JEE Mains (Apr 9) 2016 Paper

Torque on a dipole is given by: $\vec \tau= \vec M \times \vec B$
where: $M:$ Magnetic moment, $B:$ Magnetic flux density

It is given that the dipole is in equilibrium. Hence, the configuration of the magnetic field vectors and the magnetic moment is as shown in the figure so that the torque due to the two fields are opposite in direction.

$\tau_1=\tau_2$
${MB}_1\sin(30^o)={MB}_2sin(45^o)$
$B_2=\dfrac{15 \times \frac{1}{2}}{\frac{1}{\sqrt 2}}$
$B_2=10.6\ mT \approx 11\ mT$
1
#474463
Solution
A magnetic dipole is acted upon by two magnetic fields which are inclined to each other at an angle of $75^o$. One of the fields has a magnitude of $15\ mT$. The dipole attains stable equilibrium at an angle of $30^o$ with this field. The magnitude of the other field (in $mT$) is close to:
A
$1$
B
$36$
C
$11$
D
$1060$
Stopping potential is given by $eV_o=\dfrac{hc}{\lambda}-\phi$
where
$h:$ Planck's constant, $c:$ speed of light, $\lambda:$ wavelength, $\phi:$ work function

Hence,
$eV=\dfrac{hc}{\lambda_1}-\phi...........(i)$
$3eV=\dfrac{hc}{\lambda_2}-\phi...............(ii)$

From (i) and (ii),
$\phi=\dfrac{3hc}{2e\lambda_1}-\dfrac{hc}{2e\lambda_2}.........(iii)$

When photons of wavelength $\lambda_3$ is incident, stopping potential is equal to
$eV'=\dfrac{hc}{\lambda_3}-\phi.........(iv)$

From (iii) and (iv),
$V'=-\dfrac{3hc}{2e\lambda_1}+\dfrac{hc}{2e\lambda_2}+\dfrac{hc}{e\lambda_3}$
2
#474468
Solution
When photons of wavelength ${\lambda}_{1}$ are incident on an isolated sphere, the corresponding stopping potential is found to be $V$. When photons of wavelength ${\lambda}_{2}$ are used, the corresponding stopping potential was thrice that of the above value. If light of wavelength ${\lambda}_{3}$ is used then find the stopping potential for this case:
A
$\dfrac { hc }{ e } \left[ \dfrac { 1 }{ { \lambda }_{ 3 } } +\dfrac { 1 }{ { \lambda }_{ 2 } } -\dfrac { 1 }{ { \lambda }_{ 1 } } \right]$
B
$\dfrac { hc }{ e } \left[ \dfrac { 1 }{ { \lambda }_{ 3 } } +\dfrac { 1 }{ {2 \lambda }_{ 2 } } -\dfrac { 3}{ { 2\lambda }_{ 1 } } \right]$
C
$\dfrac { hc }{ e } \left[ \dfrac { 1 }{ { \lambda }_{ 3 } } - \dfrac { 1 }{ { \lambda }_{ 2 } } -\dfrac { 1 }{ { \lambda }_{ 1 } } \right]$
D
$\dfrac { hc }{ e } \left[ \dfrac { 1 }{ { \lambda }_{ 3 } } +\dfrac { 1 }{ 2{ \lambda }_{ 2 } } -\dfrac { 1 }{ { \lambda }_{ 1 } } \right]$
Applying the lens equation, we get

$\dfrac { 1 }{ v } +\dfrac { 1 }{ 10 } =\dfrac { 1 }{ { f }_{ 1 } }$   ... (1)

For mirror,

$\dfrac { 1 }{ 25.4 } -\dfrac { 1 }{ 45.8-32.2-v } =\dfrac { 1 }{ { f }_{ 2 } }$

$\dfrac { 1 }{ 25.4 } -\dfrac { 1 }{ 13.6-v } =\dfrac { 1 }{ { f }_{ 2 } }$    .... (2)

