Past NEET Papers

Hint

\(\text { (c) Moment of force, } \vec{\tau}=\vec{r} \times \vec{F}\)

\(\begin{aligned}
&\begin{aligned}
\vec{\tau} &=\left(\vec{r}-\vec{r}_{0}\right) \times \vec{F} \\
\vec{r}-\vec{r}_{0} &=(2 \hat{i}+0 \hat{j}-3 \hat{k})-(2 \hat{i}-2 \hat{j}-2 \hat{k}) \\
&=0 \hat{i}+2 \hat{j}-\hat{k}
\end{aligned} \\
&\vec{\tau}=(0 \hat{i}+2 \hat{j}-\hat{k})(4 \hat{i}+5 \hat{j}-6 \hat{k}) \\
&\vec{\tau}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
0 & 2 & -1 \\
4 & 5 & -6
\end{array}\right|=-7 \hat{i}-4 \hat{j}-8 \hat{k}
\end{aligned}\)

Hint

(b) \(|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|=|\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}|\)
Squaring on both sides
\(
\begin{aligned}
&|\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}|^{2}=|\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}|^{2} \\
&\Rightarrow \overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{A}}+2 \overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{B}} \\
&\quad=\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{A}}-2 \overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{B}} \\
&\Rightarrow 4 \overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=0 \Rightarrow 4 \mathrm{AB} \cos \theta=0 \\
&\Rightarrow \cos \theta=0 \Rightarrow \theta=90^{\circ}
\end{aligned}
\)

Hint

(b) Two vectors are
\(
\begin{aligned}
&\overrightarrow{\mathrm{A}}=\cos \omega t \hat{i}+\sin \omega t \hat{j} \\
&\overrightarrow{\mathrm{B}}=\cos \frac{\omega t}{2} \hat{i}+\sin \frac{\omega t}{2} \hat{j}
\end{aligned}
\)
For two vectors \(\overrightarrow{\mathrm{A}}\) and \(\overrightarrow{\mathrm{B}}\) to be orthogonal \(\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}=0\)
\(
\begin{aligned}
\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}} &=0=\cos \omega t \cdot \cos \frac{\omega t}{2}+\sin \omega t \cdot \sin \frac{\omega t}{2} \\
&=\cos \left(\omega t-\frac{\omega t}{2}\right)=\cos \left(\frac{\omega t}{2}\right)
\end{aligned}
\)
So, \(\frac{\omega t}{2}=\frac{\pi}{2} \therefore \mathrm{t}=\frac{\pi}{\omega}\)

Hint

\(\begin{aligned}
&\text { (d) } \quad \vec{v}_{a v}=\frac{\Delta \vec{r}(\text { displacement })}{\Delta t(\text { time taken })} \\
&=\frac{(13-2) \hat{i}+(14-3) \hat{j}}{5-0}=\frac{11}{5}(\hat{i}+\hat{j})
\end{aligned}\)

Hint

(d) Vector triple product
\(
\begin{aligned}
&\vec{A} \times(\vec{B} \times \vec{C})=\vec{B}(\vec{A} \cdot \vec{C})-\vec{C}(\vec{A} \cdot \vec{B})=0 \\
&\Rightarrow \vec{A} \|(\vec{B} \times \vec{C})
\end{aligned}
\)

Hint

(c) If two non-zero vectors are represented by the two adjacent sides of a parallelogram, then the resultant is given by the diagonal of the parallelogram passing through the point of intersection of the two vectors. Using the law of vector addition, \(\vec{d}+\vec{e}=\vec{f}\)

Hint

\(\begin{aligned}
&\text { (b) } \quad|\vec{A} \times \vec{B}|=\sqrt{3}(\vec{A} \cdot \vec{B}) \\
&\Rightarrow A B \sin \theta=\sqrt{3} A B \cos \theta \\
&\Rightarrow \tan \theta=\sqrt{3} \Rightarrow \theta=60^{\circ}
\end{aligned}\)

Hint

\(\begin{aligned}
&\text { (c) }|\vec{A}+\vec{B}|^{2}=|\vec{A}-\vec{B}|^{2} \\
&=|\vec{A}|^{2}+|\vec{B}|^{2}+2 \vec{A} \cdot \vec{B}=A^{2}+B^{2}+2 A B \cos \theta \\
&=|\vec{A}-\vec{B}|^{2}=|\vec{A}|^{2}+|\vec{B}|^{2}-2 \vec{A} \cdot \vec{B} \\
&=A^{2}+B^{2}-2 A B \cos \theta \\
&\text { So, } A^{2}+B^{2}+2 A B \cos \theta \\
&=A^{2}+B^{2}-2 A B \cos \theta \\
&4 A B \cos \theta=0 \Rightarrow \cos \theta=0 \\
&\theta=90^{\circ} \\
&\text { So, angle between } A \& B \text { is } 90^{\circ} .
\end{aligned}\)

Hint

(b) For two vectors to be perpendicular to each other
\(
\begin{aligned}
&\vec{A} \cdot \vec{B}=0 \\
&(2 \hat{i}+3 \hat{j}+8 \hat{k}) \cdot(4 \hat{j}-4 \hat{i}+\alpha \hat{k})=0 \\
&-8+12+8 \alpha=0 \\
&\alpha=-\frac{4}{8}=-\frac{1}{2}
\end{aligned}
\)

Hint

(d) \((\vec{B} \times \vec{A}) \cdot \vec{A}=\vec{C} \cdot \vec{A}=C A \cos 90^{\circ}=0\)
The dot product of any two perpendicular vectors is zero.

