Solution :

(a) Positive because; force of gravity and the displacement are in same direction i.e vertically downwards

Solution :

(i) We have, $\mathrm{F}=2000 \mathrm{~N}$

$$ \mathrm{S}=7.5 \mathrm{M} $$

$\theta=180^{\circ}$ because the braking force acts opposite to displacement or motion.

$\therefore$ Work done by the applied force $=\mathrm{W}=\mathrm{FS} \cos 180^{\circ}$

$$ =(2000)(7.5)(-1)=-15000 \mathrm{~J} $$

(ii) Work done on the road is zero because there is not displacement of the road.

Solution :

We have area $\mathrm{A}=1 \mathrm{sq} . \mathrm{km}=1000 \mathrm{~m} \times 1000 \mathrm{~m}$

$$ \begin{gathered} =10^{6} \mathrm{~m}^{2} \\ \mathrm{~d}=25 \mathrm{~mm}=25 \times 10^{-3} \mathrm{~m} \end{gathered} $$

Volume of water collected due to rainfall $=\mathrm{A} . \mathrm{d}=25 \times 10^{3} \mathrm{~m}^{3}$

Mass of water $=$ Volume $\times$ density

$$ =25 \times 10^{6} \mathrm{~kg} $$

Fore applied to raise water $=\mathrm{F}=\mathrm{mg}=25 \times 10^{7} \mathrm{~N}$

Distance moved $=\mathrm{h}=1 \mathrm{~km}=10^{3} \mathrm{~m}$

$\therefore$ Work done $=\mathrm{Fh}=25 \times 10^{7} \times 10^{3}=2.5 \times 10^{11} \mathrm{~J}$

Solution :

Total mass of man and bricks $=60+30=90 \mathrm{~kg}$

Let $\mathrm{F}$ be force applied by the man along the plane as shown in Fig.

For dynamic equilibrium $\mathrm{F}=\mathrm{mg} \sin \theta$

$$ =90 \times 9.8 \times \frac{1}{10}=88.2 \mathrm{~N} $$

$\mathrm{W}=$ The work done $=\mathrm{F} \times \mathrm{s}$

$$ \begin{aligned} & =88.2 \times 20 \\ & =1764 \mathrm{~J} \end{aligned} $$

Solution :

We have, $\mathbf{F}=(7 \mathbf{i}+2 \mathbf{j}-3 \mathbf{k}) \mathrm{N}$

$$ \mathbf{S}=(2 \mathbf{i}+8 \mathbf{j}+\mathrm{n} \mathbf{k}) \mathrm{m} $$

Work done $=\mathrm{W}=$ F.S

$$ \begin{aligned} & =(7 \mathbf{i}+2 \mathbf{j}-3 \mathbf{k}) \cdot(2 \mathbf{i}+8 \mathbf{j}+\mathrm{n} \mathbf{k}) \\ & =(7)(2)+(2)(8)-(3)(\mathrm{n}) \\ & =30-3 \mathrm{n} \end{aligned} $$

Given $\mathrm{W}=0$; therefore,

$$ 30-3 n=0 $$

or

$$ \mathrm{n}=10 $$

$\therefore \mathbf{S}=(2 \mathbf{i}+2 \mathbf{j}+10 \mathbf{k}) \mathrm{m}$

and

$$ \begin{aligned} & |\mathbf{S}|=\sqrt{(2)^{2}+(8)^{2}+(10)^{2}} \\ & =\sqrt{4+64+100}=\sqrt{168} \\ & =2 \sqrt{42} \mathrm{~m} \end{aligned} $$

Solution :

For the block using free body diagram; shown in Fig. we have

$$ \begin{aligned} & \mathrm{T}-\mathrm{mg}=\mathrm{ma} \\ & \mathrm{T}=\mathrm{m}(\mathrm{g}+\mathrm{a}) \\ & =2(10+2)=24 \mathrm{~N} \end{aligned} $$

$S=$ The distance travels in $4 \mathrm{~s}$

$$ \begin{aligned} & =u t+\frac{1}{2} a t^{2} \\ & =0+\frac{1}{2}(2)(4)^{2}=16 \mathrm{~m} \end{aligned} $$

$\therefore$ Work done is $4 \mathrm{~s}=\mathrm{FS}=$ T.S. $=(24)(16)=384 \mathrm{~J}$

$\mathrm{S} _{4}=$ Distance in 4 th is given by

$\mathrm{S} _{\mathrm{n}}=\mathrm{u}+\frac{\mathrm{a}}{2}(2 \mathrm{n}-1)$

$\mathrm{S} _{4}=(0)+\frac{2}{2}(7)=7 \mathrm{~m}$

$\therefore$ Work done in $4^{\text {th }}$ second $=\mathrm{W}^{\prime}=(24)(7)=168 \mathrm{~J}$

Solution :

We have $\mathbf{F}=(\mathbf{i}+2 \mathbf{j}+3 \mathbf{k}) \mathrm{N}$

$\mathbf{S}=-2 \mathbf{k m}$ (In negative z-direction)

$\therefore \mathrm{W}=$ F.S

$$ \begin{aligned} & =(\mathbf{i}+2 \mathbf{j}+3 \mathbf{k}) \cdot(-2 \mathbf{k}) \\ & =-6 \mathrm{~J} \end{aligned} $$

Solution :

Work done $=$ Area under $\mathrm{F}$ vs $\mathrm{S}$ graph

$$ \begin{aligned} & =\operatorname{Ar}(\mathrm{OABC})+\operatorname{Ar}(\mathrm{CDE}) \\ & =\frac{1}{2}(\mathrm{AB}+\mathrm{OC})(\perp \text { lar distance between } \mathrm{AB} \text { and } \mathrm{OC})-\frac{1}{2}(\mathrm{CE}) \mathrm{DE} \\ & =\frac{1}{2}(2+5) \times 4-\frac{1}{2} \times 1 \times 4 \\ & =14-2=12 \mathrm{~J} \end{aligned} $$

Please note that the net work done is algebraic sum of the areas of different parts. For the part of the graph below displacement axis; the work is to be taken as negative.

Solution :

When the spring is relaxed; the force $\mathrm{F}$ is zero. As the extension increases; the force required $(\mathrm{k} x)$ to produce extension $x$ also increases. So the force is a variable force. The work done has to calculated by integration.

For an instantaneous extension $x$; the force $\mathrm{F}=\mathrm{k} x$

Work done to produce a further extension $\mathrm{d} x$ i.e., $x$ to $x+\mathrm{d} x$ is $\mathrm{dW}=\mathrm{Fd} x=\mathrm{k} x \mathrm{~d} x$

Work done for an increase in length from 0 to ’ $a$ ’ is given by

$\mathrm{W}=\int \mathrm{dW}=\mathrm{k} \int _{0}^{\mathrm{a}} x \mathrm{~d} x$

$$ =\mathrm{k}\left|\frac{x^{2}}{2}\right| _{0}^{\mathrm{a}}=\frac{1}{2} \mathrm{ka}^{2} $$

Note: This work remains stored in the spring as potential energy.

Solution :

We have $\mathrm{F}=2 x^{2}-3 x+4$

Work done for a displacement $\mathrm{d} x$ is

$$ \mathrm{dW}=\mathrm{Fd} x=\left(2 x^{2}-3 x+4\right) \mathrm{d} x $$

Work done for displacement from $x=0$ to $x=3 \mathrm{~m}$ is

$$ \begin{gathered} \mathrm{W}=\int \mathrm{dW}=\int _{x=0}^{3 \mathrm{~m}}\left(2 x^{2}-3 x+4\right) \mathrm{d} x=\left|2 \frac{x^{3}}{3}-3 \frac{x^{2}}{2}+4 x\right| _{0}^{3} \\ =18-\frac{27}{2}+12=30-\frac{27}{2}=\frac{33}{2} \mathrm{~J}=16.5 \mathrm{~J} \end{gathered} $$

Solution :

Given $x=\frac{1}{3} \mathrm{t}^{3}$

The instantaneous velocity, $\mathrm{v}=\frac{\mathrm{d} x}{\mathrm{dt}}=\mathrm{t}^{2}$

The instantaneous acceleration, $\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{t}^{2}\right)=2 \mathrm{t}$

The instantaneous force $\mathrm{F}$ on the particle, from Newton’s $2^{\text {nd }}$ law is

$\mathrm{F}=\mathrm{ma}=2(2 \mathrm{t})=4 \mathrm{t}$

$\mathrm{W}=\int \mathrm{F} d x=\int \mathrm{F} \frac{\mathrm{d} x}{\mathrm{dt}} \cdot \mathrm{dt}=\int \mathrm{Fvdt}$

$=\int _{\mathrm{t}=0}^{4}(4 \mathrm{t})\left(\mathrm{t}^{2}\right) \mathrm{dt}$

$=\int _{\mathrm{t}=0}^{4} 4 \mathrm{t}^{3} \mathrm{dt}=\left|4 \cdot \frac{\mathrm{t}^{4}}{4}\right| _{0}^{4}$

$=\left[(4)^{4}-0\right]=256 \mathrm{~J}$

Solution :

Work done in case I;

$$ \mathrm{W} _{1}=\mathrm{F} \cos \alpha \cdot \mathrm{S}=10^{2} \mathrm{~N}\left(\frac{40}{50}\right) \times 10 \mathrm{~m}=10^{3} \mathrm{~J} $$

Work done in case II;

$$ \begin{aligned} & \mathrm{W} _{2}=\mathrm{F} \cos \beta \mathrm{S}=10^{2} \times \frac{30}{50} \times 10=\frac{1000 \times 3}{5}=600 \mathrm{~J} \\ \therefore & \frac{\mathrm{W} _{1}}{\mathrm{~W} _{2}}=\frac{1000}{600}=5: 3 \end{aligned} $$

Solution :

Let the mass $\mathrm{m}$ be at A on the surface of the earth. As the mass is moved upwards; the total distance $\mathrm{R}$ involved being large; the acceleration due to gravity does not remain constant. So the force is variable. Hence the work done is calculated by integration.

When the body is at distance ’ $x$ ’ from the centre of the earth, i.e. at point $L$, we have

$$ \mathrm{F} _{\mathrm{gra}}=\frac{\mathrm{GmM}}{x^{2}} $$

Work done to move the body from a distance $x$ to a distance $x+\mathrm{d} x$ or through $\mathrm{d} x$ is given by

$$ \mathrm{dW}=-\mathrm{Fd} x=-\frac{\mathrm{GmM}}{x^{2}} \mathrm{~d} x $$

Work done to move the body from A to $\mathrm{B}$ is

$$ \begin{aligned} \mathrm{W}=\int \mathrm{dW} & =-\mathrm{GmM} \int _{\mathrm{R}}^{2 \mathrm{R}} x^{-2} \mathrm{~d} x \\ & =\mathrm{GmM}\left|x^{-2+1}\right| _{\mathrm{R}}^{2 \mathrm{R}} \\ & =\mathrm{GmM}\left|\frac{1}{x}\right| _{\mathrm{R}}^{2 \mathrm{R}} \\ & =\mathrm{GmM}\left|+\frac{1}{2 \mathrm{R}}-\frac{1}{\mathrm{R}}\right| \\ & =-\left(\frac{\mathrm{GmM}}{2 \mathrm{R}}\right) \end{aligned} $$

$$ \mathrm{W}=-\left[\frac{\mathrm{GmM}}{2 \mathrm{R}}\right] $$

Negative sign indicates that work is done against the gravitational force of earth.

Solution :

We have $\mathrm{k}=40 \mathrm{Nm}^{-1}$

Initially $x _{1}=2 \mathrm{~cm}=\frac{2}{100} \mathrm{~m}$

Work done for compression through $2 \mathrm{~cm}$

or $\mathrm{W} _{1}=\frac{1}{2} \mathrm{k} x _{1}^{2}$

$$ =\frac{1}{2} \times 40 \times \frac{2}{100} \times \frac{2}{100}=8 \times 10^{-3} \mathrm{~J} $$

Work done for a compression from $0 \mathrm{~cm}$ to $4 \mathrm{~cm}$ is

$$ \begin{aligned} \mathrm{W} _{2}=\frac{1}{2} \mathrm{k} & x _{2}^{2} \\ & =\frac{1}{2} \times 40 \times \frac{4}{100} \times \frac{4}{100}=32 \times 10^{-3} \mathrm{~J} \end{aligned} $$

$\therefore$ Work for a compression from $2 \mathrm{~cm}$ to $4 \mathrm{~cm}$

$$ \begin{aligned} & =\mathrm{W} _{2}-\mathrm{W} _{1}=(32-8) \times 10^{-3} \\ & =.024 \mathrm{~J} \end{aligned} $$

Solution :

A careful observation of the graph shows the relative values of $\mathrm{E}$, the total energy and the potential energy of the particle.

In the regions $-\frac{\mathrm{b}}{2}<x<-\frac{\mathrm{a}}{2}$ and $+\frac{\mathrm{a}}{2}<x<+\frac{\mathrm{b}}{2}$; the PE of the particle exceeds the total energy ’ $\mathrm{E}$ ‘. Hence the particle cannot exist in these two regions. For all other regions; the PE is less than E and hence the particle can exist.

Hence option (4) above is correct.

Solution :

The total energy of a particle is equal to the sum of its PE and KE.

$\therefore \mathrm{E}=\mathrm{PE}+\mathrm{KE}$

$$ =\frac{1}{2} \mathrm{k} x^{2}+\frac{1}{2} \mathrm{mv}^{2} $$

Where $x$ and $\mathrm{v}$ denote instantaneous displacement and instantaneous speed respectively.

Given; P.E. $\mathrm{v}(x)=\frac{1}{8} \mathrm{k} x^{2}$

We know, $\mathrm{PE}$ is maximum when $\mathrm{KE}$ is zero.

$\therefore(\mathrm{PE}) _{\max }=\mathrm{E}$ when $\mathrm{K}=0$

$$ =\frac{1}{2} \mathrm{k} x _{\max }^{2} $$

$\therefore x _{\text {max }}=\sqrt{\frac{(\mathrm{PE}) _{\text {max }} \times 2}{k}}=\sqrt{\frac{2 \mathrm{E}}{\mathrm{k}}}$

$$ = \pm \sqrt{\frac{2 \times 1}{2}}=1 \mathrm{~m} \text { or }-1 \mathrm{~m} $$

Hence the particle must turn back at $x=+1$ and $x=-1 \mathrm{~m}$.

Solution:

$\mathrm{d}=$ Depth of rain water received $=8 \mathrm{~cm}=0.08 \mathrm{~m}$

Total volume of water received in rainfall $=\mathrm{A} \times \mathrm{d}$

$$ \begin{aligned} & =1000 \times 1000 \times \frac{8}{100} \\ & =8 \times 10^{4} \mathrm{~m}^{3} \end{aligned} $$

$$ \rho=10^{3} \mathrm{~kg} / \mathrm{m}^{3} $$

$\therefore$ Mass of water received as rainfall $=\mathrm{V} \rho$

$$ \begin{aligned} & =8 \times 10^{4} \times 10^{3} \\ & =8 \times 10^{7} \mathrm{~kg} \end{aligned} $$

W.D. by the gravitational force

$=$ Decrease in gravitational potential energy

$=\mathrm{mgh}=8 \times 10^{7} \times 10 \times 250$

Solution :

Given $\mathrm{m}=1 \mathrm{~g}=10^{-3} \mathrm{~kg}$,

$$ \begin{aligned} & \mathrm{h}=1 \mathrm{~km}=1000 \mathrm{~m} \\ & \mathrm{v}=50 \mathrm{~ms}^{-1} \end{aligned} $$

$\mathrm{E} _{\mathrm{e}}=$ The total initial energy of rain drop

$$ \begin{aligned} & =\mathrm{mgh}+0 \\ & =10^{-3} \times 10 \times 10^{3}=10 \mathrm{~J} \end{aligned} $$

$\mathrm{E} _{\mathrm{f}}=$ The total final energy of rain drop (i.e. when rain drop is just about to hit ground).

$$ \begin{aligned} & =\frac{1}{2} \mathrm{mv}^{2}+0 \\ & =\frac{1}{2} \times 10^{-3} \times(50)^{2}=1.25 \mathrm{~J} \end{aligned} $$

Let $\mathrm{W}$ be work done on the rain drop by resistive force. Then

$$ \begin{aligned} & =\mathrm{E} _{\mathrm{e}}+\mathrm{E} _{\mathrm{f}}=\mathrm{W} \\ & =10+1.25=\mathrm{W} \end{aligned} $$

or

$$ \mathrm{W}=-8.75 \mathrm{~J} $$

Solution :

The initial KE of the automobile is partially used to work against friction and is partly converted to potential energy. We have

Initial $\mathrm{KE}=\mathrm{PE}$ gained + work done against friction.

We have,

$$ \mathrm{u}=54 \mathrm{~km} / \mathrm{hr}=54 \times \frac{5}{18}=15 \mathrm{~m} / \mathrm{s} $$

Component of gravitational force down the plane $=\mathrm{mg} \sin \theta=\mathrm{mg} \sin 30^{\circ}$

Normal reaction $\mathrm{N}=\mathrm{mg} \cos 30$

$\therefore \mathrm{F} _{\text {friction }}=\mu \mathrm{N}=0.2 \mathrm{mg} \cos 30^{\circ}$

Net force down the plane $=\mathrm{F} _{\mathrm{fr}}+\mathrm{mg} \sin \theta$

$$ \begin{aligned} & =\mu \mathrm{mg} \cos \theta+\mathrm{mg} \sin \theta \\ & =\mathrm{mg}(\sin \theta+\mu \cos \theta) \end{aligned} $$

Let ’ $\ell$ ’ be the distance moved by the automobile up the plane before coming to rest; Then

$$ \frac{1}{2} m u^{2}=m g(\sin \theta+\mu \cos \theta) \ell $$

or

$$ \ell=\frac{\mathrm{u}^{2}}{2 \mathrm{~g}(\sin \theta+\mu \cos \theta)} $$

$$ \begin{aligned} & =\frac{15 \times 15}{2 \times 10(\sin 30+0.2 \cos 30)} \\ & =\frac{15 \times 15}{2 \times 10(0.5+0.1732)} \end{aligned} $$

$$ \begin{aligned} & =\frac{225}{13.464} \mathrm{~m} \\ & =16.7 \mathrm{~m} \end{aligned} $$

Solution :

We have, $\mathrm{m}=5 \mathrm{~kg}$

$$ \begin{aligned} \mathbf{u} & =(3 \mathbf{i}-4 \mathbf{j}) \mathrm{m} / \mathrm{s} \\ \therefore|\mathbf{u}|= & \sqrt{4 _{x}^{2}+4 _{y}^{2}}=5 \mathrm{~m} / \mathrm{s} \\ & \mathbf{v}=(2 \mathbf{j}-6 \mathbf{k}) \mathrm{m} / \mathrm{s} \\ \therefore|\mathbf{v}|= & \sqrt{(2)^{2}+(-6)^{2}} \\ & =\sqrt{40} \mathrm{~m} / \mathrm{s} \end{aligned} $$

$\therefore$ Change in $\mathrm{KE}=\frac{1}{2} \mathrm{mv}^{2}-\frac{1}{2} \mathrm{mu}^{2}$

$$ \begin{aligned} & =\frac{1}{2} \mathrm{~m}\left[(\sqrt{40})^{2}-(5)^{2}\right] \\ & =\frac{1}{2} \times 5 \times 15=37.5 \mathrm{~J} \end{aligned} $$

Solution :

Fig. (b)

Case (i)

Mass, the FBD is as shown. The equation of motion of the two masses is:

For $4 \mathrm{~kg}$ mass : $\mathrm{mg}-\mathrm{T}=\mathrm{ma} \hspace{40mm} . . . . . . (1)$

For $5 \mathrm{~kg}$ mass $\mathrm{T}=\mathrm{Ma} \hspace{40mm} . . . . . . (2)$

From (1) and (2) we have

$$ \begin{aligned} & \mathrm{mg}=(\mathrm{m}+\mathrm{M}) \mathrm{a} \\ & \text { or } \quad \mathrm{a}=\left(\frac{\mathrm{m}}{\mathrm{m}+\mathrm{M}}\right) \mathrm{g} \\ & =\frac{4 \times 10}{9}=\frac{40}{9} \mathrm{~m} / \mathrm{s}^{2} \end{aligned} $$

After $4 \mathrm{~s} ; \mathrm{v}=\mathrm{u}+\mathrm{at}=0+\frac{40}{9} \times 4$

$$ =\frac{160}{9} \mathrm{~m} / \mathrm{s} $$

KE of system $=\frac{1}{2}(m+M) v^{2}$

$$ \begin{aligned} & =\frac{1}{2} \times 9 \times \frac{160}{9} \times \frac{160}{9} \\ & =\frac{12800}{9} \mathrm{~J} \end{aligned} $$

Case (ii)

The $5 \mathrm{~kg}$ mass will move downwards and $4 \mathrm{~kg}$ upwards. Let $\mathrm{T}$ be the tension in the string. Using free body diagram(FBD) we get

$\mathrm{Mg}-\mathrm{T}=\mathrm{Ma} _{1} \hspace{40mm} . . . . . . (1)$

and $\quad \mathrm{T}-\mathrm{mg}=\mathrm{ma} _{1} \hspace{40mm} . . . . . . (2)$

(1) $+(2)$ gives

$(M-m) g=(M+m) a _{1}$

or $\quad a _{1}=\left(\frac{5-4}{5+4}\right) g=\frac{10}{9} \mathrm{~m} / \mathrm{s}^{2}$

$$ (M-m) g=(M+m) a _{1} $$

$$ \mathrm{a} _{1}=\left(\frac{5-4}{5+4}\right) \mathrm{g}=\frac{10}{9} \mathrm{~m} / \mathrm{s}^{2} $$

Velocity after $4 \mathrm{~s}=$ at $(\because \mathrm{u}=0)$

$$ =\frac{40}{9} \mathrm{~m} / \mathrm{s} $$

$\therefore$ KE of system at $\mathrm{t}=4 \mathrm{~s}=\frac{1}{2}(\mathrm{~m}+\mathrm{M}) \mathrm{v}^{2}$

$$ =\frac{1}{2} \times 9 \times \frac{40}{9} \times \frac{40}{9}=\frac{800}{9} \mathrm{~J} $$

Solution :

Let $\mathrm{F}$ be the resistive force offered to the shots.