Since there are 3 variables and only two equations (1) and (2) therefore only by substituting $f_{1}$ & $f_{2}$ from options correct values can be verified.
Eliminating $v$ from the equation (1), we get a relation between ${ f }_{ 1 }$ and ${ f }_{ 2 }$. This relation is satisfied only by option A.
3
#474472
Solution
To find the focal length of a convex mirror, a student records the following data:
The focal length of the convex lens is ${f}_{1}$ and that of mirror is ${f}_{2}$. Then taking index correction to be negligibly small, ${f}_{1}$ and ${f}_{2}$ are close to
A
${ f }_{ 1 }=7.8cm\quad \quad { f }_{ 2 }=12.7cm$
B
${ f }_{ 1 }=12.7cm\quad \quad { f }_{ 2 }=7.8cm$
C
${ f }_{ 1 }=7.8cm\quad \quad { f }_{ 2 }=25.4cm$
D
${ f }_{ 1 }=15.6cm\quad \quad { f }_{ 2 }=25.4cm$
$f=\left(\dfrac{c+v_r}{c-v_s}\right)f_o$
where
$f:$ Observed frequency, $f_o:$ source frequency
$c:$ speed of sound
$v_r:$ velocity of the receiver relative to the medium (positive if the receiver is moving towards the source)
$v_s:$ velocity of the source relative to the medium (positive if the source is moving away from the receiver)