Hint

\(\begin{aligned}
&\text { (c) }|\vec{A} \times \vec{B}|=A B \sin \theta \\
&\vec{A} \cdot \vec{B}=A B \cos \theta \\
&|\vec{A} \times \vec{B}|=\sqrt{3} \vec{A} \cdot \vec{B} \Rightarrow A B \sin \theta=\sqrt{3} A B \cos \theta \\
&\text { or, } \tan \theta=\sqrt{3}, \quad \therefore \theta=60^{\circ} \\&\therefore|\vec{A}+\vec{B}|=\sqrt{A^{2}+B^{2}+2 A B \cos 60^{\circ}} \\
& =\sqrt{A^{2}+B^{2}+A B}
\end{aligned}\)

Hint

(b) \(\vec{P}=\) vector sum \(=\vec{A}+\vec{B}\)
\(\vec{Q}=\) Vector differences \(=\vec{A}-\vec{B}\)
Since \(\vec{P}\) and \(\vec{Q}\) are perpendicular
\(
\begin{aligned}
&\therefore \vec{P} \cdot \vec{Q}=0 \Rightarrow(\vec{A}+\vec{B}) \cdot(\vec{A}-\vec{B})=0 \\
&\Rightarrow A^{2}=B^{2} \Rightarrow|\vec{A}|=|\vec{B}|
\end{aligned}
\)

Hint

(b) Angle between two vectors is given as from dot product \(\vec{A} \cdot \vec{B}=|\vec{A}||\vec{B}| \cos \theta\)
\(
\begin{aligned}
\cos \theta &=\frac{\vec{A} \cdot \vec{B}}{A B} \\
\text { Here, } \vec{A} &=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}} \\
\vec{B} &=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}-5 \hat{\mathrm{k}} \\
\therefore \quad A &=\sqrt{(3)^{2}+(4)^{2}+(5)^{2}}=\sqrt{50} \\
B &=\sqrt{(3)^{2}+(4)^{2}+(-5)^{2}}=\sqrt{50} \\
\text { and } \vec{A} \cdot \vec{B} &=(3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}) \cdot(3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}-5 \hat{\mathrm{k}}) \\
&=9+16-25=0 \\
\therefore \quad |\vec{A}||\vec{B}|\cos \theta &=\frac{0}{\sqrt{50} \cdot \sqrt{50}}=0 \\
\Rightarrow \quad \theta &=90^{\circ}
\end{aligned}
\)

Hint

(a) As we know,
\(
\begin{aligned}
&P=\vec{F} \cdot \vec{v}=(6 \hat{i}-4 \hat{j}+3 \hat{k}) \cdot(20 \hat{i}+15 \hat{j}-5 \hat{k}) \\
&=6 \times 20-4 \times 15-3 \times 5=45 \mathrm{~J} / \mathrm{s}
\end{aligned}
\)
Instantaneous power can be calculated by
\(\mathrm{P}_{\text {ins }}=\mathrm{F} \cdot \cos \theta v=\vec{F} \cdot \vec{v}\)
which is the scalar product of force and velocity vector.

Hint

(c) \(v=\omega \times r\)

\(\begin{aligned}
&=(3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+\hat{\mathbf{k}}) \times(5 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}) \\
&=\left|\begin{array}{ccc}
\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\
3 & -4 & 1 \\
5 & -6 & 6
\end{array}\right| \\
&=\hat{\mathbf{i}}\left|\begin{array}{ll}
-4 & 1 \\
-6 & 6
\end{array}\right|-\hat{\mathbf{j}}\left|\begin{array}{ll}
3 & 1 \\
5 & 6
\end{array}\right|+\hat{\mathbf{k}}\left|\begin{array}{cc}
3 & -4 \\
5 & -6
\end{array}\right| \\
=&(-24+6) \hat{\mathbf{i}}-(18-5) \hat{\mathbf{j}}+(-18+20) \hat{\mathbf{k}} \\
&=-18 \hat{\mathbf{i}}-13 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}
\end{aligned}\)

Hint

(b) We know that, \(R^{2}=A^{2}+B^{2}+2 A B \cos \theta\)
\((24)^{2}=(12)^{2}+(18)^{2}+2(12)(18) \cos \theta\)
\(
\cos \theta=\frac{108}{432} \Rightarrow \theta=75^{\circ} 52^{\prime}
\)

Hint

\(
\begin{aligned}
&\text { (a) } \hat{r}=0.5 \hat{i}+0.8 \hat{j}+c \hat{k} \\
&|\hat{r}|=1=\sqrt{(0.5)^{2}+(0.8)^{2}+c^{2}} \\
&(0.5)^{2}+(0.8)^{2}+c^{2}=1 \\
&c^{2}=0.11 \Rightarrow c=\sqrt{0.11}
\end{aligned}
\)
A unit vector is a vector that has a magnitude of one. It is a vector divided by its magnitude. Unit vector for
\(\vec{A}\) is \(\hat{A}: \hat{A}=\frac{\vec{A}}{A}\)
Unit vector gives direction.

Hint

\(\begin{aligned}
&\text { (c) } \quad \vec{F}=-3 \hat{i}+\hat{j}+5 \hat{k} ; \vec{r}=7 \hat{i}+3 \hat{j}+\hat{k} \\
&\text { Torque }(\vec{\tau})=\vec{r} \times \vec{F} \\
&=(7 \hat{i}+3 \hat{j}+\hat{k}) \times(-3 \hat{i}+\hat{j}+5 \hat{k}) \\
&=7 \hat{k}+35(-\hat{j})-9(-\hat{k})+15 \hat{i}-3 \hat{j}+(-\hat{i}) \\
&=14 \hat{i}-38 \hat{j}+16 \hat{k}
\end{aligned}\)

Hint

(c) Since displacement is along the \(y\)-direction, hence displacement \(\vec{s}=10 \hat{j}\).
Work done \(=\vec{F} \cdot \vec{s}\)
\(
=(-2 \hat{i}+15 \hat{j}+6 \hat{k}) \cdot 10 \hat{j}=150 J
\)

Hint

From the properties of vector product, the cross product of any vector with zero is a null vector or zero vector.

Hint

(b) Note that \((\vec{B} \times \vec{A}) \perp \vec{A}\). Hence their dot product is zero.