Work done against the force $=$ Initial K.E.

$\therefore$ F. $\mathrm{S} _{1}=\frac{1}{2} \mathrm{mV} _{1}^{2}$

F. $\mathrm{S} _{2}=\frac{1}{2} \mathrm{mV} _{2}^{2}$

$\Rightarrow \frac{\mathrm{V} _{1}^{2}}{\mathrm{~V} _{2}^{2}}=\frac{\mathrm{S} _{1}}{\mathrm{~S} _{2}}=\frac{4}{9}$

$\frac{\mathrm{V} _{1}}{\mathrm{~V} _{2}}=2: 3$

Solution :

Given $\mathrm{m}=200, \mathrm{gm}=0.2 \mathrm{~kg}$

$$ \mathrm{OA}=\mathrm{OB}=1 \mathrm{~m} $$

$$ \angle \mathrm{AOB}=30^{\circ} $$

With $\mathrm{BC} \perp \mathrm{OA}$; we have

$$ \begin{aligned} & \frac{\mathrm{OC}}{\mathrm{OB}}=\cos 30^{\circ} \\ \therefore \quad \mathrm{OC} & =\mathrm{OB} \cos 30^{\circ}=\frac{\sqrt{3}}{2} \mathrm{~m} \quad[\mathrm{OB}=1 \mathrm{~m}] \\ & =0.866 \mathrm{~m} \end{aligned} $$

Vertical height through which the bob is raised $=\mathrm{h}(1-0.866) \mathrm{m}$

$$ =0.134 \mathrm{~m} $$

Potential energy gained by the bob from $A$ to $B$

$$ \begin{aligned} & =\mathrm{mgh}=0.2 \times 10 \times 0.134 \\ & =0.268 \mathrm{~J} \end{aligned} $$

This PE is converted to $\mathrm{KE}$ as the bob goes back to $\mathrm{A}$.

$\therefore \mathrm{KE}$ at the lowest point $=\mathrm{PE}$ at $\mathrm{B}$

$$ =0.268 \mathrm{~J} $$

Solution :

We have $\mathrm{m}=1 \mathrm{~kg}, \mathrm{u}=2 \mathrm{~ms}^{-1} ; \mathrm{k}=0.5 \mathrm{~J}$

Initial $\mathrm{KE}=\frac{1}{2} \mathrm{mu}^{2}=\frac{1}{2} \times 1 \times(2)^{2}=2 \mathrm{~J}$

Frictional force over $x=0.1 \mathrm{~m}$ to $2.01 \mathrm{~m}$ range is $\mathrm{F} _{\mathrm{r}}=-\frac{\mathrm{k}}{x}$

As the force is variable; work against friction can be calculated by integration.

We have, $\mathrm{W}=\int _{x=0.1 \mathrm{~m}}^{2.01 \mathrm{~m}} \mathrm{~F} _{\mathrm{r}} \mathrm{d} x=\int _{x=0.1 \mathrm{~m}}^{x=2.01 \mathrm{~m}} \frac {\mathrm{k}}{\mathrm{x}} \mathrm{d}{x}$

$$ \begin{aligned} & =-0.5\left|\log _{\mathrm{e}} x\right| _{x=0.1 \mathrm{~m}}^{x=2.01 \mathrm{~m}} \\ & =-0.5 \log _{\mathrm{e}} \frac{2.01}{0.1} \\ & =-0.5 \times 2.303\left[\log _{10} 20.01\right] \\ & =-0.5 \times 2.303 \times 1.303 \quad \text { [Using tables] } \\ & =-1.5 \mathrm{~J} \end{aligned} $$

$\therefore \mathrm{K} _{\text {final }}-\mathrm{K} _{\text {initial }}=$ Work done

$\Rightarrow \mathrm{K} _{\text {final }}=\mathrm{K} _{\text {initial }}+$ Work done

$$ =2.0 \mathrm{~J}-1.5 \mathrm{~J}=0.5 \mathrm{~J} $$

Also, $\mathrm{K} _{\text {final }}=\frac{1}{2} \mathrm{mv}^{2}=0.5 \mathrm{~J}$

$\therefore \mathrm{v}^{2}=\sqrt{\frac{2 \times 0.5}{1}}=1 \mathrm{~ms}^{-1}$

Solution :

We have $\mathrm{m}=50.0 \mathrm{~g}=\frac{50}{1000} \mathrm{~kg}=\frac{1}{20} \mathrm{~kg}$

$$ \mathrm{u}=200 \mathrm{~ms}^{-1} $$

Initial $\mathrm{KE}=\frac{1}{2} \mathrm{mu}^{2}=\frac{1}{2} \times \frac{1}{20} \times 200 \times 200=1000 \mathrm{~J}$

Final $\mathrm{KE}=10 \%$ of initial $\mathrm{KE}=\frac{10}{100} \times 1000$

$$ =100 \mathrm{~J} $$

$\therefore \frac{1}{2} \mathrm{mv}^{2}=100$

$\mathrm{v}=\sqrt{\frac{2 \times 100}{\frac{1}{20}}}=\sqrt{4000}$

$=63.2 \mathrm{~m} / \mathrm{s}$

It is important to note that a $90 \%$ decreases in KE reduces the speed by $\left(\frac{200-63.2}{200} \times 100\right) \%$ or $68 \%$.

Solution :

Case (i)

When the surface is smooth; the entire KE of the block is stored in the spring as elastic PE. If $x _{0}$ is the maximum compression produced in the spring; we have

$$ \begin{aligned} & \frac{1}{2} \mathrm{k} x _{0}^{2}=\frac{1}{2} \mathrm{mu}^{2} \\ & x _{0}=\sqrt{\frac{\mathrm{m}}{\mathrm{k}}} \cdot \mathrm{u} \end{aligned} $$

Case (ii)

When the surface is rough; work has to be done to overcome friction and to cause a compression in the spring.

For a compression ’ $x$ ’ in the spring, we have $\mu=a x$.

$\therefore$ Force of friction for a compression $x$ is

$$ \mathrm{F} _{\mathrm{fr}}=\mu \mathrm{N}=\mu \mathrm{Mg}=(\mathrm{a} x) \mathrm{Mg}=(\mathrm{aMg}) x $$

Work done against friction for a maximum compression $x _{1}$ (say) is

$$ \begin{aligned} & \mathrm{W} _{\mathrm{fr}}=\int _{0}^{x _{1}} \mathrm{~F} _{\mathrm{fr}} \mathrm{d} x=\mathrm{aMg} \int _{0}^{x _{1}} x \mathrm{~d} x \\ & =\mathrm{aMg}\left|\frac{x^{2}}{2}\right| _{0}^{x _{1}}=\frac{1}{2} \mathrm{aMg} x _{1}^{2} \end{aligned} $$

Energy stored in the spring $=\frac{1}{2} \mathrm{k} x _{1}^{2}$

From law of conservation of energy;

$$ \frac{1}{2} \mathrm{Mu}^{2}=\frac{1}{2} \mathrm{k} x _{1}^{2}+\frac{1}{2} \mathrm{a} M g x _{1}^{2} $$

or

$$ \begin{aligned} & \mathrm{Mu}^{2}=\mathrm{k} x _{1}^{2}+\mathrm{aMg} x _{1}^{2} \\ & =(\mathrm{k}+\mathrm{aMg}) x _{1}^{2} \end{aligned} $$

$$ \therefore \quad x _{1}=\left[\sqrt{\frac{\mathrm{M}}{\mathrm{k}+\mathrm{aMg}}}\right] \mathrm{u} $$

Solution :

As the car collides with the spring, it does work against the force of friction and for compression in the spring. Let $x _{0}$ be the maximum compression produced in the spring.

Work done for compression $=\frac{1}{2} \mathrm{k} x _{0}^{2}$

Work done against friction $=(\mu \mathrm{mg}) x _{0}$

By work energy theorem;

Change in $\mathrm{KE}=$ Work done in compressing spring + Work done against friction.

$\therefore \frac{1}{2} \mathrm{mu}^{2}=\frac{1}{2} \mathrm{k} x _{0}^{2}+\mu \mathrm{mg} x _{0}$

or $\quad \mathrm{k} x _{0}^{2}+2 \mu \mathrm{mg} x _{0}-\mathrm{mu}^{2}=0$

Using quadratic formula with $\mathrm{a}=\mathrm{k} ; \mathrm{b}=2 \mu \mathrm{mg}$ and $\mathrm{c}=-\mathrm{mu}^{2}$

$$ \begin{aligned} x _{0}= & \frac{-2 \mu \mathrm{mg} \pm\left[4 \mu^{2} \mathrm{~m}^{2} \mathrm{~g}^{2}+4 \mathrm{mku}^{2}\right]^{1 / 2}}{2 \mathrm{k}} \\ & =\frac{-\mu \mathrm{mg}+\left[\mu^{2} \mathrm{~m}^{2} \mathrm{~g}^{2}+\mathrm{mku}^{2}\right]^{1 / 2}}{\mathrm{k}} \quad \text { [Neglecting-ve value] } \\ \therefore x _{0} & =\frac{(-0.5)(1000)(10)+\left[(0.25)(10)^{6}(100)+(1000)\left(6.25 \times 10^{3}\right)(5)^{2}\right]}{6.25 \times 10^{3}} \quad[\mathrm{u}=18 \mathrm{~km} / \mathrm{hr}=5 \mathrm{~m} / \mathrm{s}] \\ & =1.35 \mathrm{~m} \end{aligned} $$

On a smooth road,

$\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{2} u x _{0}^{2}$

or $\quad x _{0}=\sqrt{\frac{\mathrm{mv}^{2}}{\mathrm{k}}}$

$=\sqrt{\frac{1000 \times 25}{6250}}$

$=2 \mathrm{~m}$

Solution :

We have, power $\mathrm{P}=10000 \mathrm{MW}$

$$ =10^{10} \mathrm{~W} $$

Energy generated per second $=10^{10} \mathrm{~W}$

1 day $=24 \times 60 \times 60 \mathrm{~s}$

$\therefore$ Total energy produced per day $=24 \times 60 \times 60 \times 10^{10} \mathrm{~J}$

$\mathrm{E}=36 \times 24 \times 10^{12} \mathrm{~J}$

Let $\mathrm{m}$ be the mass converted into energy each day.

We have, $\mathrm{mc}^{2}=\mathrm{E}$

or

$$ \begin{aligned} & \mathrm{m}=\frac{\mathrm{E}}{\mathrm{c}^{2}}=\frac{36 \times 24 \times 10^{12}}{\left(3 \times 10^{8}\right)^{2}} \\ & =\frac{36 \times 24}{9} \times 10^{-4} \mathrm{~kg} \\ & =9.6 \times 10^{-3} \mathrm{~kg}=9.6 \mathrm{~g} \end{aligned} $$

Solution :

We have,

$\mathrm{M} _{\mathrm{w}}=1000 \mathrm{~kg}$

$\theta _{\mathrm{i}}=20^{\circ} \mathrm{C} ; \theta _{\mathrm{f}}=100^{\circ} \mathrm{C}$

$\mathrm{C}=4200 \mathrm{~J} \mathrm{~kg}^{-10} \mathrm{C}^{-1}$

$\therefore$ Heat gained by water $=\mathrm{M} _{\mathrm{w}} \mathrm{C}\left(\theta _{\mathrm{f}}-\theta _{\mathrm{i}}\right)$

$$ \begin{aligned} & =1000 \times 4200 \times(100-20) \\ & =336 \times 10^{6} \mathrm{~J} \end{aligned} $$

If ’ $m$ ’ is the equivalent mass, we have

$$ \begin{aligned} & \mathrm{mc}^{2}= 336 \times 10^{6} \\ & \mathrm{~m}=\frac{336 \times 10^{6}}{\left(3 \times 10^{8}\right)^{2}} \\ &= 37.3 \times 10^{-10} \mathrm{~kg} \\ &= 3.73 \times 10^{-9} \mathrm{~kg} \end{aligned} $$

Solution :

Total mass of the reacting particles $=2 \times \mathrm{m}\left({ } _{1} \mathrm{H}^{2}\right)=2 \times 2.0141=4.0282 \mathrm{u}$

Total mass of the product particles $=\mathrm{m}\left({ } _{2} \mathrm{He}^{3}\right)+\mathrm{m}\left({ } _{0} \mathrm{n}^{1}\right)$

$=(3.0160+1.0087) \mathrm{u}$

$=4.0247 \mathrm{u}$

$\therefore \Delta \mathrm{M}=$ Mass lost during the reaction $=(4.0282-4.0247) \mathrm{u}$

$=0.0035 \mathrm{u}$

$\mathrm{E}=$ The energy released $=(\Delta \mathrm{M}) \times \mathrm{c}^{2}=0.035 \times 931.5 \mathrm{MeV}$

$$ =3.2602 \quad \mathrm{MeV} $$

Solution :

Give $\mathrm{F} _{\text {res }}=5000 \mathrm{~N}$

$$ \mathrm{v}=18 \mathrm{~km} \quad \mathrm{hr}^{-1}=18 \times \frac{5}{18}=5 \mathrm{~ms}^{-1} $$

Power $=\mathbf{F} \cdot \mathbf{v}=\mathbf{F} _{\text {app }} \cdot \mathbf{v}\left[\mathbf{F} _{\text {app }}=\mathbf{F} _{\text {res }}\right.$ for steady motion $]$

$$ \begin{aligned} & =5000 \times 5=25000 \mathrm{~W} \\ & =25 \mathrm{~kW} \end{aligned} $$

Solution :

We have $\mathrm{m}=1000$ metric ton $=10^{6} \mathrm{~kg}$

Frictional resistance $=0.4 \mathrm{~kg} /$ metric ton.

$\therefore$ Total frictional force $\mathrm{F} _{\mathrm{fr}}=1000 \times 0.4=400 \mathrm{~kg}$ wt

$$ =400 \times \mathrm{gN}=4000 \mathrm{~N} $$

Gravitational force down the plane $=\mathrm{mg} \sin \theta$

$$ \begin{aligned} & =10^{6} \times 10 \times \frac{1}{200} \mathrm{~N} \\ & =5 \times 10^{4} \mathrm{~N}=50000 \mathrm{~N} \end{aligned} $$

Net opposing force $=\mathrm{F} _{\mathrm{fr}}+\mathrm{mg} \sin \theta$

$$ =4000+50000=54000 \mathrm{~N} $$

$\mathrm{V}=54 \mathrm{~km} \quad \mathrm{hr}^{-1}=54 \times \frac{5}{18}=15 \mathrm{~ms}^{-1}$

The power of the engine $=\mathrm{P}=$ F.v. $=54000 \times 25 \mathrm{~W}$

$$ =810 \mathrm{~kW} $$

Solution :

We have $\mathbf{F}=(2 \mathbf{i}+3 \mathbf{j}) \mathrm{N}$

$$ \begin{gathered} \boldsymbol{x}=\left(2 \mathrm{t}^{2}+3 \mathrm{t}+5\right) \mathbf{i} \\ \therefore \mathbf{v}=\frac{\mathrm{d} \boldsymbol{x}}{\mathrm{dt}}=(4 \mathrm{t}+3) \mathbf{i}, \mathrm{m} / \mathrm{s} \\ \text { At } \mathrm{t}=2 \mathrm{~s} ; \mathbf{v}=[4(2)+3] \mathbf{i}=11 \mathbf{i}, \mathrm{m} / \mathrm{s} \end{gathered} $$

Instantaneous power $\mathrm{P}=\mathbf{F} . \mathbf{v}$

$$ \begin{aligned} & =(2 \mathbf{i}+3 \mathbf{j})(11 \mathbf{i}) \\ & =22 \mathbf{J} / \mathrm{s}=22 \mathrm{~W} \end{aligned} $$

Solution :

We have $\mathrm{a} _{\mathrm{c}}=25 \mathrm{rt}^{2} \mathrm{~m} / \mathrm{s}^{2} ; \mathrm{r}=5 \mathrm{~m}$

Also, $\mathrm{a} _{\mathrm{c}}=\frac{\mathrm{v}^{2}}{\mathrm{r}}$

$\therefore \frac{\mathrm{v}^{2}}{\mathrm{r}}=25 \mathrm{rt}^{2}$

or $\quad \mathrm{v}^{2}=25 \mathrm{r}^{2} \mathrm{t}^{2}$

$\therefore \mathrm{v}=5 \mathrm{rt}$

The tangential acceleration $=\mathrm{a} _{\mathrm{T}}=\frac{\mathrm{dv}}{\mathrm{dt}}=5 \mathrm{r}=25 \mathrm{~m} / \mathrm{s}^{2}$

The tangential force $=\mathrm{F} _{\mathrm{T}}=\mathrm{a} _{\mathrm{T}}=10 \times 25=250 \mathrm{~N}$

Power is delivered only by tangential force;

$$ \begin{aligned} \mathrm{P}=\mathbf{F} _{\mathrm{t}} \cdot \mathbf{v} & =(250)(5 \mathrm{rt}) \\ & =250(5 \times 5 \times 2) \quad[\text { At } \mathrm{t}=2 \mathrm{~s}] \\ & =12500 \mathrm{~W} \\ & =12.5 \mathrm{~kW} \end{aligned} $$

Solution :

We have $\mathbf{F}=\frac{\mathrm{d} \mathbf{p}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{m} \mathbf{v})=\mathbf{v} \frac{\mathrm{dm}}{\mathrm{dt}}$

$$ =\mathrm{v} \cdot \frac{\mathrm{d}}{\mathrm{dt}} \quad[\text { Volume } \times \text { density of wind }] $$

$$ =\mathrm{v} \frac{\mathrm{d}}{\mathrm{dt}}(\mathrm{A} x \times \rho) $$

Where $\mathrm{A}$ is area of close section of blades. Therefore

$$ \mathrm{F}=\mathrm{vA} \rho \frac{\mathrm{d} x}{\mathrm{dt}}=\mathrm{A} \rho \mathrm{v}^{2} $$

The instantaneous power $=\mathrm{P}=\mathbf{F} \cdot \mathbf{v}=\left(\mathrm{A\rho v}^{2}\right) \mathrm{v}=\mathrm{A} \rho \mathrm{v}^{3}$

$\therefore$ Power $\mathrm{P} \propto \mathrm{v}^{3}$

Solution :

