# JEE Mains (Apr 3) 2016 Paper

Given sample data:
$x_1=90,\ x_2=91,x_3=95,x_4=92$
Mean, $\bar x=\dfrac{\sum_1^4 x_i}{4}$
$\bar x=92$

Error in measurement, ${\Delta x}_i=|x_i-\bar{x}|+r$
where $r:$ error due to least count.
$r=0.5$
$\Delta x_1=2, \Delta x_2=1, \Delta x_3=3, \Delta x_4=0$
$\Delta \bar x = \dfrac{\sum_1^4 {\Delta x}_i}{4}$
$\Delta \bar x=\dfrac{2.5+1.5+3.5+0.5}{4}=2$
1
#472810
Solution
A student measures the time period of $100$ oscillations of a simple pendulum four times. The data set is $90$s, $91$s, $95$s and $92$s. If the minimum division in the measuring clock is $1$s, then the reported mean time should be.
A
$92\pm 2$s
B
$92\pm 5.0$s
C
$92\pm 1.8$s
D
$92\pm 3$s
The angular momentum of a particle of mass $m$ with a velocity $\vec{v}$ and position vector $\vec {r}$.
$\vec{L} = m (\vec{r} \times \vec{v})$
The magnitude can be given as, $L = mvr$
From C to D the $\vec{r}$ is $(\dfrac{R}{\sqrt{2}} + a)$
So, $\vec{L}$ from C to D is = $mv (\dfrac{R}{\sqrt{2}}+ a)$
Hence option B is incorrect.
2
#472813
Solution
A particle of mass m is moving along the side of a square of side 'a', with a uniform speed v in the x-y plane as shown in the figure. Which of the following statements is false for the angular momentum $\overrightarrow{L}$ about the origin?
A
$\overrightarrow{L}=\displaystyle -\frac{mv}{\sqrt 2}R$ $\hat{k}$ when the particle is moving from A to B.
B
$\overrightarrow{L}=\displaystyle mv\left[\displaystyle\frac{R}{\sqrt 2}-a\right]\hat{k}$ when the particle is moving from C to D.
C
$\overrightarrow{L}=\displaystyle mv\left[\displaystyle\frac{R}{\sqrt 2}+a\right]\hat{k}$ when the particle is moving from B to C.
D
Both A and B are incorrect
All the potential energy is lost by dissipation due to work done by frictional force.
$P.E =$ Work done by friciton
Work done by friction from P to Q $= \mu mg (\cos\theta) PQ =2\sqrt3 \mu mg$
Work done by friction from Q to R $=\mu mg\times QR$
$2mg = 2\sqrt{3}\mu mg + \mu mg\times QR$
$2 = 2\sqrt 3 \mu + \mu QR$ - (1)
Since, equal energies are lost along $PQ$ and $QR$,
Work done by friction is the same on both path lengths,
$\mu mg \cos \theta PQ = \mu mg QR$
$PQ \cos\theta = QR$ - (2)
From (1) and (2) we get,
$\mu = 0.29, QR = 3.5$
3
#472854
Solution
A point particle of mass m, moves along the uniformly rough PQR as shown in the figure. The coefficient of friction, between the particle and the rough track equals $\mu$. 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 QR, 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 respectively close to.
A
$0.2$ and $6.5$m
B
$0.2$ and $3.5$m
C
$0.29$ and $3.5$m
D
$0.29$ and $6.5$m
The net work done by the man will be $1000$ times the work done in lifting $10 kg$ to a height of $1m$
Net work done $= 1000\times10\times 9.8\times1$  $J$
Let's assume $x kg$ of fat is burnt in doing this work. Energy balance will give the following equation,
$x\times\dfrac{20}{100}\times 3.8\times 10^7 =$ Net work done
$x = 12.8947\times 10^{-3}$ $kg$
4
#472817
Solution
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 $g=9.8ms^{-2}$.
A
$2.45\times 10^{-3}$kg
B
$6.45\times 10^{-3}$kg
C
$9.89\times 10^{-3}$kg
D
$12.89\times 10^{-3}$kg
The image attached here is the front view of the roller. The rails intersect the plane of the paper on both sides where normal reactions are drawn.
We know that the roller does not lift off from the rails.
So, $N_1 \cos\theta \times r_1 = N_2\cos\theta\times r_2$
$\therefore N_1 r_1 = N_2 r_2$
As the roller moves forward, $r_1$ becomes smaller.
So, $N_1$ increases. Hence, $N_1 > N_2$
Net force in the horizontal direction becomes
$N_1 \sin\theta - N_2 \sin\theta = (N_1 - N_2)\sin\theta$ towards the left.
Since, $N_1 > N_2$ net force is towards the left and hence the roller moves towards left.
5
#472820
Solution
A roller is made by joining together two cones at their vertices O. It is kept on two rails AB and CD which are placed asymmetrically(see figure), with its axis perpendicular to CD and its centre O at the centre of line joining AB and CD(see figure). It is given a light push so that it starts rolling with its centre O moving parallel to CD in the direction shown. As it moves, the roller will tend to.
A
Turn left
B
Turn right
C
Go straight
D
Turn left and right alternately
For a satellite in orbit, the velocity can be evaluated from the following equation,
$\dfrac{GMm}{r^2} = \dfrac{mv_o^2}{r}$
$v_o=\sqrt{\dfrac{GM}{r}}$
Here, $r = R+h$