Hint

(a)
In figure shown, \(\vec{A}+\vec{B}=\vec{C}\)
Also, \(\quad|\vec{A}|=3,|\vec{B}|=4,|\vec{C}|=5\)
As \(\vec{A}+\vec{B}=\vec{C}\)
So, \(\quad 5^{2}=3^{2}+4^{2}+2 \cdot 4 \cdot 3 \cos \theta\)
\(
\cos \theta=0 \Rightarrow \theta=\frac{\pi}{2}
\)
\(\Rightarrow \vec{A}\) is perpendicular to \(\vec{B}.\)

Hint

(d) Given:
\(
\begin{array}{ll}
x=5 t-2 t^{2} & y=10 t \\
v_{x}=\frac{d x}{d t}=5-4 t & v_{y}=\frac{d y}{d t}=10 \\
a_{x}=\frac{d v_{x}}{d t}=-4 & a_{y}=\frac{d v_{y}}{d t}=0 \\
\vec{a}=a_{x} i+a_{y} j & \vec{a}=-4 i \mathrm{~m} / \mathrm{s}^{2}
\end{array}
\)
Hence, acceleration of particle at \((t=2 \mathrm{~s})=-4 \mathrm{~m} / \mathrm{s}^{2}\)

Hint

(d) Let \(\theta\) be the angle that the particle makes with \(x\)-axis.
From figure below, \(\tan \theta=\frac{3}{\sqrt{3}}=\sqrt{3}\) \(\Rightarrow \theta=\tan ^{-1}(\sqrt{3})=60^{\circ}\)
If a vector \(\vec{R}\) in \(x-y\) plane then its orthogonal vector \(R_{x}=R \cos \theta\) and \(R_{y}=R \sin \theta\).
And \(\frac{R \sin \theta}{R \cos \theta}=\tan \theta=\frac{R_{y}}{R_{x}}\)

Hint

\(\begin{aligned}
&\text { (b) } \vec{F}=6 t \hat{i}+4 t \hat{j} \\
&F_{x}=6 t, F_{y}=4 t \\
&a_{x}=\frac{6 t}{3}=2 t, a_{y}=\frac{4 t}{3} \\
&\frac{d v_{x}}{d t}=2 t^{2} \\
&v_{x} d V_{x}=\int_{0}^{t} 2 t^{2} d t \\
&\Rightarrow \quad v_{x}=\frac{2}{3} t^{3}=\frac{2}{3} \cdot 3^{3}=18 \\
&\quad \frac{d V_{y}}{d t}=\frac{4}{3} t \Rightarrow \quad \int_{0}^{V_{y}} d V_{y}=\frac{4}{3} \int_{0}^{t} t d t
\end{aligned}\)

\(\begin{aligned}
&\Rightarrow \quad V_{y}=\frac{4}{3} \frac{t^{2}}{2}=\frac{4}{3} \cdot \frac{3^{2}}{2}=6 \\
&\Rightarrow \quad \mathrm{V}=18 \vec{i}+6 \vec{j}
\end{aligned}\)

Hint

(a) Given,
\(\mathrm{V}\)elocity \(=10 \mathrm{~m} / \mathrm{s} \quad \alpha=60^{\circ}\)
\(\mathrm{L}^{2}=\mathrm{x}^{2}+\mathrm{y}^{2}\)
On differentiating we get,
\(
0=2 x \frac{d x}{d t}+2 y \frac{d y}{d t}
\)
The Velocity of \(\mathrm{A}=\frac{\mathrm{dx}}{\mathrm{dt}}=-10 \mathrm{~m} / \mathrm{s}\)
Thus the velocity of particle \(B\) is:
\(
\frac{\mathrm{dy}}{\mathrm{dt}}=\frac{-\mathrm{x}}{\mathrm{y}} \frac{\mathrm{dx}}{\mathrm{dt}}=\cot 60^{0} \times 10=\frac{1}{\sqrt{3}} \times 10=5.8 \mathrm{~m} / \mathrm{s}
\)

Hint

(d) Position vector,
\(\vec{r}=(a \cos \omega t) \hat{i}+(a \sin \omega t) \hat{j}\)
Velocity vector,
\(
\begin{aligned}
&\vec{v}=\frac{d(\vec{r})}{d t}=\frac{d}{d t}\{(a \cos \omega t) \hat{i}+(a \sin \omega t) \hat{j}\} \\
&=(-a \omega \sin \omega t) \hat{i}+(a \omega \cos \omega t) \hat{j} \\
&=\omega[(-a \sin \omega t) \hat{i}+(a \cos \omega t) \hat{j}]
\end{aligned}
\)
Slope of position vector \(=\frac{a \sin \omega t}{a \cos \omega t}=\tan \omega t\)
Slope of velocity vector, \(=\frac{-a \cos \omega t}{a \sin \omega t}=\frac{-1}{\tan \omega t}\)
\(\therefore\) velocity is perpendicular to the displacement.

Hint

(a) Speed of the bullet \((\mathrm{v})=1000 \mathrm{~m} / \mathrm{s}\) and horizontal distance of the target \((\mathrm{s})=100 \mathrm{~m}\). Time taken to cover the horizontal distance \((t)\) \(=\frac{100}{1000}=0.1 \mathrm{sec}\).
During this time, the bullet will fall down vertically due to gravitational acceleration.
\(\therefore\) height \((h)=u t+\frac{1}{2} g t^{2}\)

\(\begin{aligned}
&=(0 \times 0.1)+\frac{1}{2} \times 10(0.1)^{2} \\
&=0.05 \mathrm{~m}=5 \mathrm{~cm}
\end{aligned}\)

Hint

(b) Trajectory is identical for both horizontal
range \(=\frac{u^{2} \sin 2 \theta}{g}\)
So that, \(\frac{g_{\text {planet }}}{g_{\text {earth }}}=\frac{\left(u_{\text {planet }}\right)^{2}}{\left(u_{\text {earth }}\right)^{2}}\)
Therefore, \(g_{\text {planet }}=\left(\frac{3}{5}\right)^{2}\left(9.8 \mathrm{~m} / \mathrm{s}^{2}\right)\) \(=3.5 \mathrm{~m} / \mathrm{s}^{2}\)

Hint

(b) At point B the direction of velocity component of the projectile along with the \(y\)-component reverses, while the \(x\)-component remains unchanged.
Hence, \(\vec{V}_{B}=2 \hat{i}-3 \hat{j}\)

Hint

(b) Horizontal range,
\(R=\frac{u^{2} \sin 2 \theta}{g}\)
Maximum height,
\(
h_{m}=\frac{u^{2} \sin ^{2} \theta}{2 g}
\)
According to the condition,
\( R=h_{m}\)
\(
\begin{aligned}
&\Rightarrow \quad \frac{u^{2} \sin 2 \theta}{g}=\frac{u^{2} \sin ^{2} \theta}{2 g} \\
&\Rightarrow 2 \sin \theta \cos \theta=\frac{\sin ^{2} \theta}{2}
\end{aligned}
\)
\(
\begin{aligned}
& 2 \cos \theta=\frac{\sin \theta}{2} \\
\Rightarrow & \cot \theta=\frac{1}{4} \\
\Rightarrow & \tan \theta=4 \\
\Rightarrow & \theta=\left[\tan ^{-1}(4)\right]
\end{aligned}
\)