$f=\left(\dfrac{330+30}{330-30}\right) \times 540=648\ Hz$
4
#474475
Solution
Two engines pass each other moving in opposite directions with uniform speed of $30 {m}/{s}$. One of them is blowing a whistle of frequency $540 Hz$. Calculate the frequency heard by driver of second engine before they pass each other. Speed of sound is $330 {m}/{sec}$.
A
$648 Hz$
B
$540 Hz$
C
$270 Hz$
D
$450 Hz$
For a pnp transistor, emitter and collector junction is taken from p terminal and base is taken from n terminal.
Multimeter can detect a short circuit (low resistance) and an open circuit (high resistance).
Low resistance is detected when positive of multimeter is connected at p-junction (1 or 3) and negative of multimeter is connected at n-terminal (2). Similarly, high resistance is detected when positive of multimeter is connected at n-junction (2) and negative of multimeter is connected at p-terminal (1 or 3).
Hence, d is the correct option.
5
#474478
Solution
An unknown transistor needs to be identified as a $npn$ or $pnp$ type. A multimeter, with +ve and -ve terminals, is used to measure resistance between different terminals of transistor. If terminal $2$ is the base of the transistor then which of the following is correct for a $pnp$ transistor?
A
+ve terminal 1, -ve terminal 2, resistance high
B
+ve terminal 2, -ve terminal 3, resistance low
C
+ve terminal 3, -ve terminal 2, resistance high
D
+ve terminal 2, -ve terminal 1, resistance high
OR gate has the property that output is one when any one of the inputs is one and zero only when all inputs are zero. From the truth table, it can be seen that the above two properties are satisfied.
6
#474479
Solution
The truth table given in figure represents:
A
AND - Gate
B
NOR - Gate
C
NAND - Gate
D
OR - Gate
Let $L$ be the inductance and $R$ be the resistance then the current $I$ can be found by using Kirchoff's law as:
$\quad L\dfrac { dI }{ dt } -IR={ V }_{ 0 }\sin { \Omega t } \\ \Rightarrow \dfrac { dI }{ dt } -\dfrac { IR }{ L } =\dfrac { { V }_{ 0 }\sin { \Omega t } }{ L } \\ \Rightarrow \dfrac { d }{ dt } (I{ e }^{ \frac { -Rt }{ L } })=\dfrac { { V }_{ 0 }\sin { \Omega t } }{ L } { e }^{ \frac { -Rt }{ L } }\\ \Rightarrow I{ e }^{ \frac { -Rt }{ L } }=\int _{ 0 }^{ t }{ \dfrac { { V }_{ 0 }\sin { \Omega t } }{ L } { e }^{ \frac { -Rt }{ L } } } dt\quad \\ \Rightarrow I=-\dfrac { { V }_{ 0 }L }{ { R }^{ 2 }+{ \Omega }^{ 2 }{ L }^{ 2 } } (\dfrac { R }{ L } \sin { \Omega t } +\Omega \cos { \Omega t } )\\ \Rightarrow I=-\dfrac { { V }_{ 0 } }{ \sqrt { { R }^{ 2 }+{ \Omega }^{ 2 }{ L }^{ 2 } } } \sin { (\Omega t+\phi ) }$
Where, $\tan { \phi } =\dfrac { L\Omega }{ R }$
So, it is clearly seen that the solution is dictated by the forcing function which is sinusoidal. Therefore, the steady state solution will be sinusoidal in nature.
So, the correct answer is option B.
7
#474483
Solution
A series LR circuit is connected to a voltage source with $V\left( t \right) ={ V }_{ 0 }\sin { \Omega t }$. After very large time, current $I\left( t \right)$ behaves as $\left( { t }_{ 0 }\gg \frac { L }{ R } \right)$
A
B
C
D
A microwave oven heats food by passing microwave radiation through it. Water, fat and other substances in the food absorb energy from the microwaves. Many molecules (such as those of water) are electric dipoles, meaning that they have a partial positive charge at one end and a partial negative charge at the other, and therefore rotate as they try to align themselves with the alternating electric field of the microwaves. Rotating molecules hit other molecules and put them into motion, thus dispersing energy. This energy, when dispersed as molecular vibration in solids and liquids (i.e. as both potential energy and kinetic energy of atoms), is heat
Hence, microwave oven acts on the principle of giving rotational energy to the water molecules.
8
#474486
Solution
Microwave oven acts on the principle of:
A
giving rotational energy to water molecules
B
giving vibrational energy to water molecules
C
transferring electrons from lower to higher energy levels in water molecule
D
giving translational energy to water molecules
Since area of triangle $csa$ is $\dfrac{1}{4}$ of total area of ellipse, therefore:
Area of $cdas$ = $\dfrac{1}{3}$Area of $abcs$
Now that from Kepler's second law areal velocities of the planets are constant which essentially means planets cover equal area in equal time interval.
Hence,
Time taken in covering path $abc$ and path $cda$ will be in proportion to their respective enclosed areas.
$\implies$ $t_{1} = 3t_{2}$

9
#474490
Solution
Figure shows elliptical path $abcd$ of a planet around the sun $S$ such that the area of triangle csa is $\dfrac{1}{4}$ the area of the ellipse. (See figure) With $db$ as the  major axis, and $ca$ as the minor axis. If ${t}_{1}$ is the time taken for planet to go over path $abc$ and ${t}_{2}$ for path taken over $cda$ then
A
${t}_{1}=4{t}_{2}$
B
${t}_{1}=2{t}_{2}$
C
${t}_{1}={t}_{2}$
D
${t}_{1}=3{t}_{2}$
For capacitor is series, equivalent capacitance is given by:
$C_{S}=\dfrac{C_1C_2}{C_1+C_2}$

For capacitor is parallel, equivalent capacitance is given by:
$C_{P}=C_1+C_2$

In the given configuration, equivalent capacitance is given by
$C_{eq}=C_s+C$, where $C= 4\ \mu F$
and $C_s=\dfrac{C \times C}{C+C}=\dfrac{C}{2}$

$\therefore C_{eq}=C+\dfrac{C}{2}=6\ \mu F$
10
#474493
Solution
Three capacitors each of $4\ \mu F$ are to be connected in such a way that the effective capacitance is $6\ \mu F$. This can be done by connecting them: