Grass-G/jee-main-questions huggingface dataset:
Subject: Physics — Elasticity — Extension in a wire of non-uniform cross-section
Difficulty: Moderate

Question: A slightly tapering wire of length[IMAGE] and end
radii is subjected to tension F. Y is the
Young'sModulus. The extension produced in the wire is.,d FkksM+s Vsifjx rkj ftl rFkk fljks dh
f= ij F ruko 
rkj


Correct option: 1

Solution:
Subject: Physics — Moving charges and — Current carrying conductor in magnetic field
Difficulty: Moderate

Question: A small circular loop of conducting wire has radius a and
carries current I. It is placed in a uniform magnetic field B
perpendicular to its plane such that when rotated slightly about its
diameter and released, it starts performing simple harmonic motion of
time period T. If the mass of the loop is $\\mathrm{m}\$ then

1) $T = \sqrt{\frac{\pi m}{IB}}$
2) $T = \sqrt{\frac{2m}{IB}}$
3) $T = \sqrt{\frac{\pi m}{2IB}}$
4) $T = \sqrt{\frac{2\pi m}{IB}}$

Correct option: 4

Solution:
$\tau = MBsin\theta = I\alpha$
$\pi R^{2}IB\theta = \frac{{mR}^{2}}{2}\alpha$.
$\omega = \sqrt{\frac{2\pi IB}{m}} = \frac{2\pi}{T};T = \sqrt{\frac{2\pi m}{IB}}$
Subject: Physics — Centre of Mass — Variable mass system
Difficulty: Tough

Question: In the fig. shown a cart moves on a smooth horizontal surface
due to an external constant force of magnitude F. The initial mass of
the cart is M_0and velocity is zero. Sand falls on to the cart with
negligible velocity at constant rate and
sticks to the cart. The velocity of the cart at time t is

1) $\frac{Ft}{M_{0} + \mu t}$
2) $\frac{F}{\mu}\mathcal{l}_{n}\frac{m_{0} + \mu t}{m_{0}}$
3) $\frac{Ft}{M_{0}}$
4) $\frac{Ft}{M_{0} + \mu t}e^{\mu t}$

Correct option: 1

Solution:
[IMAGE] Formula $F = m\frac{dv}{dt} + (V - u)\frac{dm}{dt}$
Here, u = velocity of s and = O
m = M_o[IMAGE] t = mass at time t
and [IMAGE] $\therefore$ $F = \left( M_{0} + \mu t \right)\frac{dv}{dt} + v\mu$
$(F - \mu v)dt = \left( M_{0} + \mu t \right)dv$
$\int_{0}^{t}\mspace{2mu}\frac{dt}{M_{0} + \mu t} = \int_{0}^{v}\mspace{2mu}\frac{dv}{F - \mu v}$
$\frac{1}{\mu}\left\lbrack \mathcal{l}n\left( M_{0} + \mu t \right) \right\rbrack_{0}^{t} = - \frac{1}{\mu}\mathcal{\lbrack l}n(F - \mu v)\rbrack_{0}^{v}$
$\mathcal{l}n\frac{\left( M_{0} + \mu t \right)}{M_{0}}\mathcal{= l}n\left\{ \frac{F}{F - \mu v} \right)$
$F - \mu v = \frac{M_{0}F}{M_{0} + \mu t} \Rightarrow v = \frac{Ft}{M_{0} + \mu t}$
Subject: Physics — Center of mass — Collision
Difficulty: Tough

Question: Three particles each of mass m can slide on fixed frictionless
horizontal circular tracks in the same horizontal plane as shown in the
figure. The coefficient of restitution being e=0.5. Assuming that
m_2and m_3 are at rest initially and lie along a radial line before
impact and the spring is initially unstretched, then maximum extension
in spring in subsequent motion.