Note: In the case of projectile motion range \(R\) is \(n\) times the maximum height \(h_{m}\)
\(
\begin{aligned}
&\text { i.e., } R=n h_{m} \Rightarrow \frac{u^{2} \sin 2 \theta}{g}=n \frac{u^{2} \sin ^{2} \theta}{2 g} \\
&\Rightarrow \operatorname{tan} \theta=\frac{4}{n}, \text { Angle of projection } \theta=\tan ^{-1}\left(\frac{4}{n}\right)
\end{aligned}
\)

Hint

For the maximum range of the projectile, \(\theta\) will be \(45^{\circ}\) by the law of projectile motion.
So, maximum range, \(R_{\max }=\frac{u^{2}}{g}\)
Given, \(\quad u=20 \mathrm{~ms}^{-1}\) and \(\mathrm{g}=10 \mathrm{~ms}^{-2}\)
\(
\begin{aligned}
&R_{\max }=\frac{(20)^{2}}{10}=\frac{400}{10} \\
&R_{\max }=40 \mathrm{~m}
\end{aligned}
\)

Hint

(c) Maximum height,
\(\mathrm{H}=\frac{u^{2} \sin ^{2} 45^{\circ}}{2 g}=\frac{u^{2}}{4 g}\) ..(1)
Range, \(\mathrm{R}=\frac{u^{2} \sin 90^{\circ}}{g}=\frac{u^{2}}{g}\)
\(\therefore \frac{R}{2}=\frac{u^{2}}{2 g}\) …… (2)
According to the rcquircd condition,
\(
\therefore \tan \alpha=\frac{\mathrm{H}}{\mathrm{R} / 2}
\)
Put the volues from equation (1) and (2)
\(
\Rightarrow \frac{\left(\frac{u^{2}}{4 g}\right)}{\left(\frac{u^{2}}{2 g}\right)} \tan \alpha \quad \therefore \alpha=\tan ^{-1}\left(\frac{1}{2}\right)
\)

Hint

(b) Clearly, there is no change in momentum along \(x\)-axis, {latex]=m v \cos \theta-m v \cos \theta=0[/latex]
Momentum changed only in vertical direction or \(y\)-axis.
So, \(\Delta \mathrm{P}=\Delta \mathrm{P}_{\text {vertical }}\)
\(\Rightarrow \mathrm{P}_{\text {final }}=\mathrm{P}_{\text {initial }}\)
\(=m v \sin \theta-(-m v \sin \theta)\)
\(=2 m v \sin \theta=2 m v \times \sin 45^{\circ}\)
\(=2 m v \times \frac{1}{\sqrt{2}}=\sqrt{2} m v\)
Hence, resultant change in momentum \(=\sqrt{2} m v\)

Hint

(d) Horizontal range for projection angle
\(
\left(45^{\circ}-\theta\right) \text { is, } \mathrm{R}_{1}=\frac{u^{2} \sin 2(45-\theta)}{g}
\)
Horizontal range projection angle \(\left(45^{\circ}+\theta\right)\) is, \(\mathrm{R}_{2}=\frac{u^{2} \sin 2(45+\theta)}{g}\) According to the condition,
\(
\begin{aligned}
&\Rightarrow \quad \frac{R_{1}}{R_{2}}=\frac{u^{2} \sin 2(45-\theta)}{u^{2} \sin 2(45+\theta)}=\frac{\sin (90-2 \theta)}{\sin (90+2 \theta)} \\
&\Rightarrow \frac{R_{1}}{R_{2}}=\frac{\cos 2 \theta}{\cos 2 \theta}=\frac{1}{1} \\
&\text { So, } R_{1}: R_{2}: 1: 1
\end{aligned}
\)
The angle of elevation ( \(\phi)\) of the highest point of the projectile and the angle of projection \(\theta\) are related to each other as \(\tan \phi=\frac{1}{2} \tan \theta\)

Hint

(a) Maximum height in case of the projectile is given by
\(
\begin{aligned}
&h_{m}=\frac{u^{2} \sin ^{2} \theta}{2 g} \Rightarrow \frac{h_{1m}}{h_{2m}}=\frac{\sin ^{2} \theta_{1}}{\sin ^{2} \theta_{2}} \\
&=\frac{\sin ^{2} 30^{\circ}}{\sin ^{2} 60^{\circ}}=\frac{\left(\frac{1}{2}\right)^{2}}{\left(\frac{\sqrt{3}}{2}\right)^{2}}=\frac{1}{3}
\end{aligned}
\)
(b) Range \(R=\frac{u^{2} \sin 2 \theta}{g}\)
\(
\begin{aligned}
&\Rightarrow \frac{R_{1}}{R_{2}}=\frac{\sin \left(2 \times 30^{\circ}\right)}{\sin \left(2 \times 60^{\circ}\right)} \\
&=\frac{\sin 60^{\circ}}{\sin 120^{\circ}}=\frac{\frac{\sqrt{3}}{\sqrt{3}}}{\frac{2}{2}}=1 \Rightarrow R_{1}=R_{2}
\end{aligned}
\)
(c) Time of flight
\(
\begin{aligned}
T_{f} &=\frac{2 u \sin \theta}{g} \Rightarrow \frac{T_{1f}}{T_{2f}}=\frac{\sin \theta_{1}}{\sin \theta_{2}} \\
&=\frac{\sin 30^{\circ}}{\sin 60^{\circ}}=\frac{1 / 2}{\sqrt{3} / 2}=\frac{1}{\sqrt{3}}
\end{aligned}
\)
Hence, their horizontal ranges will be equal. So, answer (b) is correct.

Hint

(c) Horizontal range is same when angle of projection with the horizontal is \(\theta\) and \(\left(90^{\circ}-\theta\right)\). So, the ratio is 1:1.