For satellite to escape, total energy of satellite must be zero. Let escape velocity of satellite at height h be $v_e$.
$E = PE + KE =0$
$\dfrac{-GMm}{r}+\dfrac{1}{2}mv_e^2 = 0$
$v_e=\sqrt{\dfrac{2GM}{r}}$

Difference in velocities is given by
$\Delta v = v_e-v_o$
$\Delta v = \sqrt{\dfrac{GM}{r}}(\sqrt 2 -1)$
$\Delta v \approx \sqrt {\dfrac{GM}{R} } (\sqrt 2 -1)$ as $h<<R$

But $g=\dfrac{GM}{R^2}$

$\therefore \Delta v \approx \sqrt {gR}(\sqrt 2 -1)$
6
#472824
Solution
A satellite is revolving in a circular orbit at a height 'h' from the earth's surface(radius of earth R$;$ h$<<$R). The minimum increase in its orbital velocity required, so that the satellite could escape from the earth's gravitational field, is closed to(Neglect the effect of atmosphere.)
A
$\displaystyle\sqrt{2gR}$
B
$\displaystyle\sqrt{gR}$
C
$\displaystyle\sqrt{gR/2}$
D
$\displaystyle\sqrt{gR}(\sqrt 2-1)$
The time period of a pendulum is given by,
$t = 2\pi\sqrt{\dfrac{l}{g}}$
$\Delta t = \dfrac{2\pi}{\sqrt g}\dfrac{\Delta l}{2\sqrt{l}}$
$\Delta l = \alpha \Delta \theta$
$\therefore \Delta t = \dfrac{\pi}{\sqrt{gl}}\alpha\Delta T$
$\therefore \dfrac{\Delta t_1}{\Delta t_2} = \dfrac{\Delta T_1}{\Delta T_2}$
$\therefore \dfrac{12}{4} = \dfrac{40 - T}{T - 20}$
$\therefore 3T - 60 = 40 - T$
$\therefore T = 25^\circ$
7
#472829
Solution
A pendulum clock loses $12$s a day if the temperature is $40^o$C and gains $4$s a day if the temperature is $20^o$C. The temperature at which the clock will show correct time, and the co-efficient of linear expansion $(\alpha)$ of the metal of the pendulum shaft are respectively.
A
$25^o$C$; \alpha=1.85\times 10^{-5}/^o$C
B
$60^o$C $;\alpha=1.85\times 10^{-4}/^o$C
C
$30^o$C$;\alpha=1.85\times 10^{-3}/^o$C
D
$55^o$C$;\alpha=1.85\times 10^{-2}/^o$C
For a polytropic process given as $PV^n = constant$
The specific heat is given as,

$C = \dfrac{R}{\gamma - 1} + \dfrac{R}{1 - n}$

$C = C_v \dfrac{\gamma - n}{1 - n}$

$C = \dfrac{C_p - C_vn}{1 - n}$

$\therefore n = \dfrac{C - C_p}{C - C_v}$
8
#472834
Solution
An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity C remains constant. If during this process the relation of pressure P and volume V is given by $PV^n=$constant, then n is given by (Here $C_P$ and $C_V$ are molar specific heat at constant pressure and constant volume, respectively).
A
$n=\displaystyle\frac{C_P}{C_V}$
B
$n=\displaystyle\frac{C-C_P}{C-C_V}$
C
$n=\displaystyle\frac{C_P-C}{C-C_V}$
D
$n=\displaystyle\frac{C-C_V}{C-C_P}$
We need to write the relation between P and V and maximize the product PV.

$P=-\dfrac { { P }_{ 0 } }{ { V }_{ 0 } } V+3{ P }_{ 0 }$

$PV=(-\dfrac { { P }_{ 0 } }{ { V }_{ 0 } } V+3{ P }_{ 0 })V$

$nRT=PV=-\dfrac { { P }_{ 0 } }{ { V }_{ 0 } } { V }_{ }^{ 2 }+3{ P }_{ 0 }{ V }_{ }$

Differentiating wrt V and putting it equal to 0, we get,

$V=\dfrac { 3{ V }_{ 0 } }{ 2 }$

$P=\dfrac { 3{ P }_{ 0 } }{ 2 }$

Hence, $T=\dfrac { 9{ P }_{ 0 }{ V }_{ 0 } }{ 4nR }$
9
#472836
Solution
'n' moles of an ideal gas undergoes a process A$\rightarrow$B as shown in the figure. The maximum temperature of the gas during the process will be.
A
$\displaystyle\frac{9P_0V_0}{4nR}$
B
$\displaystyle\frac{3P_0V_0}{2nR}$
C
$\displaystyle\frac{9P_0V_0}{2nR}$
D
$\displaystyle\frac{9P_0V_0}{nR}$
In SHM, $V=\sqrt { { A }^{ 2 }-{ x }^{ 2 } }$

$V=\sqrt { { A }^{ 2 }-\dfrac { 4{ A }^{ 2 } }{ 9 } } =\sqrt { \dfrac { 5{ A }^{ 2 } }{ 9 } }$

$3V=\sqrt { { A' }^{ 2 }-\dfrac { 4{ A }^{ 2 } }{ 9 } }$

Dividing the above two equations give $A'=\dfrac { 7A }{ 3 }$
10
#472838
Solution
A particle performs simple harmonic motion with amplitude A. Its speed is tripled at the instant that it is at a distance $\displaystyle\frac{2A}{3}$ from equilibrium position. The new amplitude of the motion is.
$\displaystyle\frac{A}{3}\sqrt{41}$
$\displaystyle 3A$
$\displaystyle A\sqrt 3$
$\displaystyle\frac{7A}{3}$