# JEE Mains (Apr 10) 2016 Paper

Flux through the loop,
$\phi=\vec B. \vec S$
$\phi =B_0\pi r^2e^{-t/\tau}$
$\varepsilon =\cfrac{d\phi}{dt}=\cfrac{B_0\pi r^2}{T}e^{-t/r}$
Heat $\displaystyle Q=\int^{\infty}_0\cfrac{\varepsilon ^2}{R}dt$
$\displaystyle Q= \int^{\infty}_0 \cfrac{B_o^2\pi^2r^4e^{-2t/\tau}}{R}$
$=\left[ \cfrac{B_o^2 \pi^2 r^4 e^{-2t/\tau}}{2R} \right]_{t=0}^{\infty}$
$=\cfrac{\pi^2r^4B_o^2}{2\tau R}$
1
#474764
Solution
A conducting metal circular-wire-loop of radius r is placed perpendicular to a magnetic field which varies with time as $B =B_0e^{-t/\tau}$, where $B_0$ and $\tau$ are constants, at time t = 0. If the resistance of the loop is R then the heat generated in the loop after a long time $(t \rightarrow \infty)$ is
A
$\cfrac{\pi^2r^4B^4_0}{2\tau R}$
B
$\cfrac{\pi^2r^4B^2_0}{2\tau R}$
C
$\cfrac{\pi^2r^4B^2_0R}{\tau}$
D
$\cfrac{\pi^2r^4B^2_0}{\tau R}$
Considering a spherical gaussian surface and applying Gauss law,
$E(4\pi { r }^{ 2 })=\dfrac { \int _{ 0 }^{ r }{ \rho (4\pi { r }^{ 2 })dr } }{ { \epsilon }_{ 0 } }$

$E=\cfrac { \int _{ 0 }^{ r }{ \rho { r }^{ 2 }dr } }{ { \epsilon }_{ 0 }{ r }^{ 2 } }$

It is given that $\dfrac{dV}{dr}$ is constant.
But $E=-\dfrac{dV}{dr}$
$\implies E$ is constant
$\implies \int \rho r^2 dr \propto r^2$
$\implies \rho r^2 \propto r$
$\implies \rho \propto \dfrac{1}{r}$
2
#474770
Solution
Within a spherical charge distribution of charge density $\rho(r)$, N equipotential surfaces of potential $V_0, V_0 + \Delta V, V_0 + 2\Delta V, ..... ,V_0 + N\Delta V\,\, (\Delta V > 0)$, are drawn and have increasing radii $r_0, r_1, r_2,.....r_N$, respectively. If the difference in the radii of the surfaces is constant for all values of $V_0$ and $\Delta V$ then :
A
$\rho (r) =$ constant
B
$\rho(r)\propto \dfrac{1}{r^2}$
C
$\rho(r) \propto \dfrac{1}{r}$
D
$\rho(r)\propto 1$
$Y=\dfrac{Fl}{\pi r^2\Delta l}$

$= \dfrac{50\times \pi \times 10^3\times 1}{\pi \times (0.005)^2\times 0.01\times 10^{-3}}$

$=2\times 10^{14}N/m^2$
3
#474776
Solution
A thin 1 m long rod has a radius of 5 mm.1 A force of 50 $\pi kN$ is applied at one end to determine its Young's modulus. Assume that the force is exactly known. If the least count in the measurement of all lengths is 0.01 mm, which of the following statements is false ?
A
The maximum value of Y that can be determined is $10^{14} N/m^2$.
B
$\frac{\Delta Y}{Y}$ gets minimum contribution . from the uncertainty in the length.
C
$\frac{\Delta Y}{Y}$ gets its maximum contribution from the uncertainty in strain
D
The figure of merit is the largest for the length of the rod.
For just complete rotation,
$v=\sqrt{Rg}$ at top point
$\Rightarrow w=\dfrac{v}{R} = \sqrt{\dfrac{g}{R}}=\sqrt{\dfrac{10}{1.25}}$
$\omega (rpm)=\dfrac{60}{2\pi}\sqrt{\dfrac{10}{1.25}}=27$
4
#474780
Solution
Concrete mixture is made by mixing cement, stone and sand in a rotating cylindrical drum. If the drum rotates too fast, the ingredients remain stuck to the wall of the drum and proper mixing of ingredients does not take place. The maximum rotational speed of the drum in revolutions per minute(rpm) to ensure proper mixing is close to : (Take the radius of the drum to be 1.25 in and its axle to be horizontal)
A
27.0
B
0.4
C
1.3
D
8.0
The small signal model of a BJT is shown in the figure. From this figure, ${ R }_{ out }={ r }_{ o }$ since the voltage across B and E will be 0 and hence current will only flow through ${ r }_{ o }$.

Input resistance will be ${ R }_{ in }={ r }_{ \pi }$ since current will flow only through ${ r }_{ \pi }$.

