# JEE Mains (Apr 11) 2015 Paper

The gravitational potential by uniform sphere is given by
$V=-\dfrac{GM}{R^3}(3R^2-r^2)\; r<R \\ V=-\dfrac{GM}{r}\ \; r>R$
where $M= \dfrac{4}{3}\pi R^3 \rho$
$V=-\dfrac{4G\pi\rho}{3}(3R^2-r^2)\; r<R \\ V=-\dfrac{4G\pi \rho R^3}{3r}\ \; r>R$
The gravitational potential is minimum at center due to negative sign and Potential inside sphere is parabolic curve and potential outside sphere is hyperbolic curve of distance from center. The potential graph is shown below.
1
#309954
Solution
Which of the following most closely depicts the correct variation of the gravitation potential $V(r)$ due to a large planet of radius $R$ and uniform mass density? (figures are not drawn to scale)
A
B
C
D
Consider small element of length $d\vec{l}$subtending small angle $d\theta$ at origin of wire:
The magnetic force on the element is $F=Id\vec{l}\times \vec{B}= IdlB \hat{r}$ radially outward direction
The resultant of tension must balance magnetic force as wire element is in equilibrium

The component of tension in downward direction $T_{resultant}=2Tsin\dfrac{d\theta}{2}$

$T_{resultant}=F \Rightarrow 2Tsin\dfrac{d\theta}{2}=IdlB$

for small angle $\dfrac{d\theta}{2}$ the value of  $sin \dfrac{d\theta}{2}$ can be approximated as $sin \dfrac{d\theta}{2}= \dfrac{d\theta}{2}$ and $d\theta=\dfrac{dl}{R}$ where R is radius of the arc.

$2T\dfrac{d\theta}{2}=IdlB \Rightarrow T\dfrac{dl}{R} =IdlB$

$T=IBR$.
2
#309955
Solution
A wire carrying current I is tied between points P and Q and is in the shape of a circular arch of radius R due to a uniform magnetic field B (perpendicular to the plane of the paper, shown by xxx) in the vicinity of the wire. If the wire subtends an angle $2\theta_0$ at the centre of the circle (of which it forms an arch) then the tension in the wire is
A
$IBR$
B
$\dfrac {IBR}{sin\theta_0}$
C
$\dfrac {IBR}{2 sin\theta_0}$
D
$\dfrac {IBR\theta_0}{sin\theta_0}$

The energy of pendulum is given as
$E_1=\dfrac{1}{2}KA_1^2=45 J$
$E_2= \dfrac{1}{2}KA_2^2=15 J$
taking ratio of (i) ad (ii)
$\dfrac{A_1}{A_2}= \sqrt{3}$
For damped oscillator  $m\ddot{x}+b\dot{x}+cx=0$
amplitudes is given by $A(t)=A_\circ{}e^{-\dfrac{bt}{2m}}cos(\omega t+\phi)$
$A_1=A_\circ{}e^{-\dfrac{bt}{2m}}cos(\omega t+\phi)$
$A_2=A_\circ{}e^{-\dfrac{b(t+15T)}{2m}}cos(\omega t+\phi)$
The ratio of amplitudes is  $\dfrac{A_2}{A_1}=e^{-\dfrac{b15T}{2m} }$ where $\dfrac{b}{m}$ is damping constant.
t is equal to time taken for 15 oscillation:$t=15 T=15 s$ where T is equal to time period for 1 complete oscillation,
$\dfrac{A_1}{A_2}=e^{\dfrac{b 15}{2m} }$
$\dfrac{15b}{2m}= ln \dfrac{A_1}{A_2}$
$\dfrac{b}{m}=\dfrac{1}{15} ln 3$

