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 \(=90.0+2.13=92.1 \mathrm{~cm}\).