1) $\frac{3}{4}V_{0}\sqrt{\frac{m}{k}}$
2) $\frac{3}{4}V_{0}\sqrt{\frac{m}{5k}}$
3) $\frac{3}{4}V_{0}\sqrt{\frac{2m}{5}k}$
4) $\frac{3}{5}V_{0}\sqrt{\frac{m}{k}}$

Correct option: 2

Solution:
Velocity just after collision
$V_{0} = V_{1} + V_{2}$.....(1)
$\frac{1}{2} = \frac{- V_{1} + V_{2}}{V_{0}}$
$- V_{1} + V_{2} = \frac{V_{0}}{2}$.....(2)
From (1) & (2)
$\Rightarrow V_{1} = \frac{V_{0}}{4}\& V_{2} = \frac{3V_{0}}{4}$
When maximum extension occurs then angular speed with rest to centre
for mass m_2& m_3 are same. Using angular momentum conservation
about centre of circle.
$m\frac{3}{4}V_{0}(2R) + 0 = {mR}^{2}\omega + m(2R)^{2}\omega$
$\omega = \frac{3}{10}\frac{V_{0}}{R}$
So velocity of$m_{2} = \frac{3}{5}V_{0}$ & velocity
of$m_{3} = \frac{3}{10}V_{0}$
When extension is maximum using energy conservation.
$\frac{1}{2}m{(\frac{3}{4}V_{0})}^{2} = \frac{1}{2}m{(\frac{3}{5}V_{0})}^{2} + \frac{1}{2}m{(\frac{3}{10}V_{0})}^{2} + \frac{1}{2}kx_{max.}^{2} \Rightarrow x_{\max} = \frac{3}{4}V_{0}\sqrt{\frac{m}{5k}}$
Subject: Physics — Projectile Motion — Horizonatal projectile
Difficulty: Moderate

Question: If the projectile hits the inclined plane perpendicularly when
thrown horizontally with $v_{0}$, from a tower of height H as shown then
the value of $\frac{4gH}{{(v_{0})}^{2}}$ is 2k. Find k.


Answer: 5

Solution:
$a_{y} = gcos30^{\circ}$
$u_{x} = V_{0}cos30^{\circ}$
$u_{y} = V_{0}sin30^{\circ}$
$V_{x} = u_{x} + a_{x}t$
$0 = \frac{V_{0}\sqrt{3}}{2} - \frac{g}{2}t$
$t = \frac{v_{0}\sqrt{3}}{g}$....(i)
$S_{y} = u_{y}t + a_{y}t^{2}$
$(Hcos30^{\circ} - 0) = V_{0}sin30^{\circ}t + \frac{1}{2}gcos30^{\circ}t^{2}$....(ii)
From (i) (ii)
$\Rightarrow \frac{4gH}{{(v_{0})}^{2}} = 10$
Subject: Physics — Thermodynamics — Thermodynamic processes
Difficulty: Moderate

Question: An ideal gas obey the law [IMAGE] constant. x is
value for which it has zero molar specific heat at normal temperature.
Then find 5 x. ([IMAGE] for the gas is 2.5 R)
gS ftlds fy, ij bl j fof'k"B A rc 5x
dk eku Kkr djksa ¼xSl


Answer: 7

Solution:
For a gas at normal temperature, ij xSl ds fy,[IMAGE] Specific heat for a polytropic process
i‚ayhVªksfid çØe ds fy, fof'k"B 
[IMAGE] According to question iz"ukuqlkj C=0
Subject: Physics — Rigid Body Dynamics — Moment of inertia
Difficulty: Moderate

Question: A uniform disc of radius R lies in the x-y plane, with its
centre at origin. its moment of inertia about zaxis is equal to its
moment of inertia about line $y = x + c$. The value of c will be

1) $- \frac{R}{2}$
2) $\pm \frac{R}{\sqrt{2}}$
3) $\frac{+ R}{4}$
4) $- R$

Correct option: 2

Solution:
$I_{PQR} = \frac{1}{4}{MR}^{2} + M{(\frac{C}{\sqrt{2}})}^{2}$
But $I_{PQR} = \frac{1}{2}{MR}^{2}$
$\therefore C = \pm \frac{R}{\sqrt{2}}$
Hence (2) is correct.
Subject: Physics — Heat — Thermal expansion
Difficulty: Tough

Question: When the temperature of copper coin is raised by 80ºC, its
diameter increases by 0.2.

1) Percentage rise in the area of a face is 0.4
2) Percentage rise in the volume is 0.6
3) Coefficient of linear expansion of copper is
$0.25 \times 10^{- 4}C^{- 1}$
4) All of the above

Correct option: 4

Solution:
$100 \times \frac{\Delta D}{D} = \alpha\Delta T \times 100$
$100 \times \left( \frac{\Delta A}{A} \right) = \beta\Delta T \times 100$
$= 2\alpha\Delta T \times 100$ $= 2(0.2)$ $= 0.4\%$
$100 \times \frac{\Delta t}{t} = \alpha\Delta T \times 100$
$\alpha = \frac{\frac{\Delta d}{d}}{\Delta T} = \frac{0.2 \times 10^{- 2}}{80}\left( \frac{\Delta d}{d} \times 100 = 0.2 \right)$
$\alpha = 0.25 \times 10^{- 4}{}^{\circ}C^{- 1}$
$100 \times \frac{\Delta V}{V} = \gamma\Delta T \times 100$
$= 3(0.2) = 0.6\%$
Subject: Physics — Communication system — Modulation
Difficulty: Moderate

Question: The maximum peak to peak voltage of an AM wave is 24 mV and the
minimum peak to peak voltage is 8 mV. The modulation factor is

1) 10
2) 20
3) 25
4) 50

Correct option: 4

Solution:
Subject: Physics — Electromagnetic induction — LR circuit
Difficulty: Tough

Question: [IMAGE] As shown in the figure, a battery of emf$\varepsilon$ is connected to
an inductor L and resistance R in series. The switch is closed at t=0.
The total charge that flows from the battery, between t=0 and
$t = t_{c}$ ($t_{c}$is the time constant of the circuit) is

1) $\frac{\varepsilon L}{R^{2}}(1 - \frac{1}{e})$
2) $\frac{\varepsilon R}{eL^{2}}$
3) $\frac{\varepsilon L}{R^{2}}$
4) $\frac{\varepsilon L}{eR^{2}}$

Correct option: 4

Solution:
$q = \int_{0}^{T_{C}}idt$
$= \frac{\varepsilon}{R}{\lbrack t - \frac{e^{- t/T_{C}}}{\frac{- 1}{T_{C}}}\rbrack}_{0}^{T_{C}}$;
$= \frac{\varepsilon}{R}\lbrack T_{C} + T_{C}e^{- 1} - T_{C}\rbrack$
$= \frac{\varepsilon}{R} \times \frac{1}{e} \times \frac{L}{R}$
$= \frac{\varepsilon L}{R^{2}e}$
Subject: Physics —  — 

Question: The primary purpose of using a parabolic mirror in a reflecting
telescope is

1) To collect more light from a distant galaxy
2) To increase the magnification
3) To overcome diffraction effects
4) To correct spherical aberration

Correct option: 4

Solution:
Subject: Physics — Thermodynamics — Adaibaticproces
Difficulty: Moderate

Question: An ideal gas at a pressure of 1 atmosphere andtemperature of 27°C
is compressed adiabaticallyuntil its pressure becomes 8 times the
initialpressure. Then the final temperature is [IMAGE]