Hence, $R=\dfrac { { R }_{ out } }{ { R }_{ in } } =\dfrac { { r }_{ o } }{ { r }_{ \pi } }$

Since ${ r }_{ o }$ is very large and of the order of $100k\Omega$ and ${ r }_{ \pi }$ is of the order of$1-10k\Omega$, the ratio $R$ will be of the order of $100-1000$.
5
#474784
Solution
The ratio (R) of output resistance $r_0$, and and the input resistance $r_i$ in measurements of input and output characteristics of a transistor is typically in the range :
A
$R\sim 10^2 - 10^3$
B
$R\sim 1 - 10$
C
$R\sim 0.1 - 1.0$
D
$R\sim 0.1 - 0.01$
Electromagnetic waves travel in the direction  perpendicular to electric as well as magnetic field. Cross product of electric and magnetic field should give the direction of electromagnetic wave.
$\overset{\rightarrow}{ E } \times \overset{\rightarrow}{B}$ = $\frac{1}{\sqrt{2}}E_{yz} (x, t) (\hat{y}-\hat{z})$$\times \frac{1}{\sqrt{2}}E_{yz} (x, t) (\hat{y}+\hat{z})$ = $E_{yz}^{2} (x, t)$ $\hat{x}$
6
#474793
Solution
Consider an electromagnetic wave propagating in vacuum. Choose the correct statement :
A
For an electromagnetic wave propagating in +y direction the 1 A electric field is $\overset{\rightarrow}{ E } = \frac{1}{\sqrt{2}}E_{yz} (x, t) \hat{z}$ and the magnetic field is $\overset{\rightarrow}{B} =\frac{1}{\sqrt{2}} B_z (x, t) \hat{y}$
B
For an electromagnetic wave propagating in +y direction the electric field is $\overset{\rightarrow}{ E } = \frac{1}{\sqrt{2}}E_{yz} (x, t) \hat{y}$ and the magnetic field is $\overset{\rightarrow}{B} =\frac{1}{\sqrt{2}} B_z (x, t) \hat{z}$
C
For an electromagnetic wave propagating in +x direction the electric field is $\overset{\rightarrow}{ E } = \frac{1}{\sqrt{2}}E_{yz} (y, z, t) (\hat{y}+\hat{z})$ and the magnetic field is $\overset{\rightarrow}{B} = \frac{1}{\sqrt{2}}E_{yz} (y, z, t) (\hat{y}+\hat{z})$
D
For an electromagnetic wave propagating in +x direction the electric field is $\overset{\rightarrow}{ E } = \frac{1}{\sqrt{2}}E_{yz} (x, t) (\hat{y}-\hat{z})$ and the magnetic field is $\overset{\rightarrow}{B} = \frac{1}{\sqrt{2}}E_{yz} (x, t) (\hat{y}+\hat{z})$
Velocity is changing as per, $v =-5t +50$
$v_{t=2}=40$
$\Delta K.E=W.D=\dfrac{1}{2}(40^2-50^2)\times 10$
$=-4500\ J$
7
#474803
Solution
Velocity-time graph for a body of mass 10 kg is shown in figure. Work-done on the body in first two seconds of the motion is :
A
- 9300J
B
12000 J
C
- 4500 J
D
- 12000 J
$\displaystyle x_{cm}=\dfrac{x}{2} = \dfrac{(\rho x)\left(\dfrac{x}{2}\right)\dfrac{1}{2}+\rho y \,\dfrac{y}{2}}{\rho(x+y)}$

$\Rightarrow \dfrac{1}{2}+\dfrac{y}{x}=\dfrac{y^2}{x^2}$

$\Rightarrow \dfrac{y}{x} = \dfrac{1+\sqrt{3}}{2}=1.37$
8
#474806
Solution
In the figure shown ABC is a uniform wire. If centre of mass of wire lies vertically below point A, then $\frac{BC}{AB}$ is close to:
A
1.85
B
1.5
C
1.37
D
3
$F=\dfrac{R}{t^2}v(t)\\ \Rightarrow m\dfrac{dv}{dt}=\dfrac{R}{t^2}v(t)$

$\Rightarrow \int\dfrac{dv}{v}=\int\dfrac{Rdt}{mt^2}$

$ln \ v=-\dfrac{R}{mt}$

$\Rightarrow$ $ln \ v\propto \dfrac{1}{t}$

So, the plot of $\log v(t)$ against $\dfrac{1}{t}$ will be a straight line. Thus plotting will be most convenient.
9
#474814
Solution
A particle of mass $m$ is acted upon by a force $F$ given by the empirical law $F = \dfrac{R}{t^2}\,v(t)$. If this law is to be tested experimentally by observing the motion starting from rest, the best way is to plot :
A
$\log v(t)$ against $\dfrac{1}{t}$
B
$v(t)$ against $t^2$
C
$\log v(t)$ against $\dfrac{1}{t^2}$
D
$\log v(t)$ against $t$
Reading one $\Rightarrow$ without slab
Reading two $\Rightarrow$ with slab
Reading three $\Rightarrow$ with sawdust
10
#474817
Solution