$\begin{aligned} & \text { Given: } \mathrm{i}=\mathrm{i}_0 \mathrm{e}^{-\mathrm{t} / \mathrm{RC}} \\ & \text { Rate of change of current } \\ & =\frac{d i}{d t} \\ & =i_0\left(\frac{-1}{R C}\right) e^{-t / R C} \\ & =\frac{-i_0}{R C} \times e^{-t / R C} \end{aligned}$
On applying the conditions given in the questions, we get:
(a) At $t=0, \frac{d i}{d t}=\frac{-i_0}{R C} \times e^0=\frac{-i_0}{R C}$
(b) At $t=R C, \frac{d i}{d t}=\frac{-i_0}{R C} \times e^{-1}=\frac{-i_0}{R C e}$
(c) At $t=10 R C, \frac{d t}{d i}=\frac{-i_0}{R C} \times e^{10}=\frac{-i_0}{R C e^{10}}$

Hint

Electric current in a discharging R-C circuit is given by the below equation:
$\begin{aligned} & i=i_0 \cdot e^{-t / R C} \quad \ldots \text {..(i) } \\ & \text { Here, } \mathrm{i}_0=2.00 \mathrm{~A} \\ & R=6 \times 10^5 \Omega \\ & C=0.0500 \times 10^{-6} \mathrm{~F} \\ & =5 \times 10^{-7} \mathrm{~F} \\ & \end{aligned}$
On substituting the values of $R, C$ and $i_0$ in equation (i), we get:
$i=2.0 e^{-t / 0.3} \dots(ii)$
According to the question, we have:
(a) current at $\mathrm{t}=0.3 \mathrm{~s}$
$i=2 \times e^{-1}=\frac{2}{e} A$
(b) rate of change of current at $t=0.3 \mathrm{~s}$
$\frac{d i}{d t}=\frac{-i_0}{R C} \cdot e^{-t / R C}$
When $t=0.3 \mathrm{~s}$, we have:
$\frac{d i}{d t}=\frac{2}{0.30} \cdot e^{\left(\frac{-0.3}{0.3}\right)}=\frac{-20}{3 e} A / s$
(c) approximate current at $\mathrm{t}=0.31 \mathrm{~s}$
$\begin{aligned} & i=2 e^{\left(\frac{-0.3}{0.3}\right)} \\ & =\frac{5.8}{3 e} A(\text { approx. }) \end{aligned}$

Hint

The area bounded under the curve is

$y=3 x^2+6 x+7$
Area bounded by the curve, $x$ axis with coordinates with $x=5$ and $x=10$ is given by.
$\text { Area } \left.\left.\left.=\int_0^y d y=\int_5^{10}\left(3 x^2+6 x+7\right) d x=3 \frac{x^3}{3}\right]_5^{10}+5 \frac{x^2}{3}\right]_5^{10}+7 x\right]_5^{10}=1135 \text { sq.units. }$

Hint

The given equation of the curve is $y=\sin x$.
The required area can be found by integrating $y$ w.r.t $x$ within the proper limits.
$\begin{aligned} & \text { Area }=\int_{x_1}^{x_2} \text { y } d x=\int_0^\pi \sin \mathrm{x} d x \\ & =[-\cos x]_0^\pi \\ & =-\cos \pi-(-\cos 0) \\ & =1+1=2 \text { sq. unit } \end{aligned}$

Hint

The given function is $y=e^{-x}$.
When $x=0, y=e^{-0}=1$
When $x$ increases, the value of $y$ decrease. Also, only when $x=\infty, y=0$
So, the required area can be determined by integrating the function from 0 to $\infty$.
$\begin{aligned} & \text { Area }=\int_0^{\infty} e^{-x} d x \\ & =-\left[e^{-x}\right]_0^{\infty} \\ & =-[0-1]=1 \text { sq . unit } \end{aligned}$

Hint

Let us consider a small element of length $\mathrm{dx}$ at a distance $\mathrm{x}$ from the origin as shown in the figure given below:
$\begin{aligned} d m & =\text { mass of the element } \\ & =\rho d x \\ & =(a+d x) d x \end{aligned}$
Therefore, Mass of the rod $=\int d m$
$\begin{aligned} & =\int_0^L(a+b x) d x \\ & =\left[a x+\frac{b x^2}{2}\right]_0^L \\ & =a L+\frac{b L^2}{2} \end{aligned}$

Hint

According to the question, we have:
$\frac{d p}{d t}=(10 N)+(2 N / s) t$
Momentum is zero at time, $\mathrm{t}=0$
Now, $d p=\left[(10 \mathrm{~N})+\left(2 \mathrm{Ns}^{-1}\right) \mathrm{t}\right] \mathrm{dt}$
On integrating the above equation, we get:
$\begin{aligned} & p=\int_0^{10} d p=\int_0^{10} 10 d t+\int_0^{10}(2 t d t) \\ & \Rightarrow p=\left[10 t+2 \frac{t^2}{2}\right]_0^{10} \\ & \Rightarrow p=10 \times 10+100-0 \\ &  p=100+100=200 \mathrm{~kg} \mathrm{~m} / \mathrm{s} \end{aligned}$

Hint

Changes in a function of $y$ and the independent variable $x$ are related as follows:
$\begin{aligned} & \frac{d y}{d x}=x^2 \\ & \Rightarrow d y=x^2 d x \end{aligned}$
Integrating of both sides, we get:
$\begin{aligned} & \int d y=\int x^2 d x \\ & \Rightarrow y=\frac{x^3}{3}+c \end{aligned}$
where $c$ is a constant
Therefore, $y$ as a function of $x$ is represented by
$y=\frac{x^3}{3}+c$

Hint

(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 $=4$

Hint

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

Hint

(a) In 3472, 7 comes after the digit 4. Its value is greater than 5. So, the next two digits are neglectec and 4 is increased by one.
Therefore, the value becomes 3500.
(b) 84
(c) 2.6
(d) 28

Hint

Length of the cylinder, $l=4.54 \mathrm{~cm}$
Radius of the cylinder, $r=1.75 \mathrm{~cm}$
Volume of the cylinder, $V=\pi r^2 l=(\pi) \times(4.54) \times(1.75)^2$
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.
$\begin{aligned} & \text { Volume, } V=\pi r^2 l \\ & =(3.14) \times(1.75) \times(1.75) \times(4.54) \\ & =43.6577 \mathrm{~cm}^3 \end{aligned}$
Since the volume is to be rounded off to 3 significant digits, we have:
$V=43.7 \mathrm{~cm}^3$

Hint

$\begin{aligned} & \text { Average thickness }=\frac{2.17+2.17+2.18}{3} \\ & =2.1733 \mathrm{~mm} \end{aligned}$
Rounding off to three significant digits, the average thickness becomes $2.17 \mathrm{~mm}$.

Hint

Consider the figure shown below:
Actual effective length $=(90.0+2.13) \mathrm{cm}$
However, in the measurement $90.0 \mathrm{~cm}$, the number of significant digits is only two.
So, the effective length should contain only two significant digits.
i.e., effective length $=90.0+2.13=92.1 \mathrm{~cm}$.