1) 627°C
2) 527°C
3) 427°C
4) 327°C

Correct option: 4

Solution:
Here, [IMAGE] As changes are adiabatic,
Subject: Physics —  — 

Question: A steel wire is rigidly fixed at both ends. Its length, mass and
cross sectional area are 1m, 0.01 kg and
[IMAGE] respectively. Then the temperature of the
wire is lowered by[IMAGE]. If the transverse waves are setup by
plucking the wire at [IMAGE] from one end and assuming that wire
vibrates with minimum number of loops possible for such a case. The
frequency of vibration is 44n Hz. Find n. [Coefficient of linear
expansion of steel [IMAGE] and Young's


Answer: 5

Solution:
[IMAGE] The mechanical strain [IMAGE] The tension in wire[IMAGE] speed of wave in wire
[IMAGE] Since the wire is plucked at
[IMAGE] from one end
The wire shall oscillate in
[IMAGE] overtone (for minimum number of loops)
TOPIC:Waves in a string
SUB TOPIC: Wire fixed at both ends
LEVEL:Moderate
Subject: Physics — Kinematics — Relative motion
Difficulty: Moderate

Question: A system is shown in the figure. Block A is moving with 1 m /s
towards left. Wedge is moving with 1 m /s towards right. Then speed of
the block B will be

1) 1 m /s
2) 2 m /s
4) none of these

Correct option: 3

Solution:
Velocity of block A w.r.t. wedge is 2 m So we have
${\overset{\rightarrow}{V}}_{BW} = {\overset{\rightarrow}{V}}_{B} - {\overset{\rightarrow}{V}}_{W} \Rightarrow {\overset{\rightarrow}{V}}_{B} = {\overset{\rightarrow}{V}}_{BW} + {\overset{\rightarrow}{V}}_{W}$
So
$V_{B} = \sqrt{{(V_{W})}^{2} + {(V_{BW})}^{2} + 2V_{W} \times V_{BW}(cos120^{\circ})}$
$= \sqrt{1^{2} + 2^{2} + 2 \times 1 \times 2 \times ( - 1/2)} = \sqrt{3}m/s$
Subject: Physics — Capacitance — Equivalent capacitance
Difficulty: Tough

Question: [IMAGE] A parallel plate capacitor has plates of area A separated by distance
 between them. It is filled with adielectric which has a
dielectric constant that varies as k(x) = K(1 + $\alpha$x) where 
is the distancemeasured from one of the plates. If ($\alpha$d) << 1,
the total capacitance of the system is best given by theexpression:
lekukUrj IysVksa ls cus,d la
muds chp dh nwjh A bu IysVksa ds chp,d ijkoS|qr inkFkZ Hkjk
gqvk gS ftldk ijk oS|qrkad k(x) = K(1+ αx)
mi;qDr eku gksxk

1) $\frac{A \in_{0}K}{d}\left( 1 + \frac{\alpha^{2}d^{2}}{2} \right)b$
2) $\frac{AK \in_{0}}{d}\left( 1 + \frac{\alpha d}{2} \right)$
3) $\frac{A \in_{0}K}{d}\left( 1 + \left( \frac{\alpha d}{2} \right)^{2} \right)$
4) $\frac{AK\epsilon_{0}}{d}(1 + \alpha d)$

Correct option: 2

Solution:
Capacitance of element ?kVd Capacitance of element ?kVd 
$\sum\frac{1}{C^{'}} = \int_{0}^{d}\mspace{2mu}\frac{dx}{K\varepsilon_{0}A(1 + \alpha x)}$
$\frac{1}{C} = \frac{1}{K\varepsilon_{0}A\alpha}\mathcal{l}n(1 + \alpha d)$
Given αd << 1
$\frac{1}{C} = \frac{1}{K\varepsilon_{0}A\alpha}\left( \alpha d - \frac{\alpha^{2}d^{2}}{2} \right)$
$\frac{1}{C} = \frac{d}{K\varepsilon_{0}A}\left( 1 - \frac{\alpha d}{2} \right)$
$C = \frac{K\varepsilon_{0}A}{d}\left( 1 + \frac{\alpha d}{2} \right)$
Subject: Physics — Kinematics — Motion under gravity
Difficulty: Moderate