3
#309956
Solution
A pendulum with time period of 1s is losing energy due to damping. At certain time its energy is 45 J. If after completing 15 oscillations, its energy has become 15 J, its damping constant (in $s^{-1}$) is:
A
$\frac {1}{2}$
B
$\frac {1}{15}ln 3$
C
$\frac {1}{30}ln 3$
D
2
CM of the rod of length l: $x_{cm} = \frac{ \int_{0}^{L}\mu (x) xdx}{\int_{0}^{L}\mu (x) dx}= \frac{ \int_{0}^{L}( a + \frac{bx}{L}) xdx}{\int_{0}^{L}( a + \frac{bx}{L})dx}= \frac{a\frac{L^2}{2} + \frac{bL^2}{3}}{aL + \frac{bL}{2}}$
$\Rightarrow \frac{7}{12}L = \frac{a\frac{L^2}{2} + \frac{bL^2}{3}}{aL + \frac{bL}{2}}$
$\Rightarrow \frac{7}{4} = \frac{3a +2b}{2a +b}$
$\Rightarrow 14a + 7b = 12a + 8b$
$\Rightarrow 2a =b$
4
#309957
Solution
A uniform thin rod AB of length L has linear mass density $\mu (x)=a+\dfrac {bx}{L}$, where x is measured from A. If the CM of the rod lies at a distance of $\left (\dfrac {7}{12}L\right )$ from A, then a and b are related as
A
$2a=b$
B
$a=2b$
C
$a=b$
D
$3a=2b$
As block is floating it's weight should be equal to buoyancy force
$\rho_{wood} V_{cylinder} g=\rho_{liquid} V_{displaced} g$
$V_{displaced} =\frac{\rho_{wood} V_{cylinder} }{\rho_{liquid}}$
$V_{displaced} =\frac{\rho_{wood} V_{cylinder} }{\rho_{liquid}}$...(i)
After displacing by small distance x, the net force on cylinder will be
$F=Buoyancy-w=\rho_{liquid}( V_{displaced} +A_{cylinder}\Delta x) g- \rho_{wood} V_{cylinder} g$
$F=\rho_{liquid} V_{displaced} g+\rho_{liquid} A_{cylinder}\Delta x g-\rho_{wood} V_{wood} g$ the net force on cylinder becomes
from equation (i) $\rho_{wood} V_{cylinder} g=\rho_{liquid} V_{displaced} g$
$F=\rho_{liquid} A_{cylinder}\Delta x g$
$ma=\rho_{liquid} A_{cylinder}\Delta x g$
$a=\frac{\rho_{liquid} A_{cylinder}}{m}\Delta x$
$a = \omega^2 \Delta x$
$\omega^2 =\frac{\rho_{liquid} A_{cylinder}}{\rho_{cylinder} V_{cylinder}}$
$\omega^2 =\frac{\rho_{liquid} A_{cylinder}}{\rho_{wood} A_{cylinder} h_{cylinder}}$
$\omega^2 =\frac{\rho_{liquid}}{\rho_{wood} h_{cylinder}}$
$\omega^2 =\frac{900}{650\times .54 }$
This should be equal to angular frequency of simple pendulum
$\omega=\sqrt{\frac{g}{l}}$
$\sqrt{\frac{900}{650\times .54}}=\sqrt{\frac{g}{l}}$
$l=g\frac{650\times .54}{900}$
$l=.06\times 65$
$=39 cm$
5
#309959
Solution
A cylindrical block of wood $(density=650 kg m^{-3})$, of base area $30 cm^2$ and height 54 cm, floats in a liquid of density 900 kg $m^{-3}$. The block is depressed slightly and then released. The time period of the resulting oscillations of the block would be equal to that of a simple pendulum of length (nearly)
A
52 cm
B
26 cm
C
39 cm
D
65 cm
Ordinary DC voltmeter cannot measures alternating voltages. When used in AC circuits, these instruments record zero reading because average value of alternating voltage over a full cycle is zero. Alternating voltage across the resistance can be measured by using an AC voltmeter. Such an AC voltmeter is known as a hot wire voltmeter which measure only virtual values(r.m.s values) of the alternating voltages. Moving coil or magnet galvanometer can measure only DC values. When AC passes through a galvanometer it will not show any deflection, because the impulses to the moving parts of the galvanometer are equal and opposite. To measure AC, the heating effect of current is used because heating effect does not depend on the direction of flow of current. The hot wire voltmeter work on the principle of heating effect of current.
6
#309960
Solution
The AC voltage across a resistance can be measured using a
A
moving magnet galvanometer
B
moving coil galvanometer
C
hot wire voltmeter
D
potentiometer
Total current in resistance $R_S$$I = nI_L + I_L = (n+1) I_L$
Voltage across $R_S$ is $V_S = V_i - V_L$

$\therefore R_S =\dfrac{V_S}{I} = \dfrac{V_i - V_L}{(n+1) I_L}$
7
#309961
Solution
The value of the resistor, $R_S$, needed in the DC voltage regulator circuit shown here, equals
A
$(V_i+V_L)/(n+1)I_L$
B
$(V_i-V_L)/n I_L$
C
$(V_i+V_L)/n I_L$
D
$(V_i-V_L)/(n+1)I_L$
Magnetic field due to bar magnet of pole strength of $m$ is given as at equatorial position is given as

$\vec{B}=\dfrac{\mu_\circ{}}{4 \pi}\dfrac{m}{r^3}$

r is distance from middle point of magnet

r= 30 cm= .3 m; $\dfrac{\mu_\circ{}}{{4 \pi}}=10^{-7}$

At neutral point the magnetic field of earth and bar magnet will be equal and opposite, the net magnetic field will be zero at neutral points.