We have $\mathrm{m} _{1}=\mathrm{m} \quad \mathrm{m} _{2}=x$ (Say)

$$ \mathrm{u} _{1}=\mathrm{u} \quad \mathrm{u} _{2}=0 $$

$$ \mathrm{v} _{1}=\frac{\mathrm{u}}{2} $$

As $\mathrm{v} _{1}=\left(\frac{\mathrm{m} _{1}-\mathrm{m} _{2}}{\mathrm{~m} _{1}+\mathrm{m} _{2}}\right) \mathrm{u} _{1}$; we get

$$ =\frac{\mathrm{u}}{2}=\left(\frac{\mathrm{m}-x}{\mathrm{~m}+x}\right) \mathrm{u} $$

or

$$ \mathrm{m}+x=2 \mathrm{~m}-2 x $$

$\therefore \quad x=\frac{\mathrm{m}}{3}$

Solution :

We have $\mathrm{m} _{1}=\mathrm{m} ; \quad \mathrm{m} _{2}=4 \mathrm{~m}$

$$ \mathrm{u} _{1}=\mathrm{u}(\text { say }) \quad \mathrm{u} _{2}=0 $$

As $\quad \mathrm{v} _{2}=\left(\frac{2 \mathrm{~m} _{1}}{\mathrm{~m} _{1}+\mathrm{m} _{2}}\right) \mathrm{u}$;

We get $\mathrm{v} _{2}=\frac{2 \mathrm{~m}}{5 \mathrm{~m}} \mathrm{u}=\frac{2}{5} \mathrm{u}$

$\therefore \frac{\mathrm{KE} \text { of } \mathrm{B}(\text { after collision })}{\text { Initial } \mathrm{KE} \text { of } \mathrm{A}}=\frac{\frac{1}{2} \mathrm{~m} _{2} \mathrm{v} _{2}^{2}}{\frac{1}{2} \mathrm{~m} _{1} \mathrm{v} _{1}^{2}}$

$=\frac{(4 \mathrm{~m})\left(\frac{2}{5} \mathrm{u}\right)^{2}}{m u^{2}}$

$=\frac{16}{25}$

$=\frac{16}{25} \times 100=64 \%$

Solution :

Initial momentum $=0$

$\therefore$ Final momentum $=0$ or $\quad \mathbf{p} _{1}+\mathbf{p} _{2}=0$

$\Rightarrow\left|\mathbf{p} _{1}\right|=\left|\mathbf{p} _{2}\right|$

$\therefore \frac{\mathrm{p} _{1}}{\mathrm{p} _{2}}=1$

$\frac{\text { KE of emitted particle }}{\text { KE of residual part }}=\frac{p _{2}^{2} / 2(m / 4)}{p _{1}^{2} / 2\left(m-\frac{m}{4}\right)}=3: 1$

Solution :

We have, $\mathrm{m} _{1}=\mathrm{m} _{2}=\mathrm{m}$

$$ \mathrm{u} _{1}=\mathrm{u} ; \mathrm{u} _{2}=0 $$

From principle of conservation of linear momentum, we have

$\mathrm{m} _{1} \mathrm{u} _{1}+\mathrm{m} _{2} \mathrm{u} _{2}=\mathrm{m} _{1} \mathrm{v} _{1}+\mathrm{m} _{2} \mathrm{v} _{2}$

$\therefore \mathrm{mu}+0=\mathrm{mv} _{1}+\mathrm{mv} _{2}=\mathrm{m}\left(\mathrm{v} _{1}+\mathrm{v} _{2}\right)$

$$ \begin{equation*} \Rightarrow \mathrm{v} _{1}+\mathrm{v} _{2}=\mathrm{u} \tag{1} \end{equation*} $$

Also $\frac{\left|\mathrm{v} _{2}-\mathrm{v} _{1}\right|}{\left|\mathrm{u} _{2}-\mathrm{u} _{1}\right|}=\frac{\mathrm{v} _{2}-\mathrm{v} _{1}}{\mathrm{u}-0}=\frac{\mathrm{v} _{2}-\mathrm{v} _{1}}{\mathrm{u}}=\mathrm{e}$

$\therefore \mathrm{v} _{2}-\mathrm{v} _{1}=\mathrm{ue} \hspace{60mm} (2)$

From (1) and (2) $\frac{\mathrm{v} _{1}}{\mathrm{v} _{2}}=\frac{1-\mathrm{e}}{1+\mathrm{e}}$

$\therefore \frac{\mathrm{K} _{1}}{\mathrm{~K} _{2}}=\frac{\frac{1}{2} \mathrm{mv} _{1}^{2}}{\frac{1}{2} \mathrm{mv} _{2}^{2}}=\left(\frac{1-\mathrm{e}}{1+\mathrm{e}}\right)^{2}$

Solution :

We have $\mathrm{m} _{1}=\mathrm{m} ; \mathrm{m} _{2}=\mathrm{m}$

$$ \mathrm{u} _{1}=10 \mathrm{~m} / \mathrm{s} ; \mathrm{u} _{2}=0 $$

$\mathrm{v} _{1}=\left(\frac{\mathrm{m} _{1}-\mathrm{m} _{2}}{\mathrm{~m} _{1}+\mathrm{m} _{2}}\right) \mathrm{u} _{1}=0$

$\mathrm{v} _{2}=\left(\frac{2 \mathrm{~m} _{1}}{\mathrm{~m} _{1}+\mathrm{m} _{2}}\right) \mathrm{u} _{1}=\left(\frac{2 \mathrm{~m}}{2 \mathrm{~m}}\right) 10=10 \mathrm{~m} / \mathrm{s}$

Hence the ball A comes to rest and falls vertically downward just below B.

Time taken by A to fall down is given by $\mathrm{h}=\frac{1}{2} \mathrm{gt}^{2}$

or $\quad \mathrm{t}^{2}=\frac{2 \times 4.9}{9.8} \Rightarrow \mathrm{t}=1 \mathrm{~s}$

Horizontal distance covered by B in $1 \mathrm{~s}=\mathrm{v} _{2} \times \mathrm{t}$

$=10 \mathrm{~m}$

Hence the required distance $=10 \mathrm{~m}$

Answer: $12.12 \mathrm{~ms}^{-1} ; 9.9 \mathrm{~ms}^{-1} ; 14 \mathrm{~ms}^{-1}$

Correct answer: (3)

Solution:

Take the initial position of the mass with the spring compressed as zero gravitational potential energy position. When the block and the spring are released; the potential energy of the spring and the decrease in gravitational potential energy of the block get converted to kinetic energy of the block. The block moves down through a vertical distance $=a \sin \theta$. From law of conservation of energy the gain in K.E. equals loss in total potential energy of system.

$$ \begin{aligned} & \therefore \frac{1}{2} \mathrm{Mv}^{2}=\frac{1}{2} \mathrm{ka}^{2}+(\mathrm{Mg}) \mathrm{a} \sin \theta \\ & \quad=\frac{1}{2}\left(\frac{\mathrm{Mg}}{\mathrm{a}}\right) \mathrm{a}^{2}+(\mathrm{Mga}) \sin \theta \\ & \text { or } \quad \mathrm{v}^{2}=\mathrm{ga}+2 \mathrm{ga} \sin \theta \\ & \quad=\mathrm{ga}(1+2 \sin \theta) \\ & \mathrm{v}=\sqrt{\mathrm{ga}(1+2 \sin \theta)} \end{aligned} $$

Correct answer: (3)

Solution:

We known that the momentum acquired by the bullet and the gun have equal magnitude (Principle of conservation of linear momentum).

$\therefore$ Momentum of bullet and gun $=|\overrightarrow{\mathrm{p}}|$ each.

The kinetic energy $k$ and $K$ of bullet and gun are:

$$ \mathrm{k}=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}} $$

and $\quad \mathrm{K}=\frac{\mathrm{p}^{2}}{2 \mathrm{M}}$

$\therefore \frac{\mathrm{K}}{\mathrm{k}}=\frac{\mathrm{m}}{\mathrm{M}}$

or $\quad \mathrm{K}=\frac{\mathrm{m}}{\mathrm{M}} \mathrm{k}$

Correct answer: (2)

Solution:

The initial potential energy mgh of the particle is used in doing work against friction in the flat part BC.

The distance traveled along $\mathrm{BC}$ before coming to rest $=\mathrm{S}=\mathrm{n} \cdot \ell$

$\mathrm{W}=$ the work done against force of friction $=\mathrm{F} _{\mathrm{fr}} .\mathrm{S}$

$$ =\mu _{\mathrm{k}} \mathrm{mg} \cdot \mathrm{S}=\left(\mu _{\mathrm{k}} \mathrm{mg}\right) \mathrm{n} \ell $$

By work-energy theorem;

Work done $=$ Change in energy $\mu _{\mathrm{k}} \mathrm{mg} \mathrm{n} \ell=\mathrm{mgh}$

$$ \mathrm{n}=\frac{\mathrm{h}}{\mu _{\mathrm{k}} \ell} $$

Correct answer: (2)

Solution:

Let $m$ be the mass of the sphere and $\rho$ its density. Then density of water is $2 \rho$. Kinetic energy of ball as it hits the water surface $=\mathrm{mgh}$

$=19.6 \mathrm{mg}$ joule

Net upward force on sphere when it is inside water $=$ upthrust - weight of the ball

$$ \begin{aligned} & =\left(\frac{\mathrm{m}}{\rho} \times 2 \rho \mathrm{g}\right)-\mathrm{mg} \\ & =\mathrm{mg} \end{aligned} $$

Let the sphere go to a depth d inside water; when it is momentarily at rest. We have,

Work done $=$ Change in K.E.

$\therefore \quad \mathrm{mg} \times \mathrm{d}=\mathrm{mg} \times 19.6$

or $\mathrm{d}=19.6 \mathrm{~m}$

$\mathrm{u}=$ velocity the sphere as it hits the water $=\mathrm{u}=\sqrt{2 \mathrm{gh}}$

or $\quad \mathrm{u}=\sqrt{2 \times 9.8 \times 19.6}=\sqrt{19.6 \times 19.6}$

$=19.6 \mathrm{~m} / \mathrm{s}$ downwards

Inside water, acceleration $=9.8 \mathrm{~m} / \mathrm{s}^{2}$ upwards

Let $t _{1}$ be time taken by sphere to come to rest inside water.

Using $\mathrm{u}+\mathrm{at}=\mathrm{v}$, we get

$19.6-9.81=0$ or $\mathrm{t} _{1}=2 \mathrm{~s}$

Again time of ascent $=$ Time of descent

$$ =2 \mathrm{~s} $$

$\therefore \mathrm{t}=$ The time spend inside water $=2 \mathrm{~s}+2 \mathrm{~s}$

$$ =4 \mathrm{~s} $$

Option (2)

Correct answer: (1)

Solution:

As there are no external forces acting on the system; work done by the external forces is zero.

$$ \therefore \mathrm{W} _{\mathrm{ext}}=0 \mathrm{~J} $$

Total initial K.E. of the system $=\frac{1}{2} \mathrm{mu}^{2}+\frac{1}{2} \mathrm{mu}^{2}$

$$ \begin{aligned} & =\frac{1}{2} \times 20 \times(25)+\frac{1}{2}(20)(25) \\ & =500 \mathrm{~J} \end{aligned} $$

Final K.E. of the system $=0$ (The blocks came to rest)

By work energy thorem; work done by internal forces, $\mathrm{W} _{\text {int }}$ equals change in K.E.

$\therefore \mathrm{W} _{\text {int }}=$ Final K.E. - Initial K.E.

$=0 \mathrm{~J}-(500 \mathrm{~J})=-500 \mathrm{~J}$

The negative sign of work in due to the fact that the internal forces of action and reaction oppose the motion of masses.

Correct answer: (4)

Solution:

The net work done $(=W)$ is sum of gain in gravitational potential energy of mass and the work done against friction. $\mathrm{U}=$ gain in gravitational P.E. $=\mathrm{mgh}$

To calculate work done against friction we can assume that block moves an incline of inclination $\theta$ as shown in Fig. The work done

$$ \begin{aligned} \mathrm{W} _{1}= & \mathrm{f} _{\mu} \times \mathrm{AB}=[\mu \mathrm{mg} \cos \theta] \times \mathrm{AB} \\ & =\mu \mathrm{mg}\left[\frac{\ell}{\sqrt{\ell^{2}+\mathrm{h}^{2}}}\right] \times \sqrt{\ell^{2}+\mathrm{h}^{2}} \\ & =\mu \mathrm{mg} \ell \end{aligned} $$

$\therefore \mathrm{W}=\mathrm{u}+\mathrm{W} _{1}$

$$ =m g(h+\mu \ell) $$

Correct answer: (2)

Solution:

As the hanging block falls; the potential energy ( $\mathrm{Mgh})$ changes to K.E. of the blocks and the potential energy of the spring. The distance ’ $h$ ’ through which the block falls is equal to the extension in the spring.

From law of conservation of energy, we have

$\operatorname{Mgh}=\frac{1}{2}(\mathrm{~m}+\mathrm{M}) \mathrm{v}^{2}+\frac{1}{2} \mathrm{k} x^{2}$

Given $\mathrm{M}=10 \mathrm{~kg} ; \mathrm{h}=1 \mathrm{~m} ; \mathrm{k}=100 \mathrm{Nm}^{-1} ; x=\mathrm{h}=1 \mathrm{~m}$

$\therefore 10 \times 10 \times 1=\frac{1}{2}(5+10) \mathrm{v}^{2}+\frac{1}{2} \times 100 \times(1)^{2}$

$\mathrm{v}^{2}=\frac{100}{15}=\frac{20}{3}$

or $\quad \mathrm{v}=2 \sqrt{\frac{5}{3}} \mathrm{~m} / \mathrm{s}$

Let $\mathrm{H}$ be the maximum distance of fall of block.

In this position both masses are at rest. The $100 \%$ gravitational P.E. lost by M is converted in potential energy of spring.

$\therefore \frac{1}{2} \mathrm{kH}^{2}=\mathrm{MgH}$

or $\mathrm{H}=\frac{2 \mathrm{Mg}}{\mathrm{k}}$

$$ =\frac{2 \times 10 \times 10}{100}=2 \mathrm{~m} $$

Correct answer: (4)

Solution:

We have $\mathrm{F}=5 \mathrm{~N}, \mathrm{~m}=15 \mathrm{~kg}, \mathrm{u}=0$

$$ \mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}=\frac{5}{15}=\frac{1}{3} \mathrm{~m} / \mathrm{s}^{2} $$

$\mathrm{S} _{3}=$ Displacement in third second $=\mathrm{u}+\frac{\mathrm{a}}{2}(2 \mathrm{n}-1)$

$$ 3^{\mathrm{rd}}=0+\frac{1}{6}(5)=\frac{5}{6} \mathrm{~m} $$

Work done in the third second $=$ $F.S _{3}$

$$ =5 \times \frac{5}{6}=\frac{25}{6} \mathrm{~J} $$

$\mathrm{v}=$ Instantaneous velocity at the end of $3^{\text {rd }}$ second $=u+$ at

$$ =0+\frac{1}{3} \times 3=1 \mathrm{~m} / \mathrm{s} $$

$\therefore \mathrm{P} _{\text {inst }}=\mathbf{F . v}=5 \times 1=5 \mathrm{Js}^{-1}$ or $5 \mathrm{~W}$.

Correct answer: (4)

Solution:

As the force exerted on the vehicle is constant; the vehicle will move with a constant acceleration. Let ’ $a$ ’ be the acceleration produced.

Velocity of the vehicle at time ’ $\mathrm{t}$ ’ $=\mathrm{v}=$ at.

We have $\mathrm{u}=0$

$\therefore$ Initial K.E. $=0$

$\mathrm{KE}$ at time $\mathrm{t}=\mathrm{K}=\frac{1}{2} \mathrm{mv}^{2}$

$$ =\frac{1}{2} \mathrm{ma}^{2} \mathrm{t}^{2} $$

From work-energy theorem

Work done $=$ Change in $\mathrm{KE}$ of particle by applied force

$\therefore \mathrm{W}=\frac{1}{2} \mathrm{ma}^{2} \mathrm{t}^{2}$

The power; $\mathrm{P}$, developed by the motor is

$\mathrm{P}=\frac{\mathrm{W}}{\mathrm{t}}=\frac{1}{2} \mathrm{~m} \mathrm{a}^{2} \mathrm{t}$

$$ \begin{aligned} & =\frac{1}{2}(\mathrm{ma})(\mathrm{at}) \\ & =\frac{1}{2} \mathrm{~F} \cdot \mathrm{at} \end{aligned} $$

The $\mathrm{P}$ vs ’ $\mathrm{t}$ ’ graph is a straight line passing through the origin.

Correct answer: (2)

Solution:

At the highest point, A, in the trajectory; only horizontal component of velocity $u$ will be present i.e. $u \cos \theta$. At $P$, the instantaneous velocity v, makes an angle $\frac{\theta}{2}$ with the horizontal.

Horizontal component of $\mathrm{v}=\mathrm{v} \cos \left(\frac{\theta}{2}\right)$

horizontal component of velocity remains constant, we have,

$\mathrm{v} \cos \left(\frac{\theta}{2}\right)=\mathrm{u} \cos \theta$

$\therefore \mathrm{v}=\frac{\mathrm{u} \cos \theta}{\cos \frac{\theta}{2}}$

Work done by force of gravity.

$=$ Change in K.E. from A to B.

$=\frac{1}{2} \mathrm{mv}^{2}-\frac{1}{2} \mathrm{~m}(\mathrm{u} \cos \theta)^{2}$

$=\frac{1}{2} m u^{2} \cos ^{2} \theta\left(\frac{1}{\cos ^{2} \frac{\theta}{2}}-1\right)$

$=\frac{1}{2} m u^{2} \cos ^{2} \theta\left(\sec ^{2} \frac{\theta}{2}-1\right)$

$=\frac{1}{2} m u^{2} \cos ^{2} \theta \tan ^{2} \frac{\theta}{2}$

Option (2)

Correct answer: (4)

Solution:

Mass $\mathrm{m}$ (or body A) will be on brink of moving when the force applied by the spring (i.e. $\mathrm{kx}$ ) is just equal to the force of limiting friction between $A$ and surface in contact. Therefore

$$ \begin{equation*} \mathrm{k} x=\mu \mathrm{mg} \tag{1} \end{equation*} $$

$x$ is the elongation in the spring

$$ \begin{equation*} x=\frac{\mu \mathrm{mg}}{\mathrm{k}} \tag{2} \end{equation*} $$

The force will be minimum for $\mathrm{M}$ when it has no kinetic energy. Applying work energy theorem to M, we get

Work done $=$ Change in kinetic energy

$$ \begin{gathered} \int _{0}^{x}(\mathrm{~F}-\mu \mathrm{Mg}-\mathrm{k} x) \mathrm{d} x=0 \\ \mathrm{~F} x-\mu \mathrm{Mg} x-\frac{1}{2} \mathrm{k} x^{2}=0 \\ \text { or } \quad \mathrm{F}=\mu \mathrm{Mg}+\frac{1}{2} \mathrm{k} x \\ =\mu \mathrm{Mg}+\frac{1}{2} \mathrm{k}\left(\frac{\mu \mathrm{mg}}{\mathrm{k}}\right) \\ =\mu \mathrm{Mg}+\frac{1}{2} \mu \mathrm{mg} \\ =\mu \mathrm{g}\left(\mathrm{M}+\frac{1}{2} \mathrm{~m}\right) \end{gathered} $$

Correct answer: (2)

Solution:

Initial level of water in $A=h _{1}$

Initial level of water in $B=h _{2}$

Final level of water in $A$ and $B=\frac{h _{1}+h _{2}}{2}$

Let $\mathrm{h} _{1}>\mathrm{h} _{2}$;

Fall in water level in $A=\left(h _{1}-\frac{h _{1}+h _{2}}{2}\right)=\frac{h _{1}-h _{2}}{2}$

Rise in water level in $B=\left(\frac{h _{1}+h _{2}}{2}-h _{2}\right)=\frac{h _{1}-h _{2}}{2}$

Mass of water transferred $=\mathrm{m}=\rho \mathrm{A}\left(\frac{\mathrm{h} _{1}-\mathrm{h} _{2}}{2}\right)$

Work done $=$ Change in gravitational P.E. $=\rho A\left(\frac{h _{1}-h _{2}}{2}\right) g\left(\frac{h _{1}-h _{2}}{2}\right)$

$$ =\alpha\left(\mathrm{h} _{1}-\mathrm{h} _{2}\right)^{2} $$

Option (2)

Correct answer: (3)

Solution:

The kinetic energy of the bullet fired changes to partly kinetic and partly potential at the highest point. The body however returns to the same horizontal with K.E. equal to the initial K.E. irrespective of the angle of projection ’ $\alpha$ ‘. So $(\mathrm{KE}+\mathrm{mc} \quad \theta)$ is the total energy available for melting ice. It depends on $\theta$ but not on $\alpha$.