Question: A ball is thrown vertically upwards from the ground. It crosses a
point at the height of 25 m. twice at an interval of 4 secs. The ball
was thrown with the velocity of


Correct option: 3

Solution:
Let initial velocity of the ball = u
$\therefore 25 = ut - \frac{1}{2}gt^{2} = ut - 5t^{2}$
$\Rightarrow 5t^{2} - ut + 25 = 0$
If t_1and t_2 are the two solutions,
[IMAGE] Then, $t_{2} - t_{1} = \sqrt{{(t_{2} + t_{1})}^{2} - 4t_{1}t_{2}}$
$\Rightarrow 4^{2} = \frac{u^{2}}{25} - 4 \times \frac{25}{5}$
$\Rightarrow 16 = \frac{u^{2}}{25} - 20 \Rightarrow u^{2} = 25 \times 36$
$\Rightarrow u = 30m/s$
Subject: Physics — Thermodynamics — Molar heat capacity
Difficulty: Moderate

Question: A diatomic ideal gas is heated at constant volume until the
pressure is doubled and again heated at constant pressure until volume
is doubled. Find out the average molar heat capacity for whole process.
in K)
nkc nqxuk ugha gks tk;s
rd ml nqxuk ugha gks

1) 28.5
2) 23.5
3) 25.3
4) 23.5

Correct option: 3

Solution:
Let initial pressure, volume, temperature be P_0, V_0, T_0
indicated by state A in P-V diagram. The gas is then isochorically taken
to state B (2P_0, V_0, 2T_0) and then taken from state B to state C
(2P_0, V_0, 4T_0) isobarically
vxj bl:vkrh s P_0, V_0, T_0ekus tSlk
fd uhps P-V fp=k bls fQj levk;rfud izfØ B (2P_0, V_0,
2T_0) vkSj fQj Bls C (2P_0, V_0, 4T_0) (2P_0, V_0, 4T_0) rd
[IMAGE] Total heat absorbed by 1 mole of gas (by definition of molar heat
capacity)
dqy tks 1 eksy xSl }kjk xzg.k dh xbZ &.
$\Delta Q = C_{v}\left( 2T_{0} - T_{0} \right) + C_{p}\left( 4T_{0} - 2T_{0} \right)$
$= \frac{5}{2}RT_{0} + \frac{7}{2}R \times 2T_{0} = \frac{19}{2}{RT}_{0}$
Total change in temperature from state A to C is
 dqy ifjorZu AvoLFkk ls Crd gS &
$\Delta T = 3T_{0}$
Molar heat capacityeksyj m"ek
Subject: Physics — Heat — Thermal expansion
Difficulty: Tough

Question: When the temperature of copper coin is raised by 80oC, its
diameter increases by 0.2.

1) percentage rise in the area of a face is 0.4
2) percentage rise in the volume is 0.6
3) coefficient of linear expansion of copper is
$0.25 \times 10^{- 4}C^{- 1}$
4) All of the above
tc rkacs ds flD eku 80ºC c<

Correct option: 4

Solution:
$100 \times \frac{\Delta D}{D} = \alpha\Delta T \times 100$
$100 \times \left( \frac{\Delta A}{A} \right) = \beta\Delta T \times 100$
$= 2\alpha\Delta T \times 100$
$= 2(0.2)$
$= 0.4\%$
$100 \times \frac{\Delta t}{t} = \alpha\Delta T \times 100$
$\alpha = \frac{\frac{\Delta d}{d}}{\Delta T} = \frac{0.2 \times 10^{- 2}}{80}\left( \frac{\Delta d}{d} \times 100 = 0.2 \right)$
$\alpha = 0.25 \times 10^{- 4}\ ^{\circ}C^{- 1}$
$100 \times \frac{\Delta V}{V} = \gamma\Delta T \times 100$
$= 3(0.2) = 0.6\%$
Subject: Physics — Thermal properties of matter — Thermal expansion
Difficulty: Moderate