Hint

(d) Here, \(R_{\max }=\frac{u^{2}}{g}=16 \mathrm{~km}=16000 \mathrm{~m}\)
or
\(
\begin{aligned}
u &=\sqrt{16000 \mathrm{~g}}=\sqrt{16000 \times 10} \\
&=400 \mathrm{~ms}^{-1}
\end{aligned}
\)

Hint

When an object is projected with velocity \(u\) making an angle \(\theta\) with the horizontal direction, then the horizontal range will be
\(
R_{1}=\frac{u^{2} \sin 2 \theta}{g} \text {…(i) }
\)
when an object is projected with velocity \(u\) making an angle \(\left(90^{\circ}-\theta\right)\) with the horizontal direction, then horizontal range will be
\(
R_{2}=\frac{u^{2} \sin 2\left(90^{\circ}-\theta\right)}{g}
\)
\(
\begin{aligned}
&=\frac{u^{2}}{g} \sin \left(180^{\circ}-2 \theta\right) \\
&=\frac{u^{2}}{g} \sin 2 \theta \text {…(ii) }
\end{aligned}
\)
From Eqs. (i) and (ii), we note that \(R_{1}=R_{2}\)
Here, the projection angle is \(30^{\circ}\) and \(60^{\circ}=\left(90^{\circ}-30^{\circ}\right)\), so horizontal range is same for both angles.
\(
\therefore \quad \frac{R_{1}}{R_{2}}=1
\)

Note: Horizontal range is same when angle of projection is \(\theta\) or \(\left(90^{\circ}-\theta\right)\).

Hint

(d)
Given, the radius of the circular path \(=R\)
The time taken by the particle to complete one revolution \(=T\)
When the particle is projected with the same speed (by which it is moving in a circular orbit) at angle \(\theta\) to the horizontal, the maximum height attained is given as
\(h_{\max }=\frac{u^{2} \sin ^{2} \theta}{2 g}\) …… (i)

\(h_{\max }=4 R \quad\) (given)
Also, we know that,
speed of the particle in a circular path,
\(
u=\frac{2 \pi R}{T}
\)
Substituting the values in the Eq. (i), we get
\(
\begin{gathered}
4 R=\frac{\left(\frac{2 \pi R}{T}\right)^{2} \sin ^{2} \theta}{2 g} \\
\Rightarrow \quad \sin \theta=\left(\frac{2 g T^{2}}{\pi^{2} R}\right)^{1 / 2} \\
\Rightarrow \quad \theta=\sin ^{-1}\left(\frac{2 g T^{2}}{\pi^{2} R}\right)^{1 / 2}
\end{gathered}
\)

Hint

(a) Velocity of swimmer w.r.t. river \(\mathrm{V}_{\mathrm{SR}}=20 \mathrm{~m} / \mathrm{s}\), Velocity of river w.r.t. ground \(V_{R G}=10 \mathrm{~m} / \mathrm{s}\)

\(
\begin{aligned}
&\vec{V}_{S G}=\vec{V}_{S R}+\vec{V}_{R G} \\
&\sin \theta=\left|\frac{\vec{V}_{R G}}{\vec{V}_{S R}}\right| \Rightarrow \sin \theta=\frac{10}{20} \\
&\Rightarrow \sin \theta=\frac{1}{2} \quad \therefore \theta=30^{\circ} \text { west }
\end{aligned}
\)
Therefore to cross the river along the shortest path, the swimmer should make his strokes \(30^{\circ}\) west.

Hint

(c) Given: The position vector
\(\vec{r}=\cos \omega t \hat{x}+\sin \omega t \hat{y}\)
\(\therefore \quad\) Velocity vector, \(\vec{v}=-\omega \sin \omega t \hat{x}+\omega \cos \omega t\) \(\hat{y}\) acceleration vector,
\(\vec{a}=-\omega^{2} \cos \omega t \hat{x}-\omega^{2} \sin \omega t \hat{y}=-\omega^{2} \vec{r}\)
\(\vec{r} \cdot \vec{v}=0\) hence \(\vec{r} \perp \vec{v}\) and \(\vec{a}\) is directed towards the origin.
Note: Nothing actually moves in the direction of the angular velocity vector \(\vec{\omega}\). The direction of \(\vec{\omega}\) simply represents that the circular motion is taking place in a plane perpendicular to it.

Hint

It is clear from the diagram that the shortest distance between the ship \(A\) and \(B\) is \(P O\).

Here, \(\sin 45^{\circ}=\frac{P Q}{OQ} \Rightarrow P Q=100 \times \frac{1}{\sqrt{2}}\) \(=50 \sqrt{2} \mathrm{~m}\)
Also, \(v_{A B}=\sqrt{v_{A}^{2}+v_{B}^{2}}=\sqrt{10^{2}+10^{2}}\) \(=10 \sqrt{2} \mathrm{~km} / \mathrm{h}\)

So, the time taken for them to reach the shortest path is
\(
t=\frac{P O}{V_{A B}}=\frac{50 \sqrt{2}}{10 \sqrt{2}}=5 \mathrm{~h}
\)

Hint

(b) Here,

\(\begin{aligned} x &=4 \sin (2 \pi \mathrm{t}) \quad \ldots(i) \\ y &=4 \cos (2 \pi \mathrm{t}) \quad \ldots(\text { ii }) \end{aligned}\)
Squaring and adding equation \((i)\) and \((ii)\) \(x^{2}+y^{2}=4^{2} \Rightarrow \mathrm{R}=4\)
The motion of the particle is circular motion, acceleration vector is along \(\vec{R}\) and its magnitude \(=\frac{V^{2}}{R}\)
Velocity of particle, \(\mathrm{V}=\omega \mathrm{R}=(2 \pi)(4)=8 \pi\)

Hint

(d) Centripetal acceleration \(a_{c}=\omega^{2} r\)
\(
=\left(\frac{2 \pi}{T}\right)^{2} r=\left(\frac{2 \pi}{0.2 \pi}\right)^{2} \times 5 \times 10^{-2}=5 \mathrm{~m} / \mathrm{s}^{2}
\)

Hint

(a) Distance covered in one circular loop \(=\) \(2 \pi r\) \(=2 \times 3.14 \times 100=628 \mathrm{~m}\); Avg Speed \(=\frac{628}{62.8}=10 \mathrm{~m} / \mathrm{sec}\); Displacement in one circular loop \(=0\); Avg Velocity \(=\frac{0}{\text { time }}=0\)

Hint

(b) According to the given condition,
Let both \(y\) boy meet at \(c\) : in time \(t\) then,
\(\mathrm{AC}=v t\) and \(\mathrm{BC}=v_{1} t\)
From pythagorus theorem,
\(
\begin{aligned}
&\mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC} \\
&v^{2} t^{2}=a^{2}+v_{1}^{2} t^{2} \\
&t^{2}\left(v^{2}-v_{1}^{2}\right)=a^{2} \\
&t^{2}=\frac{a^{2}}{\left(v^{2}-v_{1}^{2}\right)} \\
&t=\sqrt{\frac{a}{\left(v^{2}-v_{1}^{2}\right)}}
\end{aligned}
\)

Hint

(a) Since speed is constant throughout the motion, so, it is a uniform circular motion. Therefore, its radial acceleration is given by
\(
\begin{aligned}
a_{r} &=r \omega^{2} \\
&=r\left(\frac{2 \pi n}{t}\right)^{2}=r \times \frac{4 \pi^{2} n^{2}}{t^{2}} \\
&=\frac{1 \times 4 \times \pi^{2} \times(22)^{2}}{(44)^{2}} \\
=\pi^{2} \mathrm{~m} / \mathrm{s}^{2} &
\end{aligned}
\)
This acceleration is directed along the radius of the circle.