Option (3).

Correct answer: (3)

Solution:

Given potential energy $\mathrm{V}=\frac{x^{4}}{4}-\frac{x^{2}}{2}$ Joule

For maximum or minimum; $\frac{\mathrm{dV}}{\mathrm{d} x}=0$

or $x^{3}-x=0$ or $x\left(x^{2}-1\right)=0 \Rightarrow x=0,-1$ or 1

$\frac{\mathrm{d}^{2} \mathrm{~V}}{\mathrm{~d} x^{2}}=3 x^{2}-1$

At $x=0 ; \frac{\mathrm{d}^{2} \mathrm{~V}}{\mathrm{~d} x^{2}}=-1<0$ which implies maximum $\mathrm{v}(x)$

At $x= \pm 1 ; \frac{\mathrm{d}^{2} \mathrm{~V}}{\mathrm{~d} x^{2}}=2>0$. Hence PE is minimum for $x= \pm 1$

$\mathrm{V} _{\text {min }}=\frac{1}{4}-\frac{1}{2}=\frac{1}{2} \mathrm{~J}$

From law of conservation of energy.

$\mathrm{V} _{\text {min }}+\mathrm{K} _{\text {max }}=$ Total energy

$\left(\frac{1}{4}-\frac{1}{2}\right)+\mathrm{K} _{\max }=2 \mathrm{~J} \Rightarrow \mathrm{K} _{\max }=2+\frac{1}{4}=\frac{9}{4} \mathrm{~J}$

$\frac{1}{2} \mathrm{mv} _{\text {max }}^{2}=\frac{9}{4}$ or $\mathrm{V} _{\text {max }}=\frac{3}{\sqrt{2}} \mathrm{~m} / \mathrm{s} \quad$ (Given $\mathrm{m}=1 \mathrm{~kg}$ )

Correct answer: (4)

Solution:

Let $\mathrm{m}$ be the mass of the body and $\mathrm{K}$ its initial kinetic energy. We have

$$ \begin{equation*} \mathrm{K}=\frac{1}{2} \mathrm{mu}^{2} \tag{1} \end{equation*} $$

The velocity of one part $=-\mathrm{u}$.

Let $x$ be speed of second part along +ve $\mathrm{x}$-axis.

Applying momentum conservation principle.

$\mathrm{mu}=-\frac{\mathrm{m}}{2} \mathrm{u}+\frac{\mathrm{m}}{2} x \Rightarrow x=3 \mathrm{u}$

$\therefore$ K.E.of $\mathrm{B}=\frac{1}{2}\left(\frac{\mathrm{m}}{2}\right) x^{2}$

$$ \begin{aligned} & =\frac{1}{2} \frac{\mathrm{m}}{2}(3 \mathrm{u})^{2}=\frac{1}{2}\left(\frac{9}{2} \mathrm{mu}^{2}\right) \\ & =\frac{9}{2} \mathrm{~K} \end{aligned} $$

Correct answer: (2)

Solution:

Momentum imparted to $\mathrm{B}$ by the cat jumping on it $=\mathrm{mu}$

Momentum imparted to B as the cat jumps out $=\mathrm{mu}$

$\therefore$ Total momentum of $\mathrm{B}=2 \mathrm{mu}$

$\mathrm{KE}$ of $\mathrm{B}=\frac{\mathrm{p} _{\mathrm{B}}^{2}}{2 \mathrm{M}}=\frac{4 \mathrm{~m}^{2} \mathrm{u}^{2}}{2 \mathrm{M}}=\frac{2 \mathrm{~m}^{2} \mathrm{u}^{2}}{\mathrm{M}}$

Similarly,

Momentum imparted to $\mathrm{A}=2 \mathrm{mu}$

Total mass of A and cat $=\mathrm{m}+\mathrm{M}$

$K E$ of $A=\frac{4 m^{2} u^{2}}{2(m+M)}=\frac{2 m^{2} u^{2}}{m+M}$

$\therefore \frac{\mathrm{K} _{\mathrm{A}}}{\mathrm{K} _{\mathrm{B}}}=\frac{\mathrm{M}}{\mathrm{m}+\mathrm{M}}$

Option (2)

Correct answer: (4)

Solution:

We have power $\mathrm{P}=\mathrm{F} . \mathrm{v}=$ mav

$$ \Rightarrow \mathrm{a}=\frac{\mathrm{P}}{\mathrm{mv}} $$

$\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{P}}{\mathrm{mv}}$ or $\frac{\mathrm{vdv}}{\mathrm{vdt}}=\frac{\mathrm{P}}{\mathrm{mv}}$

or $\frac{\mathrm{vdv}}{\mathrm{ds}}=\frac{\mathrm{P}}{\mathrm{mv}}$

$\therefore \mathrm{v}^{2} \mathrm{dv}=\frac{\mathrm{P}}{\mathrm{m}} \mathrm{dS}$

Integrating $\int _{v _{1}}^{v _{2}} v^{2} d v=\frac{P}{m} \int _{0}^{s} d S$

$=\frac{1}{3}\left(\mathrm{v} _{2}^{3}-\mathrm{v} _{1}^{3}\right)=\frac{\mathrm{P}}{\mathrm{m}} \mathrm{S}$

or $\quad S=\frac{m}{3 P}\left(v _{2}^{3}-v _{1}^{3}\right)$

which is option (4).

Correct answer: (2)

Solution:

As work is done on the body by a machine delivering constant power; we have work

$$ \mathrm{W}=\mathrm{P} . \mathrm{t} $$

The work done on the body is stored in it as kinetic energy.

$\therefore \quad \frac{1}{2} \mathrm{mv}^{2}=$ P.t

or $\quad v^{2}=\frac{2 P}{m} t=\alpha t$ $\Rightarrow \mathrm{v} \alpha \mathrm{t}^{1 / 2}$ or $\mathrm{v}=\mathrm{kt}^{1 / 2}$

Where $\mathrm{k}^{2}=\frac{2 \mathrm{P}}{\mathrm{m}}$ and $\mathrm{k}$ is a constant.

Displacement dS in time interval $t$ to $t+d t$ is

$\mathrm{dS}=\mathrm{vdt}=\mathrm{kt}^{\frac{1}{2}} \mathrm{dt}$

$\therefore \mathrm{S}=\int \mathrm{d} \mathrm{S}=\int \mathrm{vdt}=\mathrm{k} \frac{\mathrm{t}^{3 / 2}}{3 / 2}$

or $\quad S \propto t^{3 / 2}$

Hence $\frac{S}{V} \alpha \frac{t^{3 / 2}}{t^{1 / 2}}$ or $\frac{S}{V} \alpha t$

$\therefore$ The graph $\frac{\mathrm{S}}{\mathrm{v}} \quad \mathrm{v} / \mathrm{s} \quad t$ should be a straight line through origin which is option (2).

Correct answer: (2)

Solution:

Let $A, v$ and $\rho$ respectively denote the area of the pipe, the speed of water and the density of water.

$\therefore$ Mass of water flowing out per sec. $=\frac{\mathrm{dm}}{\mathrm{dt}}=\mathrm{Av} \rho$

To get $n$ times water in same time

or $\quad \mathrm{A}^{\prime} \mathrm{v}^{\prime} \rho^{\prime}=\mathrm{n}(\operatorname{av} \rho)$

But $\mathrm{A}^{\prime}=\mathrm{A}$ and $\rho^{\prime}=\rho \quad$ (Same pipe is used)

$\therefore \mathrm{v}^{\prime}=\mathrm{nv}$

The force exerted by motor $=\mathrm{v}\left(\frac{\mathrm{dm}}{\mathrm{dt}}\right)$.

Since power $(\mathrm{P})$ is $\mathbf{F . v}$, we have

$\frac{F^{\prime}}{F}=\frac{v^{\prime}\left(d m^{\prime} / \mathrm{dt}\right)}{v(d m / d t)}=\frac{n v(n d m / d t)}{v(d m / d t)}=n^{2}$

$\frac{\mathrm{P}^{\prime}}{\mathrm{P}}=\frac{\mathrm{F}^{\prime} \mathrm{v}^{\prime}}{\mathrm{Fv}}=\left(\frac{\mathrm{F}^{\prime}}{\mathrm{F}}\right) \cdot\left(\frac{\mathrm{v}^{\prime}}{\mathrm{v}}\right)$

$=\left(\frac{\mathrm{n}^{2} \mathrm{~F}}{\mathrm{~F}}\right)\left(\frac{\mathrm{nv}}{\mathrm{v}}\right)=\mathrm{n}^{3}$

$\therefore \mathrm{P}^{\prime}=\mathrm{n}^{3} \mathrm{P}$

Factor by which power is increased $\mathrm{n}^{3}$.

Option (2)

Correct answer: (2)

Solution:

Case (1): The work done by the man is converted to KE of the stone.

$\therefore \mathrm{W}=\frac{1}{2} \mathrm{mv} _{1}^{2}$

Case (2): In applying the same force as in case (1), the work done gets converted partly into K.E. of the stone and partly into his own K.E.

$\therefore \frac{1}{2} \mathrm{mv} _{1}^{2}=\frac{1}{2} \mathrm{mv} _{2}^{2}+\frac{1}{2} \mathrm{MV}^{2} \hspace{50mm} . . . . . . . . (1)$

$\mathrm{V}=$ speed of man; $\mathrm{v} _{2}=$ speed of stone.

From law of conservation of linear momentum.

$$ \begin{align*} & \mathrm{mU} _{2}=\mathrm{MV} \\ & \mathrm{v} _{2}=\frac{\mathrm{M}}{\mathrm{m}} \mathrm{V} \tag{2} \end{align*} $$

From(1) $m v _{1}^{2}=m v _{2}^{2}+M\left[\frac{m}{M} \cdot v _{2}\right]^{2}$

$v _{1}^{2}=v _{2}^{2}\left[1+\frac{m}{M}\right]=v _{2}^{2}\left[\frac{M+m}{M}\right]$

$\therefore v _{2}=v _{1} \sqrt{\frac{M}{M+m}}=v _{1} \sqrt{\frac{M^{2}}{M(M+m)}}$

and $\quad V=\frac{m}{M} \quad v _{2} =\frac{m}{M} \quad v _{1} \sqrt{\frac{M^{2}}{M(M+m)}}=v _{1} \sqrt{\frac{m^{2}}{M(M+m)}}$

Velocity of the stone relative to the man

$$ \begin{aligned} =v _{r} & =v _{2}+v=\frac{v _{1}}{\sqrt{M(M+m)}}|M+m|=v _{1} \sqrt{\frac{M+m}{M}} \\ & \Rightarrow v _{r}>v _{1} \end{aligned} $$

Also Power $=$ Force $\times$ velocity

$\therefore \mathrm{P} _{2}=\mathrm{F} . \mathrm{v} _{\mathrm{r}} ; \mathrm{P} _{2}>\mathrm{P} _{1}$ and $\mathrm{P} _{1}=\mathrm{F} \cdot \mathrm{v} _{1}$

Hence option (2) is correct.

Correct answer: (1)

Solution:

The free body diagrams of point A and pulley $\mathrm{P} _{2}$ are as shown in Fig. (a) & (b).

We have $2 \mathrm{~T}=\mathrm{Mg}=50 \times 9.8$

$\therefore \mathrm{T}=\frac{490}{2}=245 \mathrm{~N}$

$\therefore \mathrm{F}=\mathrm{T}=245 \mathrm{~N}$

To move block up through $5 \mathrm{~cm}$; work done is $\mathrm{W}=\mathrm{mgh}$

$=50 \times 9.8 \times \frac{5}{100}=24.5 \mathrm{~J}$

For an upward velocity of $1 \mathrm{~m} / \mathrm{s}$ of the block; the string at A must be pulled at $2 \mathrm{~m} / \mathrm{s}$ because each side of the chord at $\mathrm{P} _{2}$ must move up by $1 \mathrm{~m}$ in one see.

$\therefore$ Power at $\mathrm{A}=$ F.v

$$ =245 \times 2=490 \mathrm{~W} $$

Correct answer: (2)

Solution:

By definition of coefficient of restitution, we have

$\mathrm{e}=\frac{\text { Relative velocity after collision }}{\text { Relative velocity before collision }}$

$\therefore \quad \frac{\mathrm{v} _{\mathrm{B}}-\mathrm{v} _{\mathrm{A}}}{\mathrm{u}-0}=\mathrm{e}$ or $\quad \mathrm{v} _{\mathrm{B}}-\mathrm{v} _{\mathrm{A}}=\mathrm{eu} \hspace{40mm}. . . . . . (1)$

$\mathrm{mv} _{\mathrm{A}}+\mathrm{mv} _{\mathrm{B}}=\mathrm{mu} \Rightarrow \mathrm{v} _{\mathrm{A}}+\mathrm{v} _{\mathrm{B}}=\mathrm{u} \hspace{40mm}. . . . . . (2)$

From (1) & (2) we have,

$$ \begin{aligned} \mathrm{v} _{\mathrm{B}} & =(1+\mathrm{e}) \frac{\mathrm{u}}{2} \\ \mathrm{v} _{\mathrm{A}} & =(1-\mathrm{e}) \frac{\mathrm{u}}{2} \\ \therefore \quad \frac{\mathrm{v} _{\mathrm{A}}}{\mathrm{v} _{\mathrm{B}}} & =\frac{1-\mathrm{e}}{1+\mathrm{e}} \end{aligned} $$

Correct answer: (1)

Solution:

$K=$ Initial $K E=\frac{1}{2} m v^{2}=\frac{p^{2}}{2 m}$

Using principle of momentum conservation; the velocity of the composite block is given by

$\mathrm{mv}=(\mathrm{m}+\mathrm{M}) \mathrm{V}$

or $\quad V=\frac{m v}{m+M}$ $K^{\prime}=$ Final K.E. $=\frac{1}{2}(\mathrm{~m}+\mathrm{M}) \mathrm{V}^{2}=\frac{\mathrm{p}^{2}}{2(\mathrm{~m}+\mathrm{M})} \quad[\because$ The momentum remains the same $]$

$\Delta \mathrm{E}=\mathrm{KE}$ lost $=\frac{\mathrm{p}^{2}}{2 \mathrm{~m}}-\frac{\mathrm{p}^{2}}{2(\mathrm{~m}+\mathrm{M})}=\frac{\mathrm{p}^{2}}{2}\left[\frac{\mathrm{M}}{\mathrm{m}(\mathrm{m}+\mathrm{M})}\right]$

$\therefore \frac{\Delta K}{K}=\frac{\frac{p^{2}}{2}\left[\frac{M}{m(m+M)}\right]}{\frac{p^{2}}{2 m}}=\frac{M}{m+M}$

Correct option is (1).

Correct answer: (2)

Solution:

Take velocity u (initial) of A as $+x$ direction.

Let $\mathrm{v} _{1}=\frac{\mathrm{u}}{\sqrt{3}}$ and $\mathrm{v} _{2}$ be velocity of $\mathrm{A}$ and $\mathrm{B}$ after collision as shown in Fig.

Applying principle of momentum conservation; Along $x$-direction, we have

$2 \mathrm{~mv} _{2 x}=\mathrm{mu} \quad\left[\mathrm{v} _{x}\right.$ is the $x$ component of $\mathrm{v} _{\mathrm{B}}$ after collision]

$$ \mathrm{v} _{x}=\frac{\mathrm{u}}{2} $$

Along y-direction $\mathrm{m} \frac{\mathrm{u}}{\sqrt{3}}+2 \mathrm{mv} _{2 \mathrm{y}}=0$

or $\quad v _{2 y}=\left|\frac{u}{2 \sqrt{3}}\right|$

$\therefore$ Velocity of $\mathrm{B}$ after collision is given by

$$ \mathrm{v} _{2}^{2}=\mathrm{v} _{2 \mathrm{y}}^{2}+\mathrm{v} _{2 x}^{2}=\frac{\mathrm{u}^{2}}{12}+\frac{\mathrm{u}^{2}}{4}=\frac{4 \mathrm{u}^{2}}{12}=\frac{\mathrm{u}^{2}}{3} $$

$\therefore \frac{\text { KE of } B}{\text { KE of } A}=\frac{\frac{1}{2}(2 m) \frac{u^{2}}{3}}{\frac{1}{2} m \frac{u^{2}}{3}}=2: 1$

Correct answer: (3)

Solution:

For elastic collision between $\mathrm{A}$ and $\mathrm{B}$;

$\mathrm{m} _{1}=\mathrm{m} ; \mathrm{u} _{1}=\mathrm{u} ; \mathrm{m} _{2}=2 \mathrm{~m} ; \mathrm{u} _{2}=0$

The velocity $\mathrm{v} _{2}$ of mass $\mathrm{B}$ after collisions is

$$ =\mathrm{v} _{2}=\frac{\mathrm{m} _{2}}{\mathrm{~m} _{1}+\mathrm{m} _{2}} \mathrm{u} _{1}+\frac{\mathrm{m} _{2}-\mathrm{m} _{1}}{\mathrm{~m} _{1}+\mathrm{m} _{2}} \mathrm{u} _{2}=\frac{2 \mathrm{~m}}{3 \mathrm{~m}} \mathrm{u}=\frac{2}{3} \mathrm{u} $$

For inelastic collision between B and C. Let V be the speed of combination of B and C after collision. Using principle of conservation of linear momentum.

$(m+2 m) V=(2 m)\left(\frac{2}{3} u\right)$

$\therefore \quad \mathrm{V}=\frac{4}{9} \mathrm{u}$

$\frac{\left(\mathrm{K} _{\mathrm{C}}\right) _{\text {final }}}{\left(\mathrm{K} _{\mathrm{A}}\right) _{\text {initial }}}=\frac{\frac{1}{2} \mathrm{~m}\left(\frac{4}{9} \mathrm{u}\right)^{2}}{\frac{1}{2} \mathrm{mu}^{2}}=16: 81$

Option (3)

Correct answer: (3)

Solution:

As the two particles have equal mass; after every elastic collision, the particles interchange their velocities. As initially; particle 2 has a speed double that of particle 1; the first collision will be at B. Distance traveled by $\mathrm{B}$ is double of that traveled by 1 . At $\mathrm{B}$ they again exchange their speeds.

So the $2^{\text {nd }}$ collision will be at $\mathrm{C}$ and the $3^{\text {rd }}$ collision will be again at ’ $\mathrm{A}$ ‘.

$\therefore \mathrm{n}=3$

So option (3) is correct.

Correct answer: (2)

Solution:

The bodies before and after collision are as shown in Fig. (a) & (b).

Total initial momentum of the system along $\mathrm{X}$-axis before collision $=\mathrm{mu}+0=\mathrm{mu}$

Total final momentum of the system along $\mathrm{X}$-axis after collision $=\mathrm{m} _{1} \mathrm{v} _{1} \cos \theta _{1}+\mathrm{m} _{2} \mathrm{v} _{2} \cos \theta _{2}$

$$ =m v _{1} \cos \theta _{1}+m v _{2} \cos \theta _{2} $$

By principle of momentum conservation.