Question: A simple seconds pendulum is constructed out of a very thin
string of thermal coefficient of linear expansion
[IMAGE] and a heavy particle attached to one end. The free end of the string is
suspended from the ceiling of an elevator at rest. The pendulum keeps
correct time at 0° C. When the temperature rises to 50° C, the elevator
operator of mass 60 kg being a student of Physics accelerates the
elevator vertically, to have the pendulum correct time. The apparent
weight (in N) of the operator when the pendulum keeps correct time at 50
°C is 330x. Find x. [IMAGE],d lSd.M ljy yksyd cgqr iryh NM+ ls cuk gS] ftldk çlkj
xq.kkad α= 20 x 
0ºC ij lgh Zr tc rki 50 ºC c<+rk gS rks 60kg
 fy¶V pkyd tks HkkSfrd foKku dk Nk=k gS] yksyd lgh
 fy, fy¶V dks Rofjr tc yksyd 50 ºC ij lgh
le; ç gks rks fy¶V pkyd dk vkHkklh Hkkj330x ( )
gksrk


Answer: 2

Solution:
[IMAGE] when the temperature is raised length changes to l(1 + α ∆T)
tc rki c<+ rc yEckbZ esa ifjorZu l(1 + α ΔT) gks A
when the lift acceleration upwards
tc fy¶V dh vksj Rofjr gSA
g_eff = g + a
new period of pendulum,
yksyd
Subject: Physics —  — 

Question: A uniform solid cylinder of mass M = 2 kg and radius R = 10 cm
is connected about an axis through the centre of the cylinder to a
horizontal spring with spring constant 4 N/m. The cylinder is pulled
back, stretching the spring 1 m from equilibrium. When released, the
cylinder rolls without slipping. What is the speed of the center of the
cylinder when it returns to equilibrium ?


Correct option: 3

Solution:
wsp = ∆k
Subject: Physics — Mechanical Properties of Fluids — Capillary action
Difficulty: Moderate

Question: When a capillary tube is dipped in a liquid, the liquid rises to
a height h in the tube. The free liquidsurface inside the tube is
hemispherical in shape. The tube is now pushed down so that the height
ofthe tube outside the liquid is less than h.

1) The liquid will come out of the tube like in a small fountain.
2) The liquid will come out of the tube slowly.
3) The liquid will fill the tube but not come out of its upper end.
4) The free liquid surface inside the tube will not be hemispherical.
tc,d s nzo uyh esa nzo h rd
uhps rkfd nzo ds ckgj uyh

Correct option: 3

Solution:
The angle of contact at the free liquid surface inside the
capillary tube will change such that the verticalcomponent of the
surface tension forces just balance the weight of the liquid column.
 vUnj nzo dh eqDr lrg ij lEidZ dks.k bl çdkj cny tk,xk fd
i`"B ruko cy dk ?kVd nzo LrEHk
Subject: Physics —  — 

Question: The variation of internal energy vs density of a diatomic gas
is as shown in figure. The molar heat capacity of the gas is

4) R

Correct option: 1

Solution:
According to graph
TV = constant
or PV^2 = constant
Subject: Physics — Thermodynamics — Adiabatic process
Difficulty: Moderate

Question: For an adiabatic process graph between PV&V for a sample of
ideal gas will be


Correct option: 2

Solution:
$PV\ \alpha\ T$
 for adiabatic process, 
${TV}^{\gamma - 1} = \ $constant
Subject: Physics — Rigid Body Dynamics — Torque
Difficulty: Moderate

Question: A thin uniform rod of mass is hinged at one
end. This rod is maintained in horizontal position by colliding very
tiny balls each of mass m/10 completely elastically 10 times per sec
striking at the opposite end as shown in figure. Find the speed of the
ball - (g = 10 m/sec^2)


Correct option: 2

Solution:
[IMAGE] Taking torque about hinge:
$MgL/2 = FL$
$mg\frac{L}{2} = 2(\frac{mV}{10})10L$
$V = 2.5m/sec$
Subject: Physics — Wave Optics — YDSE
Difficulty: Moderate

Question: A parallel beam of light of wavelength [IMAGE] is
incident on slit S. Another two slit
distance 2 m as shown in the figure. The minimum distance between
intensity will occur at O is.