Hint

In a circular motion, a particle repeats its motion after equal intervals of time. So, particle moving on a circular path is periodic but not simple harmonic, as it does not execute to and fro motion about a fixed point.

Hint

\(\begin{aligned}
&\text { (b) } \mathrm{W}_{\mathrm{mg}}=\Delta K \\
&\Rightarrow-\mathrm{mg} \ell=1 / 2 \mathrm{mv}^{2}-1 / 2 \mathrm{mu}^{2} \\
&\text { or, } m v^{2}=m\left(u^{2}-2 g \ell\right] \\
&\text { or, } v=\sqrt{u^{2}-2 g \ell} \hat{j} \\
&\vec{u}=u \hat{i} \\
&\therefore \vec{v}-\vec{u}=\sqrt{u^{2}-2 g \ell} \hat{j}-u \hat{i} \\
&\therefore|\vec{v}-\vec{u}|=\left[\left(u^{2}-2 g \ell\right)+u^{2}\right]^{1 / 2}=\sqrt{2\left(u^{2}-g \ell\right)}
\end{aligned}\)

Hint

The tangential acceleration is given by
\(a_{T}=r \alpha\) ….. (i)
From the 2nd equation of motion for rotational motion,
\(
\omega^{2}=\omega_{0}^{2}+2 \alpha \theta
\)
Here, initial angular velocity, \(\omega_{0}=0\), Final angular velocity,

\(
\begin{aligned}
\omega &=\frac{v}{r}=\frac{80}{20 / \pi} \\
&=4 \pi  \mathrm{rad} / \mathrm{s} \\
\theta &=2 \times 2 \pi  \mathrm{rad}
\end{aligned}
\)
So, angular acceleration
\(
\alpha=\frac{\omega^{2}}{2 \theta}=\frac{(4 \pi)^{2}}{2 \times(2 \times 2 \pi)}=\frac{16 \pi^{2}}{8 \pi}=2 \pi
\)
Hence, from Eq. (i), we have
\(
a_{T}=r \alpha=\frac{20}{\pi} \times 2 \pi=40 \mathrm{~m} / \mathrm{s}^{2}
\)

Hint

(c) Linear velocity \(v=r \omega\)
\(
v_{1}=\omega r_{1}, v_{2}=\omega r_{2}
\)
\([\omega\) is same in both cases because time period is same]
\(
{v_{1}}:{v_{2}}={r_{1}}:{r_{2}}={R}:{r}\)

Hint

(a) In the case of a body describing a vertical circle,
\(
\mathrm{T}-\mathrm{mg} \cos \theta=\frac{\mathrm{mv}^{2}}{l} ; T=m g \cos \theta+\frac{m v^{2}}{l}
\)
Tension is maximum when \(\cos \theta=+1\) and velocity is maximum
Both conditions are satisfied at \(\theta=0^{\circ}\) (i.e., at lowest point \(B\) ]

Hint

Let \(v_{r}=\) velocity of river
\(v_{\mathrm{br}}=\) velocity of boat in still water and
\(w=\) width of river
Time taken to cross the river \(=15 \mathrm{~min}\)
\(
=\frac{15}{60} \mathrm{~h}=\frac{1}{4} \mathrm{~h}
\)
Shortest path is taken when \(v_{b}\) is along
\(A B\). In this case,
\(v_{b r}^{2}=v_{r}^{2}+v_{b}^{2}\)
Now, \(\quad t=\frac{W}{v_{b}}=\frac{W}{\sqrt{v_{b r}^{2}-v_{r}^{2}}}\)
\(\therefore \quad \frac{1}{4}=\frac{1}{\sqrt{5^{2}-v_{r}^{2}}}\)
\(\Rightarrow \quad 5^{2}-v_{r}^{2}=16\)
\(\Rightarrow \quad v_{r}^{2}=25-16=9\)
\(\therefore \quad v_{r}=\sqrt{9}=3 \mathrm{~km} / \mathrm{h}\)

Hint

(a) Minimum speed with which the string is rotating in a vertical circle \((\mathrm{v})=\sqrt{g r}\)
The minimum speed of stone is independent of the mass of stone.

Hint

Let \(u\) be the speed of the stream and \(v\) be the speed of the person started from A. He wants to reach at point \(B\) directed opposite to \(A\).
As given, \(v\) makes an angle of \(120^{\circ}\) with direction of flow \(u\), the resultant of \(v\) and \(u\) is along \(A B\). From figure below
\(
\begin{aligned}
u &=v \sin \theta=v \sin 30^{\circ} \\
\therefore \quad u=\frac{v}{2} &=\frac{0.5}{2} \quad(\because v=0.5 \mathrm{~m} / \mathrm{s}) \\
&=0.25 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Hint

(a) \(T_{m}=\frac{m v_{m}^{2}}{R}\)
\(
v_{m}=\sqrt{\frac{T R}{m}}=\sqrt{\frac{25 \times 1.96}{0.25}}
\)
\(
=\frac{5 \times 14}{5}=14 \mathrm{~m} / \mathrm{s}
\)
Note: In the horizontal circular motion of the ball tension in the string is balanced by the centrifugal force \(\left(\frac{m v^{2}}{\ell}\right)\) and hence the maximum tension in the string will be for the maximum speed of the ball.