$\mathrm{mu}=\mathrm{mv} _{1} \cos \theta _{1}+\mathrm{mv} _{2} \cos \theta _{2}$

or $\mathrm{u}=\mathrm{v} _{1} \cos \theta _{1}+\mathrm{v} _{2} \cos \theta _{2} \hspace{40mm}. . . . . . . .(1) $

Applying momentum conservation in y direction. We have

$0=m v _{1} \sin \theta _{1}-m _{2} \sin \theta _{2} \hspace{40mm}. . . . . . . .(2)$

or $\mathrm{v} _{1} \sin \theta _{1}-\mathrm{v} _{2} \sin \theta _{2}=0 $

As the collision is elastic; we have

$\frac{1}{2} m u^{2}=\frac{1}{2} m v _{1}^{2}+\frac{1}{2} m v _{2}^{2}$

or $\mathrm{u}^{2}=\mathrm{v} _{1}^{2}+\mathrm{v} _{2}^{2} \hspace{40mm}. . . . . . . .(3)$

$(1)^{2}+(2)^{2}$ gives

$u^{2}=v _{1}^{2}+v _{2}^{2}+2 v _{1} v _{2}\left(\cos \theta _{1} \cos \theta _{2}-\sin \theta _{1} \sin \theta _{2}\right)$

$=\mathrm{v} _{1}^{2}+\mathrm{v} _{2}^{2}-2 \mathrm{v} _{1} \mathrm{v} _{2} \cos \left(\theta _{1}+\theta _{2}\right) \hspace{40mm}. . . . . . . .(4)$

(4) - (3) gives

$2 \mathrm{v} _{1} \mathrm{v} _{2} \cos \left(\theta _{1}+\theta _{2}\right)=0$

or

$$ \cos \left(\theta _{1}+\theta _{2}\right)=0 \Rightarrow \theta _{1}+\theta _{2}=90^{\circ} $$

Correct answer: (3)

Solution:

The potential energy $V(r)$ of the balls should be $\mathrm{V}(\mathrm{r}) \propto \frac{1}{\mathrm{r}}$ for $\mathrm{r}<2 \mathrm{R}$

and $V(r)=0$ for $r=2 R$

These condition are satisfied only in graph (b)

Correct answer: (4)

Solution:

Let v’ and $\theta$ denote the velocity after reflection and the angle of reflection as shown in Fig.

Before reflection $\mathrm{v} _{x}=\mathrm{v} \cos 45^{\circ}=\frac{\mathrm{v}}{\sqrt{2}}$ and $\mathrm{v} _{\mathrm{y}}=\mathrm{v} \sin 45^{\circ}=\frac{\mathrm{v}}{\sqrt{2}}$

As there is no force along horizontal; $\mathrm{v} _{x}^{\prime}=\mathrm{v} _{x}=\frac{\mathrm{v}}{\sqrt{2}}$

The surface exerts a force along the vertical.

$$ \therefore \frac{\mathrm{v} _{\mathrm{y}}^{\prime}}{\mathrm{v} _{\mathrm{y}}}=\mathrm{e} $$

Velocity after impact $=\mathrm{v}^{\prime}=\sqrt{\mathrm{v} _{x}^{\prime 2}+\mathrm{v} _{\mathrm{y}}^{\prime 2}}=\sqrt{\frac{\mathrm{v}^{2}}{2}+\frac{\mathrm{v}^{2}}{4}}$

$$ \mathrm{v}^{\prime}=\frac{\sqrt{3}}{2} \mathrm{v} $$

Also $\tan \theta=\frac{\mathrm{v} _{x}^{\prime}}{\mathrm{v} _{\mathrm{y}}^{\prime}}=\frac{\frac{\mathrm{v}}{\sqrt{2}}}{\frac{\mathrm{v}}{2}}=\sqrt{2}$

or $\quad \theta=\tan ^{-1}(\sqrt{2})$

Option (4)

Correct answer: (1)

Solution:

Fig. shows the initial position A and final position P of mass.

At $\mathrm{P}$; the centripetal force is only due to component of $\mathrm{mg}$ along PO. i.e. $\mathrm{mg} \cos \theta$.

Let $\mathrm{v}$ be speed of the mass $\mathrm{m}$ at $\mathrm{P}$.

Then $\mathrm{mg} \cos \theta=\frac{\mathrm{mv}^{2}}{\ell}$

or $\mathrm{v}^{2}=\mathrm{g} \ell \cos \theta \hspace{40mm}. . . . . . . .(1)$

By energy conservation principle;

Total initial energy $=$ Total energy at $P$.

$\frac{1}{2} \mathrm{mv} _{0}^{2}=\frac{1}{2} \mathrm{mv}^{2}+\mathrm{mg} \ell(1+\cos \theta) \hspace{40mm}. . . . . . . .(2) $

$\therefore \quad \frac{1}{2} \mathrm{mv} _{0}^{2}=\frac{1}{2} \mathrm{mg} \ell \cos \theta+\mathrm{mg} \ell(1+\cos \theta)$

[Use (1)]

$\mathrm{v} _{0}^{2}=\mathrm{g} \ell[2+3 \cos \theta] \hspace{40mm}. . . . . . . .(3) $

For the projectile to pass through $\mathrm{O}$; $\mathrm{t}$ second after it was at $\mathrm{P}$. We have form horizontal component of motion.

$\ell \sin \theta=(\mathrm{v} \cos \theta) \mathrm{t} \hspace{40mm}. . . . . . . .(4) $

From vertical component of motion

$-\ell \cos \theta=(\mathrm{v} \sin \theta) \mathrm{t}-\frac{1}{2} \mathrm{gt}^{2}\hspace{40mm}. . . . . . . .(5) $

From (4) and (5), we get

$-\ell \cos \theta=(\mathrm{v} \sin \theta)\left[\frac{\ell \sin \theta}{v \cos \theta}\right]-\frac{1}{2} \mathrm{~g}\left[\frac{\ell \sin \theta}{\mathrm{v} \cos \theta}\right]^{2}$

Using $\mathrm{v}^{2}=\mathrm{g} \ell \cos \theta$ we get $\tan \theta=\sqrt{2}$

$\therefore \cos \theta=\frac{1}{\sqrt{3}}$ Substituting value of $\cos \theta$ in eqn(3), we have

$\mathrm{v}=[\mathrm{g} \ell(2+\sqrt{3})]^{1 / 2}$

Correct answer: (1)

Solution:

Velocity of the bullet at the highest point $=u \cos \theta$

$$ =50 \cos \theta $$

Let $\mathrm{v}$ be the velocity acquired by bob when bullet gets embedded in to it. From the law of conservation of linear momentum, we get

$\mathrm{M}(50 \cos \theta)=(\mathrm{M}+3 \mathrm{M}) \mathrm{v}$

$$ \begin{equation*} \Rightarrow \mathrm{v}=\left(\frac{50}{4}\right) \cos \theta \tag{1} \end{equation*} $$

In Fig.; A is position of bob + bullet. The mass becomes $4 \mathrm{M}$ when bullet has got embedded. The bob moves in a vertical circle. $\mathrm{B}$ is position of bob + bullet when string has turned by $120^{\circ}$. Let $\mathrm{v} _{\mathrm{B}}$ of velocity of bob + bullet in position $B$.

From law of conservation of energy;

if $\mathrm{v} _{\mathrm{B}}$ is velocity at $\mathrm{B}$; we have

$$ \begin{equation*} \mathrm{v} _{\mathrm{B}}^{2}-\mathrm{v} _{\mathrm{A}}^{2}=2 \mathrm{~g}\left(\ell+\ell \cos 60^{\circ}\right) \tag{2} \end{equation*} $$

In position $B$ the centripetal force is due to component $4 \mathrm{mg} \cos 60^{\circ}$ of

weight in the radial direction. Therefore

$\frac{4 \mathrm{Mv} _{\mathrm{B}}^{2}}{\ell}=4 \mathrm{Mg} \cos 60^{\circ}$

$\frac{10}{3} \mathrm{M}, \mathrm{v} _{\mathrm{A}}=\frac{60}{4} \cos \theta$

$\therefore \quad \mathrm{v} _{\mathrm{B}}^{2}=\mathrm{g} \ell \cos 60=10 \times \frac{10}{3} \times \frac{1}{2}=\frac{50}{3} \mathrm{~m} / \mathrm{s} \hspace{40mm}. . . . . . . .(3)$

From eqns (2) and (3) we have

$\mathrm{v} _{\mathrm{A}}^{2}=\mathrm{v} _{\mathrm{B}}^{2}-2 \mathrm{~g}\left(\ell+\ell \cos 60^{\circ}\right)$

$$ \begin{align*} & =\frac{50}{3}+2 \times 10\left(\frac{10}{3}+\frac{10}{3} \times \frac{1}{2}\right) \quad\left[\mathrm{a}=-10 \mathrm{~m} / \mathrm{s}^{2}\right] \\ & =\frac{50}{3}+20 \times \frac{3}{2} \times \frac{10}{3} \\ & =\frac{50}{3}+100=\frac{350}{3} \tag{4} \end{align*} $$

Equating (1) and (4) we have,

$$ \begin{aligned} & \left(\frac{50}{4} \cos \theta\right)^{2}=\frac{350}{3} \\ & \cos ^{2} \theta=\frac{350}{3} \times \frac{16}{2500}=\frac{5600}{7500} \\ & \cos ^{2} \theta=\frac{56}{75} \end{aligned} $$

$$ \begin{aligned} \cos \theta & =\sqrt{\frac{56}{75}}=\sqrt{.7466} \\ & \simeq 0.86 \\ \therefore \theta & =30^{\circ} \end{aligned} $$

Correct answer: (1)

Solution:

Height above $\mathrm{XY}$ at $\mathrm{P}=5 \mathrm{R}$

Height above $\mathrm{XY}$ at $\mathrm{Q}=\mathrm{R}$

$\therefore$ Height of $\mathrm{P}$ above $\mathrm{Q}=4 \mathrm{R}$

Let $\mathrm{v}$ be the velocity of the block at $\mathrm{Q}$, we have gain in K.E. = Loss in gravitational P.E.

$\frac{1}{2} \mathrm{mv}^{2}=\mathrm{mg}(4 \mathrm{R}) \quad[\mathrm{KE}$ gained $=\mathrm{PE}$ lost $]$

$\therefore \mathrm{v}^{2}=8 \mathrm{gR} \hspace{40mm}. . . . . . . .(1)$

For circular motion at $\mathrm{Q}$, we have

$\frac{\mathrm{mv}^{2}}{\mathrm{R}}=\mathrm{N} \Rightarrow \mathrm{N}=\frac{\mathrm{m}(8 \mathrm{gR})}{\mathrm{R}}=8 \mathrm{mg}$

$\therefore$ For exerted by the track on the block $=8 \mathrm{mg}$

Correct answer: (1)

Solution:

Let ‘h’ denote the vertical distance covered by the ball from the highest point to position $\mathrm{P}$ where $\angle \mathrm{HOP}=\theta$

We have $\mathrm{h}=\left(\mathrm{R}+\frac{\mathrm{d}}{2}\right)(1-\cos \theta)$

The velocity v of the ball at $P$ in given by

$$ \begin{equation*} \mathrm{v}^{2}=2 \mathrm{gh}=2 \mathrm{~g}\left(\mathrm{R}+\frac{\mathrm{d}}{2}\right)(1-\cos \theta) \tag{1} \end{equation*} $$

Let $\mathrm{N} _{\mathrm{A}}$ denote the normal reaction of A on the ball away from the centre at point $\mathrm{P}$. We have,

$$ \begin{align*} & m g \cos \theta-N _{A}=\frac{m^{2}}{\left(R+\frac{d}{2}\right)} \\ & \quad=\frac{m}{\left(R+\frac{d}{2}\right)}\left[2 g\left(R+\frac{d}{2}\right)(1-\cos \theta)\right] \tag{1} \\ & \quad=2 m g(1-\cos \theta) \\ & \therefore \quad N _{A}=-2 m g+2 m g \cos \theta+m g \cos \theta \\ & \quad=m g(3 \cos \theta-2) \end{align*} $$

The ball will lose contact with the sphere $A$ when $\mathrm{N} _{\mathrm{A}}=0$

or $ \quad 3 \cos \theta-2=0$

or $ \quad \cos \theta=\frac{2}{3}$

$\Rightarrow \theta=\cos ^{-1}\left(\frac{2}{3}\right)$

Option (1)

Correct answer: (3)

Solution:

Let the sphere move towards left with a uniform acceleration ’ $a$ ‘.

In Fig.; A is initial position of mass and $B$ its position when it has slided down by an angle $\theta$ on the sphere. The forces acting on $m$ are as shown in Fig. Note $F _{\text {fic }}=m a$ is the fictitious force on mass $m$. $R$ is the force of normal reaction between $m$ and the sphere.

Let ’ $\mathrm{v}$ ’ denote tangential velocity of mass $m$.

We have tangential acceleration.

$$ \mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}} $$

Equation of motion is

$\mathrm{m} \frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{ma} \cos \theta+\mathrm{mg} \sin \theta$

Multiplying by v, we get

$\mathrm{mv} \frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{ma}\left(\mathrm{R} \frac{\mathrm{d} \theta}{\mathrm{dt}}\right) \cos \theta+\mathrm{mg}\left(\mathrm{R} \frac{\mathrm{d} \theta}{\mathrm{dt}}\right) \sin \theta \quad\left[\mathrm{Use} \quad \mathrm{v}=\mathrm{R} \frac{\mathrm{d} \theta}{\mathrm{dt}}\right]$

Divide the equation by $\frac{\mathrm{m}}{\mathrm{dt}}$, we get

$\mathrm{vdv}=\mathrm{aR} \cos \theta \mathrm{d} \theta+\mathrm{gR} \sin \theta \mathrm{d} \theta$

Integrating both sides

$$ \begin{equation*} \frac{\mathrm{v}^{2}}{2}=\mathrm{aR} \sin \theta-\mathrm{gR} \cos \theta+\mathrm{K} \tag{1} \end{equation*} $$

Where $\mathrm{K}$ is constant of integration

At $\theta=0 ; \mathrm{v}=0$

$\therefore \mathrm{K}=\mathrm{gR}$ $\therefore \frac{\mathrm{v}^{2}}{2}=\mathrm{aR} \sin \theta-\mathrm{gR} \cos \theta+\mathrm{gR}$

[From(1)]

or $\quad \mathrm{v}^{2}=2 \mathrm{aR} \sin \theta-2 \mathrm{gR} \cos \theta+2 \mathrm{gR}$

$=2 \mathrm{R}(\mathrm{a} \sin \theta-\mathrm{g} \cos \theta+\mathrm{g})$

or

$$ \mathrm{KE}=\frac{1}{2} \mathrm{mv}^{2}=\mathrm{mR}(\mathrm{a} \sin \theta-\mathrm{g} \cos \theta+\mathrm{g}) $$

Correct answer: (3)

Solution:

Let $\mathrm{v}$ be the velocity of the ball at $\mathrm{E}$; the upper most point of the loop.

Then $\mathrm{KE}$ at $\mathrm{E}=\mathrm{PE}$ lost over a height $\mathrm{h}$

or $\quad \frac{1}{2} \mathrm{mv}^{2}=\mathrm{mgh} \quad$ or $\mathrm{mv}^{2}=2 \mathrm{mgh} \hspace{40mm}. . . . . . . .(1) $

The forces acting on $\mathrm{m}$; at $\mathrm{E}$ are as shown in Fig.2. $\mathrm{R}$ is the reaction force which $\mathrm{m}$ exerts on loop. For dynamic equilibrium.

$\mathrm{mg}+\mathrm{R}=\frac{\mathrm{mv}^{2}}{\mathrm{r}} \quad[\mathrm{R}=$ Normal reaction $]$

or $\mathrm{mg}+\mathrm{R}=\frac{2 \mathrm{mgh}}{\mathrm{r}} \quad$ [Using (1)]

For ’ $h$ ’ to be minimum; the normal reaction $\mathrm{R}$ at top of the track should be zero. Therefore

$$ \frac{2 \mathrm{mgh} _{\min }}{\mathrm{r}}=\mathrm{mg} $$

or $\mathrm{h} _{\text {min }}=\frac{\mathrm{r}}{2}$

Option(3)

Correct answer: (4)

Solution:

We have $\overrightarrow{\mathrm{F}}=2 x \hat{\mathrm{j}}$ newton

Total work done in moving the particle along the closed path is

$$ \begin{aligned} \mathrm{W}= & \int _{\mathrm{ABCDA}} \mathbf{F} \cdot \mathrm{d} \boldsymbol{x}=\int _{0}^{2}(2 x \mathbf{j}) \cdot(\mathrm{d} x \mathbf{i})+\int _{0}^{2}(2 x \mathbf{j}) \cdot \mathrm{d} x \mathbf{j}+\int _{0}^{2}(2 x \mathbf{j}) \cdot \mathrm{d} x(-\mathbf{i})+\int _{0}^{2}(2 x \mathbf{j}) \cdot \mathrm{d} x(-\mathbf{j}) \\ & =0+2\left|\frac{x^{2}}{2}\right| _{0}^{2}+0-2\left|\frac{x^{2}}{2}\right| _{2}^{0} \\ & =0+2\left(\frac{4}{2}\right)-2\left(-\frac{4}{2}\right) \\ & =4+4=8 \mathrm{~J} \end{aligned} $$

As the work done by the force along a closed path is non-zero; the force in non-conservative.

Option (4)

Correct answer: (1)

Solution:

Let $x$ be the instantaneous elongation in length of spring (equal to vertical distance moved down by mass $\mathrm{m}$ ) and $\mathrm{v}$ instantaneous speed acquired by $\mathrm{m}$. The loss in gravitational potential energy is converted into

(i) Kinetic energy of mass $m$ and

(ii) Energy stored in the spring

Therefore,

$\operatorname{mg} x=\frac{1}{2} \mathrm{k} x^{2}+\frac{1}{2} \mathrm{mv}^{2}$

or $\mathrm{mv}^{2}=2 \mathrm{mg} x-\mathrm{k} x^{2}$

Differentiating w.r.t. $x$, we get

$\mathrm{m} \cdot 2 \mathrm{v} \frac{\mathrm{dv}}{\mathrm{d} x}=2 \mathrm{mg}-2 \mathrm{k} x$

$\mathrm{v}$ is maximum when $\frac{\mathrm{dv}}{\mathrm{d} x}=0$

or $\mathrm{mg}=\mathrm{k} x$ or $x=\frac{\mathrm{mg}}{\mathrm{k}} \hspace{40mm}. . . . . . . .(2) $

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

$\mathrm{mg}\left(\frac{\mathrm{mg}}{\mathrm{k}}\right)=\frac{1}{2} \mathrm{k}\left(\frac{\mathrm{mg}}{\mathrm{k}}\right)^{2}+\frac{1}{2} \mathrm{mv} _{\max }^{2}$

$\Rightarrow \frac{1}{2} \frac{\mathrm{m}^{2} \mathrm{~g}^{2}}{\mathrm{k}}=\frac{1}{2} \mathrm{mv} _{\max }^{2}$

or $\quad \mathrm{v} _{\text {max }}=\sqrt{\frac{\mathrm{m}}{\mathrm{k}}} \mathrm{g}$

Correct answer: (1)

Solution:

Let the mass be released a distance $\ell$ above $\mathrm{B}$ on the incline. $\mathrm{C}$ is position of mass when its is again momentarily at rest.

Then total distance covered by the block $=(\ell+2) \mathrm{m}$

Vertical distance covered by the block $=(\ell+2) \sin 30^{\circ}$

Work done by gravity on the block $=\mathrm{mg}(\ell+2) \sin 30$

or $\quad \mathrm{W}=\frac{1}{2} \mathrm{mg}(\ell+2)$

The spring constant of the spring $=\mathrm{k}=\frac{\mathrm{F}}{x}$

$$ =\frac{20 \mathrm{~N}}{\frac{20}{100} \mathrm{~m}}=100 \mathrm{Nm}^{-1} $$

$\mathrm{U}=$ The potential energy stored in the spring $=\frac{1}{2} \mathrm{k}(2)^{2}$

$$ =\frac{1}{2}(100)(2)^{2}=200 \mathrm{~J} $$

From law of conservation of energy; $\mathrm{U}=\mathrm{W}$

$\therefore \quad \frac{1}{2} \mathrm{mg}(\ell+2)=200 \mathrm{~J}$

$$ \begin{aligned} & (\ell+2)=\frac{200 \times 2}{\mathrm{mg}} \\ & =\frac{200 \times 2}{10 \times 10}=4 \mathrm{~m} \end{aligned} $$

Correct answer: (1)

Solution:

When the block is released, the spring pushes the block towards right. The potential energy of the spring is converted to kinetic energy till it loses contact with the spring.

Initial $\mathrm{PE}$ in the spring $=\frac{1}{2} \mathrm{ka}^{2}$

At a distance $x$ from the wall; compression is reduced to $(2 \mathrm{a}-x)$ for values of $x$ less than the natural length $2 a$ of the spring. Let ’ $v$ ’ be the speed of the block at this instant.

