*Simulator***Previous Years AIEEE/JEE Main Questions**

Q. The potential energy function for the force between two atoms in a diatomic molecule is approximately given by $\mathrm{U}(\mathrm{x})=\frac{\mathrm{a}}{\mathrm{x}^{12}}-\frac{\mathrm{b}}{\mathrm{x}^{6}}$ , where a and b are constant and x is the distance between the atoms. if the dissociation energy of the molecule is $\left[\mathrm{U}(\mathrm{x}=\infty)-\mathrm{U}_{\text {at equilibrium }}\right], \mathrm{D}$ is :
( 1)$\frac{b^{2}}{6 a}$
( 2)$\frac{\mathrm{b}^{2}}{2 \mathrm{a}}$
(3) $\frac{b^{2}}{12 a}$
(4) $\frac{\mathrm{b}^{2}}{4 \mathrm{a}}$

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**Sol.**(4) $\mathrm{U}(\mathrm{x}=\infty)=0,$ at equilibrium $\frac{\mathrm{d} \mathrm{U}}{\mathrm{dx}}=0 ;-\frac{12 \mathrm{a}}{\mathrm{x}^{13}}+\frac{6 \mathrm{b}}{\mathrm{x}^{7}}=0 \Rightarrow \mathrm{x}^{6}=\frac{2 \mathrm{a}}{\mathrm{b}}$

Q. At time t = 0s particle starts moving along the x-axis. If its kinetic energy increases uniformly with time ‘t’, the net force acting on it must be proportional to :-
(1) $\sqrt{t}$
(2) constant
(3) t
(4) $\frac{1}{\sqrt{t}}$

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**Sol.**(4) $\frac{1}{2} \mathrm{mv}^{2}=\mathrm{kt}$ $\mathrm{v}=\sqrt{\frac{2 \mathrm{kt}}{\mathrm{m}}}$ $\frac{\mathrm{d} \mathrm{v}}{\mathrm{dt}}=\sqrt{\frac{\mathrm{k}}{2 \mathrm{mt}}}$ $\mathrm{F}=\mathrm{m} \cdot \frac{\mathrm{d} \mathrm{v}}{\mathrm{dt}}$ $\mathrm{F} \propto \frac{1}{\sqrt{\mathrm{t}}}$

Q.

**This question has Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements.**If two springs $\mathrm{S}_{1}$ and $\mathrm{S}_{2}$ of force constants $\mathrm{k}_{1}$ and $\mathrm{k}_{2}$, respectively, are stretched by the same force, it is found that more work is done on spring $\mathrm{S}_{1}$ than on spring $\mathrm{S}_{2}$.**Statement-1:**If stretched by the same amount, work done on $\mathrm{S}_{1}$, will be more than that on $\mathrm{S}_{2}$**Statement-2 :**$\mathrm{k}_{1}$ < $\mathrm{k}_{2}$ (1) Statement-1 is true, Statement-2 is true and Statement-2 is not the correct explanation of Statement-1. (2) Statement-1 is false, Statement-2 is true (3) Statement-1 is true, Statement-2 is false (4) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation of statement-1.**[AIEEE-2012]****Download eSaral App for Video Lectures, Complete Revision, Study Material and much more...**