Hint

(b) Linear speed \(=\) radius \(\times\) angular speed
\(
\begin{array}{ll}
\text { or } & v=r \omega \\
\text { Here, } \quad & r=20 \mathrm{~cm}=0.20 \mathrm{~m} \\
& \omega=10 \mathrm{rad} / \mathrm{s} \\
\therefore & v=0.20 \times 10 \\
\Rightarrow & v=2 \mathrm{~m} / \mathrm{s}
\end{array}
\)

Hint

(c) On circular motion, the force acts along the radius, and displacement at a location is perpendicular to the radius i.e., \(\theta=90^{\circ}\)
As work done \(=\vec{F} \cdot \vec{S}=F S \cos 90^{\circ}=0\)

Hint

(b) The situation is depicted in the figure below.
Let \(v_{\mathrm{b}}\) be the velocity of boat, \(v_{r}\) be the velocity of river and \(v_{\text {r }}\) be the resultant velocity of boat.
From the figure and concept of relative velocity
\(
\begin{aligned}
v_{r b}^{2} &=v_{r}^{2}+v_{b}^{2} \\
\therefore \quad v_{r} &=\sqrt{v_{r b}^{2}-v_{b}^{2}} \\
&=\sqrt{10^{2}-8^{2}}=6 \mathrm{~km} \mathrm{~h}^{-1}
\end{aligned}
\)

Hint

\(\begin{aligned}
&\text { (b) Centripetal acc. }=\omega^{2} r=4 \pi^{2} f^{2} r \\
&=4 \times(3.14)^{2} \times \frac{120}{60} \times \frac{30}{100}=23.7 \mathrm{~ms}^{-2} \\
&{[\because \omega=2 \pi f]}
\end{aligned}\)

Hint

(a) Given, the distance between the two buildings \(d=100 \mathrm{~m}\)
height of each tower, \(h=200 \mathrm{~m}\)
speed of each bullet, \(v=25 \mathrm{~ms}^{-1}\)
The situation can be shown as below in the figure below
where \(x\) be the vertical distance travelled from the top of the building and \(t\) be the time at which they collide. As two bullets are fired toward each other, So, their relative velocity will be
\(
v_{\text {r }}=25-(-25)=50 \mathrm{~ms}^{-1}
\)
Then, time \(=\frac{d}{v_{\text {r }}}=\frac{100}{50}=2 \mathrm{~s}\)
The distance or height at which they collide is calculated from equation of motion,
\(
x=u t+\frac{1}{2} a t^{2}
\)
The bullet is initially a rest i.e. \(u=0\) and as it is moving under the effect of gravity \(a=-g\), so
\(
\begin{aligned}
&x=-\frac{1}{2} g t^{2} \\
&x=-\frac{1}{2} \times 10(2)^{2}=-20 m
\end{aligned}
\)
The negative sign shows that the bullets will collide \(20 \mathrm{~m}\) below the top of the tower i.e. at a height of \((200-20)=180 \mathrm{~m}\) from the ground after \(2 \mathrm{~s}\).

Hint

(b) The motion of the object shot in two cases can be depicted as shown in figure below
Using the third equation of motion,
\(v^{2}=u^{2}-2 g h\) ………. (i)
As the object stops finally, so
\( v=0\)
For inclined motion,
\(
g=g \sin \theta \text { and } h=x
\)
Substituting these values in Eq. (i), we get
\(
\Rightarrow u^{2}=2 g \sin \theta x \Rightarrow x=\frac{u^{2}}{2 g \sin \theta}
\)
For case \((i), \quad x_{1}=\frac{u^{2}}{2 g \sin 60^{\circ}}\)
For case \((ii), x_{2}=\frac{u^{2}}{2 g \sin 30^{\circ}}\)
\(
\begin{aligned}
\Rightarrow \frac{x_{1}}{x_{2}} &=\frac{u^{2}}{2 g \sin 60^{\circ}} \times \frac{2 g \sin 30^{\circ}}{u^{2}} \\
&=\frac{1}{2} \times \frac{2}{\sqrt{3}}=\frac{1}{\sqrt{3}} \text { or } 1: \sqrt{3}
\end{aligned}
\)

Hint

(b) Given, initial velocity \((u)=3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}\)
Acceleration \((a)=(0.4 \hat{\mathbf{i}}+0.3 \hat{\mathrm{j}})\)
Time \((t)=10 \mathrm{~s}\)
From first equation of motion, \(v=u+\) at
\(
\begin{aligned}
&v=3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+10(0.4 \hat{\mathbf{i}}+0.3 \hat{\mathbf{j}}) \\
&v=7 \hat{\mathbf{i}}+7 \hat{\mathbf{j}} \Rightarrow|v|=7 \sqrt{2}
\end{aligned}
\)

Hint

(a)

\(
\mathbf{R}=\mathbf{r}_{1}-\mathbf{v}_{1} t=\mathbf{r}_{2}-\mathbf{v}_{2} t
\)
i.e. \(\quad \mathbf{r}_{1}-\mathbf{r}_{2}=\left(\mathbf{v}_{2}-\mathbf{v}_{1}\right) t\)
So, \(\frac{\mathbf{r}_{1}-\mathbf{r}_{2}}{\left|\mathbf{r}_{1}-\mathbf{r}_{2}\right|}=\frac{\left(\mathbf{v}_{2}-\mathbf{v}_{1}\right) t}{\left|\mathbf{v}_{2}-\mathbf{v}_{1}\right| t}\)
\(
\frac{\mathbf{r}_{1}-\mathbf{r}_{2}}{\left|\mathbf{r}_{1}-\mathbf{r}_{2}\right|}=\frac{\left(\mathbf{v}_{2}-\mathbf{v}_{1}\right)}{\left|\mathbf{v}_{2}-\mathbf{v}_{1}\right|}
\)

Hint

(a)
The situation is depicted in the figure below.
Here, \(\quad \mathbf{v}=50 \mathrm{~km} / \mathrm{h}\) due North
\(\mathbf{v}_{2}=50 \mathrm{~km} / \mathrm{h}\) due West
From figure it indicates that angle between \(\mathbf{v}_{1}\) and \(\mathbf{v}_{2}\) is \(90^{\circ}\).
Now, \(-\mathbf{v}_{1}=50 \mathrm{~km} / \mathrm{h}\) due South
\(\therefore\) Change in velocity
\(
\begin{aligned}
&=\left|\mathbf{v}_{2}-\mathbf{v}_{1}\right|=\left|\mathbf{v}_{2}+\left(-\mathbf{v}_{1}\right)\right| \\
&=\sqrt{v_{2}^{2}+v_{1}^{2}} \\
&=\sqrt{50^{2}+50^{2}} \\
&=70.7 \mathrm{~km} / \mathrm{h}
\end{aligned}
\)