$\vec{B}_{magnet}=\vec{B}_{earth}={10^{-7}}\frac{m}{.3^3}$

$\vec{B}_{earth}=B_{H}$ from diagram

$\Rightarrow m= \dfrac{3.6\times 10^-5\times .3 ^3}{10^-7}$

$m=9.72$ A-m
8
#309962
Solution
A short bar magnet is placed in the magnetic meridian of the earth with north pole pointing north. Neutral points are found at a distance of 30 cm from the magnet on the East - West line, drawn through the middle point of the magnet. The magnetic moment of the magnet in $Am^2$ is close to (Given $\dfrac {\mu_0}{4\pi}=10^{-7}$ in SI units and $B_H=$ Horizontal component of earth's magnetic field $=3.6\times 10^{-5} Tesla.)$
A
4.9
B
14.6
C
19.4
D
9.7
Angle between $-\vec{A}\ \& \ \vec{B}$ is $(\pi - \Delta \theta)$.
$|\vec{B}-\vec{A}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2 |\vec{A}||\vec{B}| cos(\pi - \Delta \theta)}$
$=\sqrt{2|\vec{A}|^2(1 - cos \Delta \theta)}$
$=\sqrt{4|\vec{A}|^2 \sin^2(\Delta \theta /2)}$
$=2|\vec{A}|sin(\Delta \theta /2)$
$\approx |\vec{A}| \Delta \theta$
9
#309963
Solution
A vector $\vec A$ is rotated by a small angle $\Delta \theta$ radians $(\Delta \theta < < 1)$ to get a new vector $\vec B$. In that case $|\vec B-\vec A|$ is
A
$|\vec A|\left (1-\dfrac {\Delta \theta^2}{2}\right )$
B
0
C
$|\vec A|\Delta \theta$
D
$|\vec B|\Delta \theta -|\vec A|$
The dimensions of $\dfrac{\dot{Q}}{A}$ is $\dfrac{J/s}{m^2}=\dfrac{[ML^2T^{-2}]}{L^2T^{-1}}=[MT^{-3}]$
Checking units of options
A)$\eta \dfrac{S\Delta \theta}{h}=[ML^{-1}T^{-1}]\dfrac{[ML^2T^{-2}]}{[M][K]} \dfrac{[K]}{[L]}= [MT^{-3}]$ option A is correct
B) The dimensions of option B can be derived from option A by multiplying dimension of $\dfrac{1}{\rho g}$
$[MT^{-3}]\times \dfrac{1}{[ML^{-3}][LT^{-2}]}= [L^{2}T^{-1}]$
C)  The dimensions of option C can be derived from option B by multiplying dimension of $\eta^2$
$[L^{2}T^{-1}]\times \dfrac{1}{{[ML^{-1}T^{-1}]}^2}= [M^{-2}L^{4}T]$
D) The dimensions of option D can be derived from option C by multiplying dimension of ${\rho g}$
$[M^{-2}L^{4}T]\times [ML^{-3}][LT^{-2}]= [M^{-1}L^2T^{-1}]$
Hence correct answer is option A.
10
#309965
Solution
A beaker contains a fluid of density $\rho \ kg/m^3$, specific heat $S$ J /$kg^oC$ and viscosity $\eta$. The beaker is filled up to height h. To estimate the rate of heat transfer per unit area $(\overset {\cdot}{Q}/A)$ by convection when beaker is put on a hot place, a student proposed that it should depend on $\eta$, $\left (\dfrac {S\Delta \theta}{h}\right )$ and $\left (\dfrac {1}{\rho g}\right )$ when $\Delta \theta$ (in $^oC)$ is the difference in the temperature between the bottom and top of the fluid. In that situation the correct option for $(\overset {\cdot}{Q}/A)$ is
$\eta \dfrac {S\Delta \theta}{h}$
$\eta \left (\dfrac {S\Delta \theta}{h}\right )\left (\dfrac {1}{\rho g}\right )$
$\left (\dfrac {S\Delta \theta}{\eta h}\right )\left (\dfrac {1}{\rho g}\right )$
$\dfrac {S\Delta \theta}{\eta h}$