**Sol.**(2) Given same force $\mathrm{F}=\mathrm{k}_{1} \mathrm{x}_{1}=\mathrm{k}_{2} \mathrm{x}_{2}$ $\Rightarrow \frac{\mathrm{k}_{1}}{\mathrm{k}_{2}}=\frac{\mathrm{x}_{2}}{\mathrm{x}_{1}}$ $\mathrm{W}_{1}=\frac{1}{2} \mathrm{k}_{1} \mathrm{x}_{1}^{2} \& \mathrm{W}_{2}=\frac{1}{2} \mathrm{k}_{2} \mathrm{x}_{2}^{2}$ As $\frac{\mathrm{W}_{1}}{\mathrm{W}_{2}}>1$ so $\frac{\frac{1}{2} \mathrm{k}_{1} \mathrm{x}_{1}^{2}}{\frac{1}{2} \mathrm{k}_{2} \mathrm{x}_{2}^{2}}>1$ $\Rightarrow \frac{\mathrm{Fx}_{1}}{\mathrm{Fx}_{2}}>1 \Rightarrow \frac{\mathrm{k}_{2}}{\mathrm{k}_{1}}>1$ $\therefore \mathrm{k}_{2}>\mathrm{k}_{1}$ statement 2 is true OR if $\mathrm{x}_{1}=\mathrm{x}_{2}=\mathrm{x}$ $\frac{\mathrm{W}_{1}}{\mathrm{W}_{2}}=\frac{\frac{1}{2} \mathrm{K}_{1} \mathrm{x}^{2}}{\frac{1}{2} \mathrm{K}_{2} \mathrm{x}^{2}}=\frac{\mathrm{K}_{1}}{\mathrm{K}_{2}}$ $\therefore \frac{\mathrm{W}_{1}}{\mathrm{W}_{2}}=\frac{\mathrm{K}_{1}}{\mathrm{K}_{2}}<1$ $\therefore \mathrm{W}_{1}<\mathrm{W}_{2}$ statement 1 is false

Q. When a rubber-band is stretched by a distance x, it exerts a restoring force of magnitude $\mathrm{F}=\mathrm{ax}+\mathrm{bx}^{2}$ where a and b are constants. The work done in stretching the unstretched rubber-band by L is:-
(1) $\frac{\mathrm{aL}^{2}}{2}+\frac{\mathrm{bL}^{3}}{3}$
(2) $\frac{1}{2}\left(\frac{\mathrm{aL}^{2}}{2}+\frac{\mathrm{b} \mathrm{L}^{3}}{3}\right)$
(3) aL $^{2}+\mathrm{bL}^{3}$
(4) $\frac{1}{2}\left(\mathrm{aL}^{2}+\mathrm{bL}^{3}\right)$

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**Sol.**(1) Work done $=\int_{0}^{\mathrm{L}} \mathrm{Fdx}$ $=\int_{0}^{\mathrm{L}}\left(\mathrm{ax}+\mathrm{bx}^{2}\right) \mathrm{d} \mathrm{x}$ $=\frac{\mathrm{aL}^{2}}{2}+\frac{\mathrm{bL}^{3}}{3}$

Q. A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up? Fat supplies $3.8 \times 10^{7}$ J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 $\mathrm{ms}^{-2}$ :-
(1) $12.89 \times 10^{-3} \mathrm{kg}$
(2) $2.45 \times 10^{-3} \mathrm{kg}$
(3) $6.45 \times 10^{-3} \mathrm{kg}$
(4) $9.89 \times 10^{-3} \mathrm{kg}$

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**Sol.**(1) Work done against gravity = (mgh) 1000 in lifting 1000 times $=10 \times 9.8 \times 10^{3}$ $=9.8 \times 10^{4}$ Joule $20 \%$ efficiency is to converts fat into energy. $\left[20 \% \text { of } 3.8 \times 10^{7} \mathrm{J}\right] \times(\mathrm{m})=9.8 \times 10^{4}$ (Where m is mass) $\mathrm{m}=12.89 \times 10^{-3} \mathrm{kg}$

Q. A point particle of mass, moves along the uniformly rough track PQR as shown in the figure. The coefficient of friction, between the particle and the rough track equals . The particle is released, from rest, from the point P and it comes to rest at a point R. The energies, lost by the ball, over the parts, PQ and PR, of the track, are equal to each other, and no energy is lost when particle changes direction from PQ to QR. The values of the coefficient of friction $\mu$ and the distance x(=QR) are, respecitvely close to :

**[JEE-Mains-2016]**
Q. A body of mass m $=10^{-2}$ kg is moving in a medium and experiences a frictional force F = $-\mathrm{K} \mathrm{V}^{2}$. Its intial speed is $\mathrm{v}_{0}$ = 10 $\mathrm{ms}^{-1}$. If, after 10 s, its energy is , the value of k will be :-
(1) $10^{-4} \mathrm{kg} \mathrm{m}^{-1}$
(2) $10^{-1} \mathrm{kg} \mathrm{m}^{-1} \mathrm{s}^{-1}$
(3) $10^{-3} \mathrm{kg} \mathrm{m}^{-1}$
(4) $10^{-3} \mathrm{kg} \mathrm{s}^{-1}$