Hint

(c) The centripetal acceleration of a particle moving on a circular path is given by
\(
a_{c}=\frac{v^{2}}{R}
\)
In the given figure,
\(
\begin{aligned}
&a_{c}=a \cos 30^{\circ}=15 \cos 30^{\circ} \mathrm{m} / \mathrm{s}^{2} \\
&\Rightarrow \quad \frac{v^{2}}{R}=15 \cos 30^{\circ} \\
&\Rightarrow \quad v^{2}=R \times 15 \times \frac{\sqrt{3}}{2}=2.5 \times 15 \times \frac{\sqrt{3}}{2} \\
&\therefore \quad v=5.7 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Hint

(a) Given, that two stones of masses \(m\) and \(2 \mathrm{~m}\) are whirled in horizontal circles, the heavier one in a radius \(\frac{r}{2}\) and lighter one in radius \(r\) as shown in figure below.

As lighter stone is n times that of the value of heavier stone when they experience the same centripetal forces, we get
\(
\left(F_{c}\right)_{\text {heavier }}=\left(F_{c}\right)_{\text {lighter }}
\)

\(\begin{aligned}
&\Rightarrow \quad \frac{2 m(v)^{2}}{(r / 2)}=\frac{m(n v)^{2}}{r} \\
&\Rightarrow \quad n^{2}=4 \Rightarrow n=2
\end{aligned}\)

Hint

(b) When the wheel rolls on the ground without slipping and completes half rotation, point \(P\) takes a new position as \(P^{\prime}\) as shown in the figure.
Horizontal displacement, \(x=\pi R\)
Vertical displacement, \(y=2 R\)
Thus, displacement of the point \(P\) when wheel completes half rotation,
\(
\begin{aligned}
& s =\sqrt{x^{2}+y^{2}}=\sqrt{(\pi R)^{2}+(2 R)^{2}} \\
&=\sqrt{\pi^{2} R^{2}+4 R^{2}} \\
\text { but } \quad R &=1 m \\
\therefore \quad & s =\sqrt{\pi^{2}(1)^{2}+4(1)^{2}}=\sqrt{\pi^{2}+4} \mathrm{~m}
\end{aligned}
\)

Hint

(a)
Angular velocity of the particle is given by
\(
\begin{aligned}
&\omega=\frac{2 \pi}{T} \\
&\omega \propto \frac{1}{T}
\end{aligned}
\)
or \(\quad \omega \propto \frac{1}{T}\)
[T= time period of the particle ]
It simply implies that \(\omega\) does not depend on the mass of the body and radius of the circle.
\(
\therefore \quad \frac{\omega_{1}}{\omega_{2}}=\frac{T_{2}}{T_{1}}
\)
but given time period is same, i.e. \(T_{1}=T_{2}\) Hence, \(\frac{\omega_{1}}{\omega_{2}}=\frac{1}{1}= 1\)

Hint

By the concept of centrifugal force, cream is separated from milk. A mass \(m\) of milk revolving at a distance \(r\) from the axis of rotation of the centrifuge requires a centripetal force \(m r \omega^{2}\), where \(\omega\) is the angular speed of the centrifuge. If in place of this mass of milk, lighter particles of mass (cream) \(m^{\prime}\left(m^{\prime}<m\right)\) are present, then the centripetal force \(\left(m r \omega^{2}\right)\) on the milk will be greater than the centripetal force \(\left(m^{\prime} r \omega^{2}\right)\) on the cream. As a result, the cream moves towards the axis of rotation under the effect of the net force \(\left(m-m^{\prime}\right) r \omega^{2}\). When the centrifuge is stopped, the cream is found at the top and milk at the bottom.

Hint

At the highest point vertical component of velocity becomes zero.
At highest point only horizontal component of velocity remains
\(
\Rightarrow u_x=u \cos \theta
\)
\(
\begin{aligned}
& u_x=u \cos \theta=10 \cos 30^{\circ} \\
& =5 \sqrt{3} m s^{-1}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \Delta \vec{P}=\vec{P}_f-\vec{P}_i \\
& \vec{P}_f=m u \hat{i} \\
& \vec{P}_i=m u(-\hat{j}) \\
& \Delta \vec{P}=m u \hat{i}-m u(-\hat{j}) \\
& \Delta \vec{P}=m u(\hat{i}+\hat{j}) \\
& \vec{F}=\frac{\Delta \vec{P}}{\Delta t}
\end{aligned}
\)
Direction of change of momentum and direction of force acting on the player will be same, so correct answer is North east direction

Hint

\(
\begin{aligned}
H & =\frac{u^2 \sin ^2 \theta}{2 g} \\
H & =\frac{(280)^2\left(\sin ^2 30\right)}{2 \times 9.8} \\
& =\frac{280 \times 280 \times 0.5 \times 0.5}{2 \times 9.8} \\
H & =1000 \mathrm{~m}
\end{aligned}
\)

Hint

\(
\begin{aligned}
s & =u t-\frac{1}{2} g t^2 \\
& =16-\frac{1}{2} \times 10 \times 16 \\
& =-64 \mathrm{~m}
\end{aligned}
\)
Height of bridge above water surface \(=64 \mathrm{~m}\)

Hint

Time period of satellite above earth surface
\(
\begin{aligned}
& T=2 \pi \sqrt{\frac{A^3}{G M}}=2 \pi \sqrt{\frac{R^3}{G d \frac{4}{3} \pi R^3}} \\
& T=2 \pi \sqrt{\frac{3}{4 \pi G d}} \\
& T=\sqrt{\frac{3 \pi}{G d}} \quad T^2=\frac{3 \pi}{G d}
\end{aligned}
\)

Hint

\(
\text { between } 1 \text { to } 2
\)

\(
\begin{aligned}
& \left(\frac{u}{3}\right)^2=u^2-2 a \times 24 \\
& \Rightarrow 2 a(24)=\frac{8 u^2}{9} \dots(i)
\end{aligned}
\)
between 2 to 3
\(
0=\left(\frac{u}{3}\right)^2-2 a s \dots(ii)
\)
From equation (i) and (ii)
\(
\begin{aligned}
& 2 a s=\frac{2 a(24)}{8} \\
& s=3 \mathrm{~cm}
\end{aligned}
\)
Length of wooden block is \(24+3=27 \mathrm{~cm}\)