By conservation of energy, we get

$(\mathrm{PE}) _{\text {initial }}=(\mathrm{PE}) _{\text {final }}+\mathrm{KE}$

or $\quad \frac{1}{2} \mathrm{ka}^{2}=\frac{1}{2} \mathrm{k}(2 \mathrm{a}-x)^{2}+\frac{1}{2} \mathrm{mv}^{2}$

$m v^{2}=k\left[a^{2}-(2 a-x)^{2}\right]$

$\mathrm{v}=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}\left[\mathrm{a}^{2}-(2 \mathrm{a}-x)^{2}\right]}$

Correct answer: (3)

Solution:

We have total energy $=\mathrm{E}$

Also Total energy $\mathrm{E}=\mathrm{KE}+\mathrm{PE}$

When $\mathrm{PE}=\mathrm{E}$

$\mathrm{KE}=$ zero

and have $\mathrm{v}=0$

$\therefore$ Option (3) is correct.

Correct answer: (4)

Solution:

The initial K.E. of the block in used to store elastic potential energy in the block and to perform work against friction.

$$ \begin{aligned} & \therefore \quad \frac{1}{2} \mathrm{Mu}^{2}=\frac{1}{2} \mathrm{~kb}^{2}+\mu(\mathrm{Mg}) \mathrm{b} \\ & \quad \mu \mathrm{Mgb}=\frac{1}{2} \mathrm{Mu}^{2}-\frac{1}{2} \mathrm{~kb}^{2} \\ & \text { or } \quad \mu=\frac{\mathrm{Mu}^{2}-\mathrm{kb}{ }^{2}}{2 \mathrm{Mgb}} \end{aligned} $$

Correct answer: (2)

Solution:

When mass M is moved a distance ’ $a$ ’ towards wall 1 ; spring $S _{1}$ gets compressed by ’ $a$ ‘. However spring $S _{2}$ has no change in its length because mass $M _{2}$ is free to move. The energy $U _{1}$ of system is due to compression of spring $S _{1}$ only.

$$ \begin{equation*} \therefore \quad \mathrm{U} _{1}=\frac{1}{2} \mathrm{k} _{1} \mathrm{a}^{2} \tag{1} \end{equation*} $$

On being released, block $\mathrm{B}$ moves towards equilibrium position, overshoots the position and compresses

$S _{2}$ by $\frac{a}{2}$ before coming to rest. Now mass $M _{1}$ detaches itself from wall and there is no change in length of $S _{1}$. The energy of system, $U _{2}$, is now due to compression of spring $S _{2}$ only.

Therefore $\mathrm{U} _{2}=\frac{1}{2} \mathrm{k} _{2}\left(\frac{\mathrm{a}}{2}\right)^{2} \hspace{40mm}. . . . . . . . . (2)$

From law of conservation of energy $U _{1}=U _{2}$ i.e.

$$ \begin{aligned} & \frac{1}{2} \mathrm{k} _{1} \mathrm{a}^{2}=\frac{1}{2} \mathrm{k} _{2}\left(\frac{\mathrm{a}}{2}\right)^{2} \\ \therefore \quad & \frac{\mathrm{k} _{2}}{\mathrm{k} _{1}}=4 \end{aligned} $$

Correct answer: (2)

Solution:

For mass M suspended from the free lower end of the spring; if ’ $h$ ’ denotes the maximum extension; then the P.E. lost by the mass is stored as energy in the spring.

$\therefore \frac{1}{2} \mathrm{kh}^{2}=\mathrm{mgh}$

or $\mathrm{h}=\frac{2 \mathrm{mg}}{\mathrm{k}}$

$=\frac{2 \times 0.1 \times 10}{10}=\frac{2}{100} \mathrm{~m}$.

Correct answer: (3)

Solution:

The bob is let go from rest in position A.

For vertical motion of the bob, let $\mathrm{v}$ be speed of bob in position $\mathrm{B}$ and $\ell+\Delta \ell$, the instantaneous length of string.

$\mathrm{v}^{2}=2 \mathrm{~g}(\ell+\Delta \ell)$

or $\quad \mathrm{v}=\sqrt{2 \mathrm{~g}(\ell+\Delta \ell)}$

Stretching force at $\mathrm{B}=\mathrm{mg}+\mathrm{F} _{\text {centrifugal }}$

$$ \begin{aligned} & =\mathrm{mg}+\frac{\mathrm{mv}^{2}}{(\ell+\Delta \ell)}=\mathrm{mg}+\frac{\mathrm{m} 2 \mathrm{~g}(\ell+\Delta \ell)}{(\ell+\Delta \ell)} \\ & =3 \mathrm{mg} \end{aligned} $$

$\therefore \mathrm{k} \Delta \ell=3 \mathrm{mg}$

and $\quad \Delta \ell=\frac{3 \mathrm{mg}}{\mathrm{k}}$

By energy conservation $\mathrm{PE}$ lost $=\mathrm{KE}$ of the bob $+\mathrm{PE}$ in string

$$ \begin{array}{rl} \mathrm{mg}(\ell & +\Delta \ell)=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{k}(\Delta \ell)^{2} \\ \mathrm{mv}^{2} & =2 \mathrm{mg}(\ell+\Delta \ell)-\mathrm{k}(\Delta \ell)^{2} \\ & =2 \mathrm{mg}\left(\ell+\frac{3 \mathrm{mg}}{\mathrm{k}}\right)-\mathrm{k}(\Delta \ell)^{2} \\ & =2 \mathrm{mg}\left(\ell+\frac{3 \mathrm{mg}}{\mathrm{k}}\right)-\frac{(3 \mathrm{mg})^{2}}{\mathrm{k}} \\ \quad & v^2 =2 \mathrm{~g} \ell+\frac{6 \mathrm{mg}^{2}}{\mathrm{k}}-\frac{9 \mathrm{mg}^{2}}{\mathrm{k}} \\ = & 2 \mathrm{~g} \ell-\frac{3 \mathrm{mg}^{2}}{\mathrm{k}} \end{array} $$

$=2 \mathrm{~g}\left(\ell-\frac{3 \mathrm{mg}}{2 \mathrm{k}}\right)$

$\mathrm{v}=\sqrt{2 \mathrm{~g}\left(\ell-\frac{3 \mathrm{mg}}{2 \mathrm{k}}\right)}$

Option(3)

Correct answer: (4)

Solution:

The linear mass density of the chain is $\lambda=\frac{\mathrm{M}}{\mathrm{L}}$

The chain is pulled without acceleration. The length hanging will continuously change. Let y be the length hanging at an instant of time.

The force required to pull the chain up the table $=\mathrm{F}=($ Mass $) \mathrm{g}=(\lambda \mathrm{y}) \mathrm{g}=\lambda \mathrm{gy}$

Work done to pull a small length dy up the table $=\mathrm{dW}=\lambda \mathrm{gy}(-\mathrm{dy})$

$\therefore$ Total work done to pull the entire

$$ \begin{aligned} \text { chain } & =\mathrm{W}=\int _{\mathrm{L} / \mathrm{n}}^{0} \lambda \mathrm{gy}(-\mathrm{dy})=-\lambda \mathrm{g}\left|\frac{\mathrm{y}^{2}}{2}\right| _{\mathrm{L} / \mathrm{n}}^{0} \\ & =-\lambda \mathrm{g}\left[0-\frac{\mathrm{L}^{2}}{2 \mathrm{n}^{2}}\right] \\ & =\frac{\lambda \mathrm{gL}^{2}}{2 \mathrm{n}^{2}}=\left(\frac{\mathrm{M}}{\mathrm{L}}\right) \frac{\mathrm{gL}^{2}}{2 \mathrm{n}^{2}}=\frac{1}{2} \frac{\mathrm{MgL}}{\mathrm{n}^{2}} \end{aligned} $$

$\therefore \mathrm{W} \alpha \frac{1}{\mathrm{n}^{2}}$

Option (4)

Correct answer: (1)

Solution:

In Fig. shown; A and B are the two masses. The masses will have same angular velocity but different linear velocity.

$\therefore \quad \frac{\mathrm{v}}{\ell}=\frac{\mathrm{v}^{\prime}}{\ell^{\prime} / 2} \Rightarrow \mathrm{v}^{\prime}=\frac{\mathrm{v}}{2}$

Using energy conservation principle; Gain in gravitational P.E. $=$ Loss in Kinetic energy $\mathrm{mg} \ell+\mathrm{mg} \frac{\ell}{2}=\frac{1}{2} \mathrm{mv}^{2}+\frac{1}{2} \mathrm{~m}\left(\frac{\mathrm{v}}{2}\right)^{2}$

or $\quad \frac{3}{2} \mathrm{~g} \ell=\frac{\mathrm{v}^{2}}{2}+\frac{\mathrm{v}^{2}}{8}=\frac{5}{8} \mathrm{v}^{2}$

or $\quad \mathrm{v}=\frac{3}{2} \times \frac{8}{5} \mathrm{~g} \ell=\sqrt{\frac{12}{5} \mathrm{~g} \ell}$

Option (1)

Correct answer: (4)

Solution:

We have $\mathrm{F}=-\frac{\mathrm{dU}}{\mathrm{d} x}$

$\therefore \mathrm{dU}=-\mathrm{F} \mathrm{d} x$ or $\mathrm{U}(x)=-\int \mathrm{F} \mathrm{d} x$

or $\quad \mathrm{U}(x)=-\int _{0}^{x}\left(-\mathrm{k} x+\mathrm{a} x^{3}\right) \mathrm{d} x$

$$ =\frac{k x^{2}}{2}-\frac{a x^{4}}{4} $$

$\mathrm{U}(x)=0$ for $x=0$ and $x=\sqrt{\frac{2 \mathrm{k}}{\mathrm{a}}}$

$\mathrm{U}(x)$ is negative for $x>\sqrt{\frac{2 \mathrm{k}}{\mathrm{a}}}$

As $\mathrm{F}(x)=0$ at $x=0$; the slope of potential energy should be zero at $x=0$.

$\therefore$ The must appropriate option is (4).

Correct answer: (1)

Solution:

For Fig. (a); the $\mathrm{KE}$ at $\mathrm{C}$ is the difference of PE lost and the work against friction.

$\therefore \quad \frac{1}{2} \mathrm{mv} _{1}^{2}=\operatorname{mgh}-\mu \mathrm{mg} x \hspace{40mm}. . . . . . . .(1)$

$$ =\operatorname{mg}(\mathrm{h}-\mu x) $$

Similarly at F;

$$ \begin{aligned} \frac{1}{2} \mathrm{mv} _{2}^{2} & =\mathrm{mgh}-\frac{\mu}{2} \mathrm{mg} \frac{x}{2}-\frac{\mu}{4} \mathrm{mg} \frac{x}{2} \\ & =\operatorname{mgh}-\frac{\mu \mathrm{mg} x}{4}\left(1+\frac{1}{2}\right) \end{aligned} $$

$$ \begin{equation*} =\operatorname{mg}\left[\mathrm{h}-\frac{3 \mu x}{8}\right] \tag{2} \end{equation*} $$

From eqns (1) and (2) we have

$\frac{\mathrm{v} _{2}^{2}}{\mathrm{v} _{1}^{2}}=\frac{8 \mathrm{~h}-3 \mu x}{8(\mathrm{~h}-\mu x)}$

$\frac{\mathrm{v} _{2}}{\mathrm{v} _{1}}=\left[\frac{8 \mathrm{~h}-3 \mu x}{8(\mathrm{~h}-\mu x)}\right]^{1 / 2}$

Correct answer: (4)

Solution:

For a conservative force $\mathrm{F}=-\frac{\mathrm{dU}}{\mathrm{d} x}$

$\mathrm{U}(x)-\mathrm{U}(\mathrm{O})=-\int \mathrm{Fd} x=-\mathrm{k} \int _{0}^{x} x \mathrm{~d} x=-\mathrm{k} \frac{x^{2}}{2}$

Given $\mathrm{U}(\mathrm{O})=\frac{1}{2} \mathrm{ka}^{2}$

Therefore, $\mathrm{U}(x)=\frac{1}{2} \mathrm{ka}^{2}-\frac{1}{2} \mathrm{k} x^{2}=\frac{1}{2} \mathrm{k}\left(\mathrm{a}^{2}-x^{2}\right)$

The graph of $\mathrm{U}(x)$ vs $x$ is a parabola. However at $x=0, \mathrm{U} \neq 0$

$\mathrm{U}(x)=0$ at $x= \pm \mathrm{a}$, For $|x|>\mathrm{a} ; \mathrm{U}(x)$ is a-ve number.

The option is best represented in (4).

Correct answer: (3)

Solution:

As the force is variable; the work done is calculated by integration.

Initially $x=0$

[Unstretched band]

and finally $x=\mathrm{L}$

[Given]

$\therefore$ Work done $\mathrm{W}=|\mathrm{dW}|=\int \mathrm{Fd} x$

$$ \begin{aligned} & =\int _{0}^{\mathrm{L}}\left(\mathrm{a} x+\mathrm{b} x^{2}\right) \mathrm{d} x \\ & =\left|\frac{\mathrm{a}{x^2 }^{2}}{2}\right| _{0}^{\mathrm{L}}+\left|\frac{\mathrm{b} x^{2}}{3}\right| _{0}^{\mathrm{L}} \\ & =\frac{\mathrm{aL^2}}{2}+\frac{\mathrm{bL^3}}{3} \end{aligned} $$

Correct answer: (3)

Solution:

We have $\mathrm{U}=\frac{\mathrm{M}}{\mathrm{r}^{6}}-\frac{\mathrm{N}}{\mathrm{r}^{12}}$

$\therefore \quad \mathrm{F}=-\frac{\mathrm{dU}}{\mathrm{dr}}=-\frac{\mathrm{d}}{\mathrm{dr}}\left(\frac{\mathrm{M}}{\mathrm{r}^{6}}-\frac{\mathrm{N}}{\mathrm{r}^{12}}\right)$

$$ =-\frac{6 \mathrm{M}}{\mathrm{r}^{7}}+\frac{12 \mathrm{~N}}{\mathrm{r}^{13}} $$

In equilibrium; $\mathrm{F}=0$

$\therefore \quad \frac{12 \mathrm{~N}}{\mathrm{r}^{13}}=\frac{6 \mathrm{M}}{\mathrm{r}^{7}} \quad$ or $\quad \mathrm{r}^{6}=\frac{2 \mathrm{~N}}{\mathrm{M}}$

$\therefore \quad U(r)=\frac{M}{2 N / M}-\frac{N}{4 N^{2} / M^{2}}=\frac{M^{2}}{2 N}-\frac{M^{2}}{4 N}=\frac{M^{2}}{4 N}$

Correct answer: (3)

Solution:

The force applied is variable. The work done can be calculated as area under F vs $\mathrm{S}$ graph as shown in Fig. The variation from $10 \mathrm{~m}$ to $20 \mathrm{~m}$ is linear decrease from $100 \mathrm{~N}$ to $50 \mathrm{~N}$ represented by BC.

Work done $=$ Area under F vs S graph

$$ \begin{aligned} & =\operatorname{Area} \mathrm{OABCD} \\ & =\operatorname{Ar}(\mathrm{OABE})+\operatorname{Ar} \operatorname{Trap}(\mathrm{BCDE}) \\ & =100 \times 10+\frac{1}{2}(100+50) \times 10 \\ & =1000+750 \\ & =1750 \mathrm{~J} \end{aligned} $$

Option(3)

Correct answer: (3)

Solution:

We have $\mathbf{F}=\mathrm{q}(\mathbf{v} \times \mathbf{B})$

$$ \Rightarrow \mathbf{F} \perp \mathbf{v} $$

Work done by the force $=\mathrm{W}=\mathbf{F} . \mathbf{s}$ $=\mathbf{F} \cdot \mathbf{v} \mathrm{dt}$

$=\mathrm{Fv} d \mathrm{t} \cos 90 \degree =0$

$\therefore \mathrm{W} _{1}=\mathrm{W} _{2}=0$

Option (3)

Correct answer: (1)

Solution:

We have $\mathrm{F}=-\mathrm{k}(\mathrm{yi}+x \mathbf{j})$

The particle undergoes two displacements

(i) $\mathbf{S} _{1} ;(0,0)$ to $(a, 0)$

(ii) $\mathbf{S} _{2} ;(\mathrm{a}, 0)$ to $(\mathrm{a}, \mathrm{a})$

For (i) $\mathbf{S} _{1}=(\mathrm{a} \mathbf{i}+\mathrm{o} \mathbf{j})-(\mathbf{o} \mathbf{i}+\mathrm{j})=\mathrm{a} \mathbf{i}$

$$ \begin{aligned} \therefore \mathrm{W} _{1} & =\mathbf{F} _{1} \cdot \mathrm{S} _{1}=-\mathrm{k}(\mathrm{ai}+x \mathbf{j}) \cdot(\mathrm{ai}) \quad[\mathrm{y}=0 \text { along } x-\mathrm{axis}] \\ & =-\mathrm{ka}^{2} \end{aligned} $$

(ii)

$\mathbf{S} _{2}=(\mathrm{a} \mathbf{i}+\mathrm{a} \mathbf{j})-(\mathrm{a} \mathbf{i}+\mathrm{o} \mathbf{j})=\mathrm{a} \mathbf{j}$

$$ \begin{aligned} & \mathbf{F} _{2}=-\mathrm{k}(\mathrm{yi}+\mathrm{a} \mathbf{j}) \quad[\because \mathrm{n}=\text { a parallel to y direction }] \\ & \begin{aligned} \therefore \mathrm{W} _{2} & =\mathbf{F} _{2} \cdot \mathbf{S} _{2} \\ & =-\mathrm{k}[\mathrm{yi}+\mathrm{a} \mathbf{j}] \cdot(\mathrm{a} \mathbf{j}) \\ & =-\mathrm{ka}^{2} \end{aligned} \end{aligned} $$

$\therefore$ Total work $=\mathrm{W} _{1}+\mathrm{W} _{2}=-2 \mathrm{ka}^{2}$

Option (1)

Correct answer: (4)

Solution:

In the figure;

$$ \mathbf{r}=x \mathbf{i}+y \mathbf{j} $$

$$ \begin{gathered} \mathbf{F}=\frac{\mathrm{k}}{\left(x^{2}+\mathrm{y}^{2}\right)^{3 / 2}}[x \mathbf{i}+\mathrm{y} \mathbf{j}] \\ =\frac{\mathrm{k}}{\left(x^{2}+\mathrm{y}^{2}\right)^{3 / 2}} \mathbf{r} \end{gathered} $$

$\therefore \mathbf{F}$ is along $\mathbf{r}$ or the force is radial and the particle is moved along the circular path.