2) 1 mm

Correct option: 2

Solution:
[IMAGE], when n = 1
Subject: Physics — Ray Optics — Prism
Difficulty: Moderate

Question: In the shown figure a ray of light enters the face XY of a right
angered prism at grazing incident. It emerges from the adjacent face. YZ
at an angle[IMAGE]. If x is the critical angle of the
prism held in water of refractive index [IMAGE], then the
angle [IMAGE] with normal is


Correct option: 4

Solution:
[IMAGE] As at XY
[IMAGE] At YZ
[IMAGE] On solving
Subject: Physics — Newton's Laws of Motion — Newton's second law
Difficulty: Moderate

Question: System shown in figure is in equilibrium. The magnitude of change
in tension in the string just before and just after, when one of the
spring is cut. Mass of both the blocks is same and equal to m and
springconstant of both springs is k. (Neglect any effect of rotation)


Correct option: 1

Solution:
Before cutting the spring
After cutting the spring
[IMAGE] 2mg - mg = 2 ma
a = g/2
$T_{3} = mg/2$
$T_{2} - T_{3} = mg - mg/2 = mg/2$
Subject: Physics — Ray Optics — Lens maker's formula
Difficulty: Moderate

Question: A point object in air is in front of the curved surface of a
plano-convex lens. The radius of curvature of the curved surface is 30
cm and the refractive index of the lens material is 1.5, then the focal
length of the lens (in cm) is10n. Find n.


Answer: 6

Solution:
Subject: Physics — electronic device — Diodes
Difficulty: Easy

Question: In the circuit, if the forward voltage drop for the diode is
[IMAGE] the current in the circuit will be
(approximately)

1) 3.4 mA
2) 2 mA
3) 2.5 mA
4) 3 mA

Correct option: 1

Solution:
Subject: Physics —  — 

Question: In the given A.C. circuit, if battery voltage
is[IMAGE], power delivered by battery will have a power
factor of[IMAGE]. Find n.


Answer: 3

Solution:
[IMAGE] Phasor diagram
[IMAGE] TOPIC: Alternating current
SUB TOPIC: LCR circuit
LEVEL: Moderate
Subject: Physics — Dual Nature of Matter — Einstein's photoelectric equation
Difficulty: Moderate

Question: the frequency of the light used in an experiment on
photoelectric effect. If the work function is $\frac{x}{42}$ (in eV)
then value of x is. $\left( h = 6.4 \times 10^{- 34}Js \right)$

1) 33.3
2) 33.5
3) 33.6
4) 33.7

Correct option: 3

Solution:
$V = \frac{1.6}{4 \times 10^{14}}\left( v - 2 \times 10^{14} \right) = \frac{1.6}{4 \times 10^{14}}v - 0.8$
Compare it with $V = \frac{h}{e}v - \frac{\phi}{e}$
$\frac{\phi}{e} = 0.8$
$\phi = 0.8eV$
Subject: Physics — Electrostatics — Electric field due to infinite sheet of charge
Difficulty: Moderate

Question: Take two infinite planes with same surface charge density. Place
them as shown it the figure. The electric field in the region between
them is given by