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**Sol.**(1) $\frac{1}{2} \mathrm{mv}_{\mathrm{f}}^{2}=\frac{1}{8} \mathrm{mv}_{0}^{2}$ $\mathrm{v}_{\mathrm{f}}=\frac{\mathrm{v}_{0}}{2}=5 \mathrm{m} / \mathrm{s}$ $\left(10^{-2}\right) \frac{d V}{d t}=-k v^{2}$ $\int_{10}^{5} \frac{\mathrm{dv}}{\mathrm{v}^{2}}=-100 \mathrm{k} \int_{0}^{10} \mathrm{dt}$ $\frac{1}{5}-\frac{1}{10}=100 \mathrm{k}(10)$ $\mathrm{k}=10^{-4}$

Q. A time dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the work done by the force during the first 1 sec. will be :
(1) 9 J (2) 18 J (3) 4.5 J (4) 22 J

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**Sol.**(3) F = 6t = ma $\Rightarrow \mathrm{a}=6 \mathrm{t}$ $\Rightarrow \frac{\mathrm{dv}}{\mathrm{dt}}=6 \mathrm{t}$ $\int_{0}^{\mathrm{v}} \mathrm{d} \mathrm{v}=\int_{0}^{1} 6 \mathrm{t} \mathrm{dt}$ $\mathrm{v}=\left(3 \mathrm{t}^{2}\right)_{0}^{1}=3 \mathrm{m} / \mathrm{s}$ From work energy theorem $\mathrm{W}_{\mathrm{F}}=\Delta \mathrm{K.E}=\frac{1}{2} \mathrm{m}\left(\mathrm{v}^{2}-\mathrm{u}^{2}\right)$ $=\frac{1}{2}(1)(9-0)=4.5 \mathrm{J}$

Q. A particle is moving in a circular path of radius a under the action of an attractive potential $\mathrm{U}=-\frac{\mathrm{k}}{2 \mathrm{r}^{2}} .$ Its total energy is :-.
(1) $\frac{\mathrm{k}}{2 \mathrm{a}^{2}}$
(2) Zero
(3) $-\frac{3}{2} \frac{\mathrm{k}}{\mathrm{a}^{2}}$
$(4)-\frac{\mathrm{k}}{4 \mathrm{a}^{2}}$

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**Sol.**(2) $\mathrm{F}=-\frac{\partial \mathrm{u}}{\partial \mathrm{r}}=\frac{\mathrm{K}}{\mathrm{r}^{3}}$ Since it is performing circular motion $\mathrm{F}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\frac{\mathrm{K}}{\mathrm{r}^{3}}$ $\mathrm{mv}^{2}=\frac{\mathrm{K}}{\mathrm{r}^{2}}$ $\Rightarrow \mathrm{K.E.}=\frac{1}{2} \mathrm{mv}^{2}=\frac{\mathrm{K}}{2 \mathrm{r}^{2}}$ Total energy = P.E. + K.E. $=-\frac{\mathrm{K}}{2 \mathrm{r}^{2}}+\frac{\mathrm{K}}{2 \mathrm{r}^{2}}=\mathrm{Zero}$

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Previous year question

Physics work , energy and power .

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questions are pretty good and solutions seems comfortable thanks if it helped me in my exams

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Stained coffee cup

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Sleep, where you lay

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Waking up

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Powerless

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Got the taste on my tongue

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And the heat where you laid

I could stay right here and burn in it all day

Waking up

Beside you, I’m my loaded gun

I can’t contain this anymore

I’m all yours, I’ve got no control

No control

Powerless

And I don’t care, it’s obvious

I just can’t get enough of you

The pedal’s down, my eyes are closed

No control

Lost my senses

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Her…

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