$\therefore \mathbf{F} \perp \mathbf{S}$

$\therefore$ Work done $=$ zero

Option (4)

Correct answer: (4)

Solution:

We observe that $\mathrm{F} _{x}=\mathrm{k} \times 576$ for $x=1$

$\mathrm{F} _{x}=\mathrm{k} \times 144\left(=\mathrm{k} \cdot\left(\frac{576}{4}\right)=\mathrm{k} \cdot\left(\frac{576}{2^{2}}\right)\right)$ for $x=2$

$\mathrm{F} _{x}=\mathrm{k} \times 64\left(=\mathrm{k} \cdot\left(\frac{576}{9}\right)=\mathrm{k} \cdot\left(\frac{576}{3^{2}}\right)\right)$ for $x=3$

$\mathrm{F} _{x}=\mathrm{k} \times 36\left(=\mathrm{k} \cdot\left(\frac{576}{10}\right)=\mathrm{k} \cdot\left(\frac{576}{4^{2}}\right)\right)$ for $x=4$

It follows that $\mathrm{F} _{x}=\frac{576 \mathrm{k}}{x^{2}}$

$\therefore$ Required work done =$\int_{x_{1}}^{x _{2}} F _{x} d x$

$$ \begin{aligned} & =(576 \mathrm{k})\left|-\frac{1}{x}\right| \\ & =(576)\left[\frac{1}{x _{1}}-\frac{1}{x _{2}}\right]=576 \mathrm{k}\left[\frac{1}{7}-\frac{1}{17}\right] \\ & =\left(\frac{576 \mathrm{k}}{119}\right) \times 10=\left[\frac{5760 \mathrm{k}}{119}\right] \text { Joule } \end{aligned} $$

Correct answer: (1)

Solution:

The gravitational force, between the mass $m$ and the earth (Mass $=M$ ), when the mass $m$ is at a height $x$ above the surface of the earth, would be

$$ \mathrm{F}(x)=\frac{\mathrm{GMm}}{(\mathrm{R}+x)^{3}} $$

$\therefore$ The work done, in moving the object (against the gravitational force) from the surface of the earth to a height $h$ above the surface, is

$$ \begin{aligned} \mathrm{W} & =\int _{x=0}^{x=\mathrm{h}}-\mathrm{F}(x) \mathrm{d} x=-\mathrm{GMm} \int _{x=0}^{x=\mathrm{h}}(\mathrm{R}+x)^{-3} \mathrm{~d} x \\ & =-(\mathrm{GMm})\left|\frac{-(\mathrm{R}+x)^{-2}}{2}\right| _{x=0}^{x=\mathrm{h}} \\ & =\left(\frac{\mathrm{GMm}}{2}\right)\left[\frac{\mathrm{h}(\mathrm{h}+2 \mathrm{R})}{\mathrm{R}^{2}(\mathrm{R}+\mathrm{h})^{2}}\right] \end{aligned} $$

Correct answer: (2)

Solution:

Consider the object at a distance $x$ from the bottom of the plane. The forces, acting on it are as shown. The force of friction, $\mathrm{F} _{\mathrm{R}}$, at this position, would be

The external force $\mathrm{F}(x)$, has to be infinitesimally greater than $\left(\mathrm{mg} \sin \theta+\mu _{0} x \mathrm{mg} \cos \theta\right)$. Hence the work done in taking the object from the bottom of the plane, to its top, is

$$ \begin{aligned} & \mathrm{W}=\int _{x=0}^{x=\mathrm{L}} \mathrm{F}(x) \mathrm{dx} \\ & \therefore \quad \mathrm{W}=\int _{x=0}^{x=\mathrm{L}}\left[\mathrm{mg} \sin \theta+\left(\mu _{0} \mathrm{mg} \cos \theta\right) x\right] \mathrm{d} x \\ & \quad=\mathrm{mg} \sin \theta|x| _{0}^{\mathrm{L}}+\mu _{0} \mathrm{mg} \cos \theta\left|\frac{x^{2}}{2}\right| _{0}^{\mathrm{L}} \end{aligned} $$

$$ =\frac{\mathrm{mgL}}{2}\left[2 \sin \theta+\left(\mu _{0} \mathrm{mg} \cos \theta\right) \mathrm{L}\right] $$

[Note: A quick way of finding the correct option, in the problem, can be as follows: The gain in P.E. of the object, when it moves up the plane, equals $\mathrm{Mgh}=\mathrm{MgL} \sin \theta$. The work done, by the force, (when the object is moved up without imparting it any velocity) would equal this gain in P.E. if the plane were a smooth surface. We should, therefore, get $\mathrm{W}=\mathrm{MgL} \sin \theta$ when $\mu _{0}=0$. It is early to check that only option (2) satisfies this requirement].

Correct answer: (3)

Solution:

The acceleration, a ( $x$ ), imparted to the object, at a distance $x$ from its starting position, is

$$ \mathrm{a}(x)=\mathrm{k} x^{2}(\mathrm{k}=\text { constant of proportionality }) $$

The force, $\mathrm{F}(x)$, acting on the object, at this position, is

$$ \mathrm{F}(x)=\mathrm{Ma}(x)=\operatorname{Mk} x^{2} $$

$\therefore$ Work done in moving a distance $\mathrm{d} x=\mathrm{dW}=\mathrm{F}(x) \mathrm{d} x$

$\therefore$ Total work done $=\mathrm{W}=\int _{x=0}^{x=\mathrm{L}} \mathrm{F}(x) \mathrm{d} x=\frac{\mathrm{MkL}^{3}}{3}$

This work increases only the K.E. of the object. Hence if $\mathrm{v}$ is the speed of the object, at the position $x=\mathrm{L}$, we have

$$ \begin{aligned} & \quad \frac{1}{2} \mathrm{Mv}^{2}=\frac{\mathrm{MkL}^{3}}{3} \\ & \therefore \quad \mathrm{v}=\mathrm{kL} \sqrt{\frac{2}{3} \mathrm{~L}} \end{aligned} $$

Correct answer: (1)

Solution:

We have, $\mathrm{F}(x)-\left(\mu _{0} x \mathrm{Mg}\right)=\mathrm{Ma}$

The work done, by the force $\mathrm{F}(x)$, in moving a distance $\mathrm{d} x$, is $\mathrm{F}(x) \mathrm{d} x$. The total work done, in moving a distance, $\mathrm{L}$, is then

$$ \begin{aligned} \mathrm{W}= & \int _{x=0}^{x=\mathrm{L}} \mathrm{M}\left(\mathrm{a}+\mu _{0} \mathrm{~g} x\right) \mathrm{d} x=\mathrm{M}\left(\mathrm{L}+\frac{\mu _{0} \mathrm{gL^2}}{2}\right) \\ = & \frac{\mathrm{ML}}{2}\left(1+\mu _{0} \mathrm{gL}\right) \end{aligned} $$

This equals the increase in K.E. of the object. Hence the velocity v, after covering a distance L, is given by

$$ \begin{aligned} & \frac{1}{2} M v^{2}=\frac{M L}{2}\left(1+\mu _{0} g L\right) \\ & \therefore \quad v=\sqrt{L\left(1+\mu _{0} g L\right)} \end{aligned} $$

Correct answer: (3)

Solution:

The free body diagram, for the forces acting on the object, is as shown. We, therefore, have

$$ \mathrm{N}=\mathrm{mg} \cos \theta+\mathrm{F} \sin \theta $$

Also, it is the component $\mathrm{F} \cos \theta$, of the force $\mathrm{F}$, that does work in moving the object through a distance L. The work done is, therefore

$$ \mathrm{W}=\mathrm{F} \cos \theta . \mathrm{L} $$

We are given that $\sin \theta=\frac{\mathrm{h}}{\mathrm{L}}$.

Hence $\mathrm{N}=\mathrm{mg}\left(\sqrt{1-\frac{\mathrm{h}^{2}}{\mathrm{~L}^{2}}}\right)+\mathrm{F} \frac{\mathrm{h}}{\mathrm{L}}=\frac{1}{\mathrm{~L}}\left[\mathrm{mg} \sqrt{\mathrm{L}^{2}-\mathrm{h}^{2}}+\mathrm{Fh}\right]$

and $\mathrm{W}=\left\{\mathrm{F} \sqrt{1-\frac{\mathrm{h}^{2}}{\mathrm{~L}^{2}}}\right\} \mathrm{L}=\left(\mathrm{F} \sqrt{\mathrm{L}^{2}-\mathrm{h}^{2}}\right)$

Correct answer: (4)

Solution:

The force, needed over the first segment $=$ Ma.

Hence $\mathrm{W} _{1}=$ work done over the first segment $=\operatorname{Ma}(2 \ell)$

Over the second segment, if $F$ ’ is the force needed, we have

$\mathrm{F}^{\prime}-\mu \mathrm{Mg}=\mathrm{Ma}$

$\therefore \mathrm{F}^{\prime}=[\mathrm{Ma}+\mu \mathrm{Mg}]$

Hence, $\mathrm{W} _{2}=$ Work done over the second segment.

$$ =\mathrm{F}^{\prime}(\ell)=\mathrm{M}(\mathrm{a}+\mu \mathrm{g}) \ell $$

$\therefore \mathrm{W} _{1}=\mathrm{W} _{2}$ if $2 \mathrm{Ma} \ell=\mathrm{Ma} \ell+\mu \mathrm{Mg} \ell$

$\therefore \mu=\frac{\mathrm{a}}{\mathrm{g}} \simeq \frac{2}{10}=0.2$

Correct answer: (3)

Solution:

The coefficient of friction, at a point distant $x$ from the starting point, is

$$ \mu _{x}=\mu _{1}+\alpha x $$

Where $\alpha=\left(\frac{\mu _{2}-\mu _{1}}{\ell}\right)=$ constant rate of increase of $\mu$.

$\therefore$ If $\mathrm{F} _{x}$ is the force needed here, we would have

$\mathrm{F} _{x}-\mu _{x} \mathrm{Mg}=\mathrm{Ma} \quad $ or $\quad \mathrm{F} _{x}=\mathrm{M}\left(\mathrm{a}+\mu _{x} \mathrm{~g}\right)$

$\therefore$ Work done $=\int _{0}^{\ell} \mathrm{F} _{x} \mathrm{~d} x$

$$ \begin{aligned} & =(\mathrm{Ma}) \ell+\mathrm{M} \mu _{1} \mathrm{~g} \ell+\mathrm{Mag} \frac{\ell^{2}}{2} \\ & =(\mathrm{Ma}) \ell+\operatorname{Mg} \ell\left(\frac{\mu _{2}+\mu _{1}}{2}\right) \\ & =\frac{1}{2} \mathrm{M} \ell\left[2 \mathrm{a}+\left(\mu _{1}+\mu _{2}\right) \mathrm{g}\right] \end{aligned} $$

Correct answer: (1)

Solution:

We have $\mathrm{S}(\mathrm{t})=\mathrm{kt}^{4}$

$\therefore \mathrm{v}(\mathrm{t})=\frac{\mathrm{d}}{\mathrm{dt}}[\mathrm{S}(\mathrm{t})]=4 \mathrm{kt}^{3}$

$\therefore \quad \mathrm{a}(\mathrm{t})=\frac{\mathrm{d}}{\mathrm{dt}}[\mathrm{v}(\mathrm{t})]=12 \mathrm{kt}^{2}$

The force, $F(t)$, is, therefore, given by

$F(t)=m a(t)=12 k m t^{2}$

The instantaneous power, $\mathrm{P}(\mathrm{t})$, is given by

$\mathrm{P}(\mathrm{t})=[\mathrm{F}(\mathrm{t})][\mathrm{v}(\mathrm{t})]$

$\therefore \mathrm{P}(\mathrm{t})=\left(48 \mathrm{k}^{2} \mathrm{~m}\right) \mathrm{t}^{5}$

The ratio, of the instantaneous powers, at $\mathrm{t}=16 \mathrm{~s}$ and $\mathrm{t}=8(\mathrm{~s})$ is

$\frac{\mathrm{P}(16)}{\mathrm{P}(8)}=\frac{48 \mathrm{k}^{2} \mathrm{~m}(16)^{5}}{48 \mathrm{k}^{2} \mathrm{~m}(8)^{5}}=\left(\frac{16}{8}\right)^{5}$

$=2^{5}=32$

$\therefore$ The required ratio is $(32: 1)$

Correct answer: (3)

Solution:

For the inclined plane, $\sin \theta=\frac{h}{\ell}$

The various forces acting on the object, are as shown. We have

$$ \mathrm{N}=(\mathrm{mg} \cos \theta+\mathrm{F} \sin \theta) $$

Since the force $F$ just moves the object up the plane, we have $(F \sin \theta)$ as just infinitesimally greater than the sum of $(m g \sin \theta+\mu \mathrm{N})$.

The work done, $\mathrm{W}$, is given by

$$ \begin{aligned} \mathrm{W}= & (\mathrm{F} \cos \theta) \mathrm{S} \\ & =[\mathrm{mg} \sin \theta+\mu(\mathrm{mg} \cos \theta+\mathrm{F} \sin \theta)] \mathrm{S} \\ & =[\mathrm{mg}(\sin \theta+\mu \cos \theta)+\mu \mathrm{F} \sin \theta] \mathrm{S} \\ & =\left[\mathrm{mg}\left(\frac{\mathrm{h}}{\ell}+\mu\left(\frac{\sqrt{\ell^{2}-\mathrm{h}^{2}}}{\ell}\right)\right)+\mu \mathrm{F} \frac{\mathrm{h}}{\ell}\right] \mathrm{S} \\ & =\frac{\mathrm{S}}{\ell}\left[\mathrm{mgh}+\mu\left[\mathrm{mg} \sqrt{\ell^{2}-\mathrm{h}^{2}}+\mathrm{Fh}\right]\right] \end{aligned} $$

Correct answer: (2)

Solution:

For the inclined plane, we have

$$ \sin \theta=\frac{h}{\ell} $$

and $\cos \theta=\frac{\sqrt{\ell^{2}-\mathrm{h}^{2}}}{\ell}$

Further $\mathrm{N}=\mathrm{mg} \quad \cos \theta$

and $F=m g \sin \theta+\mu \quad m g \quad \cos \theta$

$=m g(\sin \theta+\mu \cos \theta)$

$$ =\frac{\mathrm{mg}}{\ell}\left[\mathrm{h}+\mu \cdot \sqrt{\ell^{2}-\mathrm{h}^{2}}\right] $$

The rate of increase of $\mu$, along the plane, is

$$ \alpha=\left(\frac{\mu _{2}-\mu _{1}}{2 L}\right) $$

Hence value of $\mu$ at the initial position

$$ \begin{aligned} \mu(L) & =\mu _{1}+\left(\frac{\mu _{2}-\mu _{1}}{2 L}\right) \cdot L \\ & =\left(\frac{\mu _{1}+\mu _{2}}{2}\right) \end{aligned} $$

The value of $\mu$ at the final position is $\mu _{2}$.

Hence average value of $\mu$, over the movement from $x=\mathrm{L}$ to $x=2 \mathrm{~L}$, is

$$ \begin{aligned} \mu _{\mathrm{AV}}= & \frac{1}{2}\left[\frac{\mu _{1}+\mu _{2}}{2}+\mu _{2}\right] \\ & =\left(\frac{\mu _{1}}{4}+\frac{3 \mu _{2}}{4}\right)=\left(\frac{\mu _{1}+3 \mu _{2}}{4}\right) \end{aligned} $$

The work done is, therefore,

$$ \begin{aligned} \mathrm{W}= & \mathrm{F} _{\mathrm{AV}} \times \mathrm{L} \\ & =\frac{\mathrm{mg}}{\ell}\left[\mathrm{h}+\left(\sqrt{\ell^{2}-\mathrm{h}^{2}}\right) \mu _{\mathrm{AV}}\right] \cdot \mathrm{L} \\ & =\frac{\mathrm{mgL}}{\ell}\left[\mathrm{h}+\left(\frac{1}{4}\right)\left(\mu _{1}+3 \mu _{2}\right) \sqrt{\ell^{2}-\mathrm{h}^{2}}\right] \end{aligned} $$

[Note: The same result is obtained by integrating $(\mathrm{F}(x) \mathrm{d} x)$ from $x=\mathrm{L}$ to $x=2 \mathrm{~L}$. Here we would take $\mu(x)=\mu _{1}+\left(\frac{\mu _{2}-\mu _{1}}{2 \mathrm{~L}}\right) x$ and $\mathrm{F}(x)=\frac{\mathrm{mg}}{\ell}\left[\mathrm{h}+\sqrt{\ell^{2}-\mathrm{h}^{2}} \mu(x)\right]$

Correct answer: (1)

Solution:

Let us assume that the displacement, $s$, of the particle, varies as $t^{n}$. Hence

$$ \begin{gathered} \mathrm{s}=\mathrm{kt}^{\mathrm{n}} \\ \therefore \quad \mathrm{v}=\frac{\mathrm{ds}}{\mathrm{dt}}=n k t^{\mathrm{n}-1} \\ \text { and } \mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{n}(\mathrm{n}-1) \mathrm{kt}^{\mathrm{n}-2} \end{gathered} $$

$\therefore$ Force, $\mathrm{F}$, acting on the particle, equals $\mathrm{n}(\mathrm{n}-1) \mathrm{mkt}^{\mathrm{n}-2}$

$\therefore$ Power $(=$ force $\times$ velocity $)=\mathrm{n}^{2}(\mathrm{n}-1) \mathrm{mk}^{2} \mathrm{t}^{(\mathrm{n}-2)+(\mathrm{n}-1)}$

Since power is constant, we have,

$2 \mathrm{n}-3=0 \quad$ or $\mathrm{n}=\frac{3}{2}$

$\therefore \quad \mathrm{a}=\mathrm{n}(\mathrm{n}-1) \mathrm{kt}^{\mathrm{n}-2}=-\frac{3}{4} \mathrm{kt}^{-1 / 2}$

It is graph (A) only that can correspond to this dependence of ’ $a$ ’ on ’ $t$ ‘.

Correct answer: (4)

Solution:

Let us assume that the displacement, $\mathrm{s}$, of the particle, varies as $\mathrm{t}^{\mathrm{n}}$. We then have

$$ \begin{gathered} \mathrm{s}=\mathrm{kt}^{\mathrm{n}} \\ \therefore \quad \mathrm{v}=\frac{\mathrm{ds}}{\mathrm{dt}}=\mathrm{knt}^{\mathrm{n}-1} \end{gathered} $$

and acceleration $=\frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{kn}(\mathrm{n}-1) \mathrm{t}^{\mathrm{n}-2}$

$\therefore$ Force $=\mathrm{mkn}(\mathrm{n}-1) \mathrm{t}^{\mathrm{n}-2}$

$\mathrm{P}=$ Power $=$ Force $\times$ Velocity

$=\mathrm{mk}^{2} \mathrm{n}^{2}(\mathrm{n}-1) \mathrm{t}^{(\mathrm{n}-1)+(\mathrm{n}-2)}$

$=\mathrm{mk}^{2} \mathrm{n}^{2}(\mathrm{n}-1) \mathrm{t}^{2 \mathrm{n}-3}$

The given graph shows that $\mathrm{P} \alpha \mathrm{t}$. Hence

$$ 2 n-3=1 \text { or } n=2 $$

$\therefore \mathrm{v}=2 \mathrm{kt}^{2-1}=2 \mathrm{kt}$

Thus $v \alpha t$. This is shown in the graph or labelled as graph D.

Correct answer: (3)

Solution:

Maximum compression takes place when the final K.E. of the system is zero; the initial K.E. has been ‘completely’ used to provide P.E. to the compressed cubes. The force (F), needed to change the lengths by an amount $\Delta \mathrm{L}$, can be written as.

$\mathrm{F}=\mathrm{k} \Delta \mathrm{L} \quad(\mathrm{k}=$ ‘compressional’ constant for the metal)

Using Hooke’s law, we can write

$\mathrm{Y}=\frac{(\mathrm{F} / \mathrm{A})}{(\Delta \mathrm{L} / \mathrm{L})}=\frac{\mathrm{F}}{\mathrm{L}^{2}} \times \frac{\mathrm{L}}{\Delta \mathrm{L}}=\frac{\mathrm{F}}{\mathrm{L} \Delta \mathrm{L}}$

$\therefore \mathrm{F}=\mathrm{YL}(\Delta \mathrm{L})=\mathrm{k} \Delta \mathrm{L}$

$\therefore \mathrm{k}=\mathrm{YL}$

Now initial K.E. of the system

$$ \begin{aligned} & =2 \times\left(\frac{1}{2} \mathrm{MV}^{2}\right)=\mathrm{a} \text { constant } \\ & =\mathrm{E} _{\mathrm{K}} \text { (say) } \end{aligned} $$

For maximum compression, this K.E gets completely converted into the P.E. of the two ‘compressed’ cubes. As $\Delta \mathrm{L}$ denotes the ‘maximum compression’ we have

Final P.E. of the system $=2 \times \frac{1}{2} k(\Delta L)^{2}=E _{k}$

$\therefore \mathrm{k}(\Delta \mathrm{L})^{2}=\mathrm{E} _{\mathrm{k}}$

or

$$ \begin{aligned} & \Delta \mathrm{L}=\sqrt{\frac{\mathrm{E} _{\mathrm{k}}}{\mathrm{k}}}=\sqrt{\frac{\mathrm{E} _{\mathrm{k}}}{\mathrm{YL}}} \quad(\because \mathrm{k}=\mathrm{YL}) \\ & =\text { Constant. } \frac{1}{\sqrt{\mathrm{Y}}} \end{aligned} $$

Thus $\Delta \mathrm{L} \alpha \mathrm{Y}^{-1 / 2}$

It is the graph, labelled as graph C, that shows this kind of dependence of $\Delta \mathrm{L}$ on Y. Hence option (3) is the correct choice.