1) $\frac{\sigma}{2\epsilon_{0}}\lbrack(1 + \sqrt{3})\overset{\hat{}}{y} - \frac{\overset{\hat{}}{x}}{2}\rbrack$
2) $\frac{\sigma}{2\epsilon_{0}}\lbrack(1 - \frac{\sqrt{3}}{2})\overset{\hat{}}{y} - \frac{\overset{\hat{}}{x}}{2}\rbrack$
3) $\frac{\sigma}{2\epsilon_{0}}\lbrack(1 + \sqrt{3})\overset{\hat{}}{y} + \frac{\overset{\hat{}}{x}}{2}\rbrack$
4) $\frac{\sigma}{\epsilon_{0}}\lbrack(1 + \frac{\sqrt{3}}{2})\overset{\hat{}}{y} + \frac{\overset{\hat{}}{x}}{2}\rbrack$

Correct option: 2

Solution:
$\overset{\rightarrow}{E} = \frac{\sigma}{2\varepsilon_{0}}\lbrack(1 - \frac{\sqrt{3}}{2})\overset{\hat{}}{y} - \frac{1}{2}\overset{\hat{}}{x}\rbrack$
Subject: Physics — Newton's Laws of Motion — Friction
Difficulty: Moderate

Question: In the figure $m_{A} = m_{B} = m_{C} = 60kg$. The co-efficient
of friction between C and ground is 0.5, B and ground is 0.3, A& B is
0.4. C is pulling the string with the maximum possible force without
moving. Then tension in the string connected to A will be

1) 120 N
2) 60 N
3) 100 N
4) zero

Correct option: 4

Solution:
Maximum frictional force between C and ground =300 Nt
Max. frictional force between B and ground =360 Nt
So man is unable to pull B Hence T=
Subject: Physics — Behaviour of perfect gases — Work done
Difficulty: Moderate

Question: One mole of an ideal gas is kept enclosed under a light piston
(area=[IMAGE] ) connected by acompressed spring (spring
constant 100 N / m). The volume of gas is [IMAGE] and
its temperature is 100K. The gas is heated so that it compresses the
spring further by 0.1 m. Then the work done by the gas in the process
is: (in Joule) (Take R =8.3 J / K -mole and suppose there is no
atmosphere).,d 100 N / m½
ls tqM+s gYds fiLVu ¼kjk
 ¼R =8.3 J ekusa fd ok;qe.My ugha

1) 1.0
2) 1.5
3) 2.0
4) 1.8

Correct option: 2

Solution:
Before heating let the pressure of gas be
[IMAGE] from the equilibrium piston.
 -
[IMAGE] Since during heating process, pwafd xeZ izfØ nkSjku,
The spring is compressed further by 0.1 m fLizax 0.1 m
Subject: Physics — Relative motion — River problem
Difficulty: Moderate

Question: A river is flowing with a speed of $1\ kmh^{- 1}$. A swimmer
wants to go to point c starting from A. He swims with a speed of
$5\ kmh^{- 1}$ at an angle $\theta$w.r.t. the river flow, at what angle
with the river bank should the swimmer swim?

1) 30º
2) 37º
3) 53º
4) 60º

Correct option: 3

Solution:
Path of swimmer relative to ground is a straight line at 45º
with bank, therefore, x and y components of swimmer's resultant
velocity must be equal.
Let velocity of swimmer$|{\overset{\rightarrow}{v}}_{m}| = v$
$V_{x} = V_{r} + V_{m,r}cos\theta$
$V_{y} = V_{m,r}sin\theta$
[IMAGE] Condition for reaching the point C,
$tan45^{\circ} = \frac{v_{y}}{v_{x}}$
$V_{y} = V_{x}$.
$(V_{r} + V_{m,r}cos\theta) = V_{m,r}sin\theta$
$1 + 5cos\theta = 5sin\theta$
On squaring$1 + 25\cos^{2}\theta + 10cos\theta = 25 - 25\cos^{2}\theta$
$50\cos^{2}\theta + 10cos\theta - 24 = 0$
On solving, we get$\theta = 53^{\circ}$