Correct answer: (2)

Solution:

Under isothermal conditions ( $\mathrm{Temp}=\mathrm{T}=$ constant , we have,

$\therefore \mathrm{PV}=\mathrm{nRT}=\mathrm{a}$ constant

The force, $\mathrm{F}$, on the piston, at any instant, is PA where $\mathrm{A}=$ area of cross section of the piston. If the piston gets displaced by an amount $\mathrm{d} x$, the work done is

$\mathrm{dW}=\mathrm{Fd} x=\mathrm{PAd} x=\mathrm{PdV}$

$(\because \mathrm{Ad} x=\mathrm{dV}=$ change in volume of the gas).

$\text { Hence, } \mathrm{W}$=$\int_{\mathrm{V}_0}^{\mathrm{V}_0^{\prime}} \mathrm{PdV}$

=$nRT \int^{V_0^’}_ {V _0} \frac{dv}{v}=nRT |ln| _{V _0}^{V _0^’}$

$=\mathrm{nRT} \ell \mathrm{n}\left(\frac{\mathrm{V} _{0}^{\prime}}{\mathrm{V} _{0}}\right)=\mathrm{nRT} \operatorname{n}\left[\frac{\mathrm{A}\left(x _{0}+\Delta x\right)}{\mathrm{A} x _{0}}\right]$

$=\operatorname{nRT} \ln \left(1+\frac{\Delta x}{x _{0}}\right)$

$\therefore$ Power $=\frac{\mathrm{W}}{\Delta \mathrm{t}}=\frac{\mathrm{nRT}}{\Delta \mathrm{t}} \ln \left(1+\frac{\Delta x}{x _{0}}\right)$

Correct answer: (1)

Solution:

For an elastic collision between spheres (1) or (2), the speed acquired by sphere (2) is given by

$\mathrm{v} _{2}=\left[\frac{2 \mathrm{M} _{1}}{\mathrm{M} _{1}+\mathrm{M} _{2}}\right] \mathrm{v}$

For this value of speed of sphere (2), the speed, acquired by sphere (3), after sphere (2) collides elastically with it, is

$v _{3}=\frac{2 M _{2}}{\left(M _{2}+M _{3}\right)} v _{2}=\frac{4 M _{1} M _{2}}{\left(M _{1}+M _{2}\right)\left(M _{2}+M _{3}\right)} v$

All other factors, except $M _{2}$ are constants. Hence, for maximum value of $v _{3}$, we need to have $\frac{d _{3}}{d M _{2}}=0$

Now $\frac{d v _{3}}{d M _{2}}=4 M _{1}\left[\frac{\left(M _{1}+M _{2}\right)\left(M _{2}+M _{3}\right) \times 1-M _{2}\left(M _{1}+M _{2}\right) \times 1+\left(M _{2}+M _{3}\right) \times 1}{\left(M _{1}+M _{2}\right)^{2}\left(M _{2}+M _{3}\right)^{2}}\right]$

$$ =\frac{4 M _{1}\left[M _{3} M _{1}-M _{2}^{2}\right]}{\left(M _{1}+M _{2}\right)^{2}\left(M _{2}+M _{3}\right)^{2}} $$

$\therefore \frac{\mathrm{dv} _{3}}{\mathrm{dM} _{2}}=0$ if $\mathrm{M} _{1} \mathrm{M} _{3}=\mathrm{M} _{2}^{2}$

$\therefore \mathrm{M} _{2}=\sqrt{\mathrm{M} _{1} \mathrm{M} _{3}}$

$=\sqrt{20 \mathrm{~m} \times 5 \mathrm{~m}}=10 \mathrm{~m}$

The maximum value of $\mathrm{v} _{3}$ corresponds to this value of $\mathrm{M} _{2}$.

Hence, $\left(v _{3}\right) _{\max }=\frac{4 \times(20 m) \times(10 m)}{(20 m+10 m)(10 m+5 m)} v$

$=4 \times \frac{200}{450} \mathrm{v}=\frac{16}{9} \mathrm{v}$

Correct answer: (2)

Solution:

Let $v _{1}$ and $v _{2}$ be the speeds of the two balls after the collision. Since momentum is conserved, we have

$$ \mathrm{m} \cdot 4 \mathrm{u}=\mathrm{m} \cdot \mathrm{v} _{1}+\mathrm{m} \cdot \mathrm{v} _{2} \text { or } \mathrm{v} _{1}+\mathrm{v} _{2}=4 \mathrm{u} $$

Also e $=$ coefficient of restitution $=\frac{\mid \text { Relative velocity after collision } \mid}{\mid \text { Relative velocity before collision } \mid}$

$\therefore \quad 0.5=\frac{\mathrm{v} _{2}-\mathrm{v} _{1}}{4 \mathrm{u}-0}$ or $\quad \mathrm{v} _{2}-\mathrm{v} _{1}=2 \mathrm{u}$

Therefore give $\mathrm{v} _{2}=3 \mathrm{u}$ and $\mathrm{v} _{1}=\mathrm{u}$

$\therefore$ Total K.E. after collision $=\frac{1}{2} \mathrm{~m}(\mathrm{u})^{2}+\frac{1}{2} \mathrm{~m}(3 \mathrm{u})^{2}=\frac{1}{2} \mathrm{~m}\left(10 \mathrm{u}^{2}\right)$

Also total K.E. before collision $=\frac{1}{2} \mathrm{~m}(4 \mathrm{u})^{2}=\frac{1}{2} \mathrm{~m}\left(16 \mathrm{u}^{2}\right)$

$\therefore \frac{\text { Loss in K.E.of the system }}{\text { Original K.E.of the system }}=\frac{\frac{1}{2} \mathrm{~m}\left(6 \mathrm{u}^{2}\right)}{\frac{1}{2} \mathrm{~m}\left(16 \mathrm{u}^{2}\right)}=\frac{3}{8}$

Correct answer: (4)

Solution:

The masses $\mathrm{m} _{1}$ and $\mathrm{m} _{2}$ being equal, their velocities, after $\mathrm{m} _{1}$ collides elastically with $\mathrm{m} _{2}$, are zero and $\mathrm{u}$, respectively.

A little later, when $\mathrm{m} _{2}$ and $\mathrm{m} _{3}$ have the same velocity, $\mathrm{v}$, and the spring has been compressed by an amount $x _{0}$, we have, by the laws of conservation of momentum and energy.

$\mathrm{m} _{2} \times \mathrm{u}=\mathrm{m} _{2} \mathrm{v}+\mathrm{m} _{3} \mathrm{v} ; \mathrm{v}=\frac{\mathrm{u}}{3}$

and $\frac{1}{2} m _{2} u^{2}=\frac{1}{2} m _{2}\left(\frac{u}{3}\right)^{3}+\frac{1}{2} m _{3}\left(\frac{u}{3}\right)^{2}+\frac{1}{2} k x _{0}^{2}$

or $\quad \mathrm{u}^{2}=\frac{\mathrm{u}^{2}}{9}+\frac{2 \mathrm{u}^{2}}{9}+\mathrm{k}^{x _{0}^{2}} /\left(\mathrm{m} _{2}\right)$

$\therefore \mathrm{k}=\frac{\mathrm{m} _{2}}{x _{0}^{2}} \cdot \frac{2}{3} \mathrm{u}^{2}=\frac{2}{3} \times \frac{50 \times 10^{-3} \mathrm{u}^{2}}{x _{0}^{2}}=\frac{1}{30} \frac{\mathrm{u}^{2}}{x _{0}^{2}}$

Correct answer: (2)

Solution:

The force being a variable one, we have to calculate the work through

$$ \begin{aligned} & \mathrm{W}=\int \mathbf{F} \cdot \mathbf{d s}=\int \mathbf{F} \cdot(\mathrm{dy} \mathbf{j}+\mathrm{dz} \mathbf{k}) \\ & =\int[(\mathrm{yz}) \mathrm{dy}+(\mathrm{yz}) \mathrm{dz}] \end{aligned} $$

For path (1)

$$ \begin{aligned} \mathrm{W} _{1}= & \left(\mathrm{W} _{1}^{\prime}\right) \text { along } \mathrm{OK}+\left(\mathrm{W} _{2}^{\prime}\right)(\text { along } \mathrm{KL}) \\ & =(\mathrm{O})+\int[(2 \mathrm{z})(0)+(2 \mathrm{z}) \mathrm{dz}]=\left|\mathrm{z}^{2}\right| _{0}^{2}=4 \mathrm{~J} \end{aligned} $$

For path (2)

$\mathrm{W} _{2}=\left(\mathrm{W} _{2}^{\prime}\right)$ along $\mathrm{OM}+\left(\mathrm{W} _{2}^{\prime \prime}\right)($ along $\mathrm{ML})$

$$ =(\mathrm{O})+\int[(2 \mathrm{y}) \mathrm{dy}+(2 \mathrm{y})(0)]=\left|\mathrm{y}^{2}\right| _{0}^{2}=4 \mathrm{~J} $$

For path (3).

Here $\mathrm{y}=\mathrm{z}$ and $\mathrm{dy}=\mathrm{dz}$ for all segments on the path OL

$\therefore \quad \mathrm{W} _{3}=\int _{0}^{2} 2 \mathrm{y}^{2} \mathrm{dy}=\left|\frac{2}{3} \mathrm{y}^{3}\right| _{0}^{2}=\frac{16}{3} \mathrm{~J}$

We thus find that the force $\mathbf{F}$ is non-conservative as $\mathrm{W} _{1}=\mathrm{W} _{2} \neq \mathrm{W} _{3}$

(The work done, in going from $\mathrm{O}$ to $\mathrm{L}$, is seen to be path dependent)

Correct answer: (3)

Solution:

Let the dart emerge with a speed $v _{1}$ after its passage through the first plate. Since this plate has acquired a velocity v, we have

$$ [0.02 \mathrm{~m}] \mathrm{V}=(0.02 \mathrm{~m}) \mathrm{v} _{1}+\mathrm{mv} $$

The dart, now moving with a speed $\mathrm{v} _{1}$, gets embedded in the second plate and their combined mass, again moves with a speed $v$. By the law of conservation of momentum, we now get

$$ (0.02 \mathrm{~m}) \mathrm{v} _{1}=(0.02 \mathrm{~m}+4.98 \mathrm{~m}) \mathrm{v}=5 \mathrm{mv} $$

$\therefore$ From the first equation, we get

$$ \begin{aligned} \quad[0.02 \mathrm{~m}] \mathrm{V} & =5 \mathrm{mv}+\mathrm{mv}=6 \mathrm{mv} \\ \therefore \mathrm{v}=\left(\frac{0.02 \mathrm{~m}}{60}\right) \mathrm{V} & =\left(\frac{\mathrm{V}}{300}\right) \end{aligned} $$

Correct answer: (3)

Solution:

At point A, the total energy of the ball

$$ =\frac{1}{2} \mathrm{mv}^{2}+\mathrm{mgh} $$

At point B, since the velocity of the ball becomes (momently) zero, its total energy is just its potential energy, i.e. $\mathrm{mgR}$.

$\therefore \quad \frac{1}{2} \mathrm{mv}^{2}+\mathrm{mgh}=\mathrm{mgR}$

$\therefore \mathrm{v}^{2}=2 \mathrm{~g}(\mathrm{R}-\mathrm{h})$

$$ =(-2 g) h+2 g R $$

$\therefore$ The graph of $\mathrm{v}^{2}$, against $\mathrm{h}$, is a straight line of slope $(-2 \mathrm{~g})$ and having an intercept $2 \mathrm{gR}$ on the $\mathrm{v}^{2}$ axis. This corresponds to graph $\mathrm{M}$.

Hence option (3) is the correct choice.

Correct answer: (2)

Solution:

Let the combined mass $\left(\mathrm{m} _{1}+\mathrm{m} _{2}\right)$ move with a velocity $\mathrm{v}$ along a direction inclined at an angle $\theta$ to the $\mathrm{y}-$ axis. By the law of conservation of momentum, we than have

$\left(\mathrm{m} _{1}+\mathrm{m} _{2}\right) \mathrm{v} \cos \alpha=\mathrm{m} _{1} \mathrm{v} _{1}$ and $\left(\mathrm{m} _{1}+\mathrm{m} _{2}\right) \mathrm{v} \sin \alpha=\mathrm{m} _{2} \mathrm{v} _{2}$

$\therefore\left(\mathrm{m} _{1}+\mathrm{m} _{2}\right)^{2} \mathrm{v}^{2}=\mathrm{m} _{1}^{2} \mathrm{v} _{1}^{2}+\mathrm{m} _{2}^{2} \mathrm{v} _{2}^{2}$

$\therefore \mathrm{v}^{2}=\frac{\mathrm{m} _{1}^{2} \mathrm{v} _{1}^{2}+\mathrm{m} _{2}^{2} \mathrm{v} _{2}^{2}}{\left(\mathrm{~m} _{1}+\mathrm{m} _{2}\right)^{2}}$

Loss in K.E.

$$ \begin{aligned} & =\frac{1}{2} \mathrm{~m} _{1} \mathrm{v} _{1}^{2}+\frac{1}{2} \mathrm{~m} _{2} \mathrm{v} _{2}^{2}-\left[\frac{1}{2}\left(\mathrm{~m} _{1}+\mathrm{m} _{2}\right) \frac{\left(\mathrm{m} _{1}^{2} \mathrm{v} _{1}^{2}+\mathrm{m} _{2}^{2} \mathrm{v} _{2}^{2}\right)}{\left(\mathrm{m} _{1}+\mathrm{m} _{2}\right)^{2}}\right] \\ & =\frac{1}{2} \frac{\left[\left(\mathrm{m} _{1}+\mathrm{m} _{2}\right)\left(\mathrm{m} _{1} \mathrm{v} _{1}^{2}+\mathrm{m} _{2} \mathrm{v} _{2}^{2}\right)-\left(\mathrm{m} _{1}^{2} \mathrm{v} _{1}^{2}+\mathrm{m} _{2}^{2} \mathrm{v} _{2}^{2}\right)\right]}{\left(\mathrm{m} _{1}+\mathrm{m} _{2}\right)^{2}} \\ & =\frac{1}{2} \frac{\mathrm{m} _{1} \mathrm{~m} _{2}}{\left(\mathrm{~m} _{1}+\mathrm{m} _{2}\right)}\left[\mathrm{v} _{1}^{2}+\mathrm{v} _{2}^{2}\right] \end{aligned} $$

Correct answer: (1)

Solution:

The free body diagram for the cart is as shown. When the cart has moved a distance, $x$, the acceleration produced in it, equals $\mathrm{a} _{0} x$. Hence we have

$\mathrm{F} \cos \theta=\mathrm{M}\left(\mathrm{a} _{0} x\right)+\mu \mathrm{N}$

and $\mathrm{N}=\mathrm{Mg}-\mathrm{F} \sin \theta$

We, therefore, get $\mathrm{F}(\cos \theta+\mu \sin \theta)=\mathrm{Ma} _{0} x+\mu \mathrm{Mg}$

$\therefore \mathrm{F}=\mathrm{F}(x)=\frac{\mathrm{M}\left[\mathrm{a} _{0} x+\mu \mathrm{g}\right]}{(\cos \theta+\mu \sin \theta)}$

$\therefore \mathrm{dW}=\frac{\mathrm{M}\left[\mathrm{a} _{0} x+\mu \mathrm{g}\right]}{(\cos \theta+\mu \sin \theta)}(\cos \theta)(\mathrm{d} x)$

$\therefore \mathrm{W}=\frac{\mathrm{M} \cos \theta}{\cos \theta+\mu \sin \theta}\left[\mathrm{a} _{0} \frac{\mathrm{L}^{2}}{2}+\mu \mathrm{gL}\right]$

$$ =\frac{\mathrm{ML} \cos \theta}{2(\cos \theta+\mu \sin \theta)}\left[\mathrm{a} _{0} \mathrm{~L}+2 \mu \mathrm{g}\right] $$

Correct answer: (1)

Solution:

The velocity, of the ball $\mathrm{m} _{1}$, just before it strikes the ball $\mathrm{m} _{2}$ (initially at rest) is $\sqrt{2 \mathrm{gh}}$.

The law of conservation of momentum yields.

$\left(\mathrm{m} _{1} \sqrt{2 \mathrm{gh}}\right)+\mathrm{m} _{2} \times 0=\left(\mathrm{m} _{1}+\mathrm{m} _{2}\right) \mathrm{v}$

Where $v$ is the velocity of the ‘combination’ just after the ‘collision’. Hence

$$ \mathrm{v}=\frac{\mathrm{m} _{1} \sqrt{2 \mathrm{gh}}}{\left(\mathrm{m} _{1}+\mathrm{m} _{2}\right)} $$

The K.E. of the ‘combination’, just after the collision, makes it rise to a height $\frac{\mathrm{h}}{\mathrm{n}}$. We, therefore, have

$$ \frac{1}{2}\left(\mathrm{~m} _{1}+\mathrm{m} _{2}\right) \mathrm{v}^{2}=\left(\mathrm{m} _{1}+\mathrm{m} _{2}\right) \mathrm{g}\left(\frac{\mathrm{h}}{\mathrm{n}}\right) $$

$\therefore \mathrm{v}^{2}=\frac{2 \mathrm{gh}}{\mathrm{n}}$

$\therefore\left(\frac{\mathrm{m} _{1}}{\mathrm{~m} _{1}+\mathrm{m} _{2}}\right)^{2}(2 \mathrm{gh})=\frac{2 \mathrm{gh}}{\mathrm{n}}$

$\therefore \frac{\mathrm{m} _{1}}{\mathrm{~m} _{1}+\mathrm{m} _{2}}=\frac{1}{\sqrt{\mathrm{n}}}$

or $\quad\left(\frac{m _{1}+m _{2}}{m _{1}}\right)=\sqrt{n}$

$\therefore \frac{\mathrm{m} _{2}}{\mathrm{~m} _{1}}=(\sqrt{\mathrm{n}}-1)$

Correct answer: (4)

Solution:

The initial horizontal (or $x$ ) and vertical (or y) component, of the velocity of the ball, are $\mathrm{v} \cos 45^{\circ}$ and $\mathrm{v} \sin$ $45^{\circ}$ respectively.

The surface being smooth, the final (i.e. after ‘reflection’) horizontal component, of the velocity of the ball, would still be $\mathrm{v} \cos 45^{\circ}$. The vertical component, however, would be (ev $\sin 45^{\circ}$ ). The vertical component, however, would be ( $\mathrm{v} \sin 45^{\circ}$ ). If therefore, the final velocity (v) of the ball makes an angle $\theta$ with the horizontal, we have

$\mathrm{v} \cos \theta=\mathrm{v} \cos 45^{\circ}$

and $\mathrm{v} \sin \theta=\mathrm{e} v \sin 45^{\circ}$

$\therefore \tan \theta=\mathrm{e} \tan 45^{\circ}=\frac{1}{\sqrt{3}} \times 1=\frac{1}{\sqrt{3}}$

$\therefore \theta=30^{\circ}$

The angle of ‘reflection’ is, therefore, $\left(90^{\circ}-30^{\circ}\right)=60^{\circ}$. Hence the difference, between the angle of ‘reflection’ and the angle of incidence, is $\left(60^{\circ}-45^{\circ}\right)=+15^{\circ}$.

The final velocity, $\mathbf{V}$ is given by

$$ \mathbf{V}=\left(\mathrm{v} \cos 45^{\circ}\right) \mathbf{i}+\left(\mathrm{ev} \sin 45^{\circ}\right) \mathbf{j} $$

The initial velocity, $\mathbf{V}$, is given by

$\mathbf{V}=\left(v \cos 45^{\circ}\right) \mathbf{i}+\left(v \sin 45^{\circ}\right)(-\mathbf{j})$

$\therefore$ Change in velocity

$$ \begin{aligned} & =\mathbf{V}-\mathbf{v}=\mathrm{v} \sin 45^{\circ}(\mathrm{e}+1) \mathbf{j} \\ & =\mathrm{v} \frac{1}{\sqrt{2}}\left(\frac{1}{\sqrt{3}}+1\right) \mathbf{j} \\ & =\frac{\mathrm{v}}{\sqrt{6}}(1+\sqrt{3}) \mathbf{j} \\ & =\left(\frac{2.732}{\sqrt{6}} \mathrm{v}\right) \mathbf{j} \simeq(1.13 \mathrm{v}) \mathbf{j} \\ & =11.3 \mathrm{~m} / \mathrm{s} \mathbf{j} \end{aligned} $$