Grass-G/jee-main-questions huggingface dataset:
Subject: Physics — Behaviour of perfect gases — Ideal gas equation
Difficulty: Moderate

Question: In a cyclic waste power plant for thermal decomposition of waste
hot gases are supplied from a "Kentuki Chamber" where gases are heated
under kentuki process [IMAGE] where P, V and T are
pressure, volume and temperature of gases in chamber and k is positive
constant. Find work done by 5 mole of hot gases when temperature is
raised from 500 K to 550 K:,d pØ IykUV fo?kVu
"dsUVqdh psEcj" ls rIr xSals çnku dh tkrh
çfØ xeZ tgk¡ P, V o T psEcj
eksyksa 500 K ls 550 K rd o`f)

1) 625 R
2) - 625 R
3) 250 R
4) - 250 R

Correct option: 4

Solution:
Subject: Physics — Mechanical Properties of Fluids — Bernoulli's equation
Difficulty: Moderate

Question: Water flows in a horizontal tube (see figure). The pressure of
water changes by 700 Nm-2 between A and B where the area of cross
section are [IMAGE] and [IMAGE], respectively. Find the rate of flow of water
through the tube. (density of water [IMAGE] )

1) $\text{27}\text{3}\text{2}\text{ cm}^{\text{3}}/\text{s}$
2) $2420{cm}^{3}/s$
3) $3020{cm}^{3}/s$
4) $1810{cm}^{3}/s$

Correct option: 1

Solution:
using equation of continuity
$40V_{A} = 20V_{B}$
$\Rightarrow 2V_{A} = V_{B}$
Using Bernoulli's equation
$P_{A} + \frac{1}{2}\rho V_{A}^{2} = P_{B} + \frac{1}{2}\rho V_{B}^{2}$
$\Rightarrow P_{A} - P_{B} = \frac{1}{2}\rho\left( V_{B}^{2} - V_{A}^{2} \right)$
$\Rightarrow \Delta P = \frac{1}{2}1000\left( V_{B}^{2} - \frac{V_{B}^{2}}{4} \right)$
$\Rightarrow \Delta P = 500 \times \frac{3V_{B}^{2}}{4}$
$\Rightarrow V_{B} = \sqrt{\frac{(\Delta P) \times 4}{1500}} = \sqrt{\frac{(700) \times 4}{1500}m/s}$
Volume flow rate $= 20 \times 100 \times V_{B} = 2732{cm}^{3}/s$
Subject: Physics — Rigid Body Dynamics — Conservation of angular momentum
Difficulty: Moderate

Question: A uniform rod of mass 200 gram and length $L = \frac{1}{3}m$ is
initially at rest in vertical position. The rod is hinged at centre such
that it can rotate freely without friction about a fixed horizontal axis
passing through its centre. Two particles of mass m=100 gram each having
horizontal velocity of equal magnitude u=17 m /s strikes the rod at top
and bottom simultaneously as shown and stick to the rod. Find the
angular speed (in rad/s) of rod when it becomes horizontal.

1) 78.50
2) 76.50
3) 77.50
4) 67.50

Correct option: 2

Solution:
From conservation of angular momentum.
$mu\frac{L}{2} + mu\frac{L}{2} = \lbrack 2m\frac{L^{2}}{12} + m\{{(\frac{L}{2})}^{2} + {(\frac{L}{2})}^{2}\}\rbrack\omega$
$muL = \lbrack\frac{{mL}^{2}}{6} + \frac{{mL}^{2}}{4} + \frac{{mL}^{2}}{4}\rbrack\omega = \frac{2{mL}^{2}}{3}\omega$
or $\omega = \frac{3u}{2L} = 76.50$
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.

1) 7
2) 3
3) 5
4) 6

Correct option: 2

Solution:
[IMAGE] Phasor diagram
[IMAGE] TOPIC: Alternating current
SUB TOPIC: LCR circuit
LEVEL: Moderate
Subject: Physics — AC — Power factor
Difficulty: Moderate

Question: In the given A.C. circuit, if battery voltage is
$V = 200\sqrt{2}sin(100\pi t)$, power delivered by battery will have a
power factor of

1) $\frac{1}{2}$
2) $\sqrt{\frac{3}{10}}$
3) $\frac{3}{\sqrt{10}}$
4) $\frac{1}{\sqrt{10}}$

Correct option: 3

Solution:
$i_{1} = \frac{200\sqrt{2}}{\sqrt{30^{2} + 40^{2}}}sin(\omega t - \tan^{- 1}(\frac{40}{30}))$
$= 4\sqrt{2}sin(100\pi t - 53^{\circ})$
Phasor diagram
$\Rightarrow tan\theta = \frac{R_{y}}{R_{x}} = \frac{1}{3} \Rightarrow cos\theta = \frac{3}{\sqrt{10}}$
Subject: Physics — Thermodynamics — Adiabatic process
Difficulty: Moderate

Question: For an adiabatic process graph between PV & V for a sample of
ideal gas will be:
vkn'kZ xSl ds,d uewus dk:)ks"k çØe


Correct option: 2

Solution:
[IMAGE] for adiabatic process,:)ks"k çØe
Subject: Physics —  — 

Question: The positions of
 are shown on the
binding energy curve as shown in figure. The energy released in the
fusion reaction will be closest to [IMAGE] will be
closest to( in MeV )


Answer: 16

Solution:
Q = -2 × 1 - 7 × 5.5 + 8 × 7 = 56 - 2 - 38.5 = 15.5 ]
Subject: Physics — Newton's Laws of Motion — Static friction
Difficulty: Moderate

Question: Initially the block is at rest acceleration of the block is

1) $2\ m/s^{2}$
2) $0\ m/s^{2}$
3) $1\ m/s^{2}$
4) $0.5\ m/s^{2}$

Correct option: 2

Solution:
[IMAGE] $N + 24 - 100 = 0$ For vertical direction
$\therefore N = 76N$
Now, $0 \leq f_{s} \leq \mu_{s}N\ $
$0 \leq f_{s} \leq 76 \times 0.5$
$0 \leq f_{s} \leq 38N$
$\therefore 32 < 38\text{ Hence }f = 32$
$\therefore\ \ $Acceleration of block is zero.
Subject: Physics — Wave Optics — YDSE
Difficulty: Moderate

Question: In the shown figure three slits, separated by
$d = \sqrt{\frac{2\lambda D}{3}}$ and illuminated by monochromatic
parallel beam of light of wavelength $\lambda$. P is a point on the line
perpendicular to the plane of slits through S_3. If I_0 is the
intensity of the wave from each slit, the intensity at p is

1) $I_{0}$
2) ${2I}_{0}$
3) ${4I}_{0}$
4) ${3I}_{0}$

Correct option: 4

Solution:
Phase difference between S_2 &
S_3$\ = \frac{2\pi}{\lambda}(\sqrt{D^{2} + d^{2}} - D)$
$= \frac{2\pi}{\lambda}(\frac{D \times d^{2}}{2D^{2}}) = \frac{2\pi}{3}$
Phase difference between S_1 &
S_3$\ = \frac{2\pi}{\lambda} = (\sqrt{D^{2} + 4d^{2}} - D)$
$= \frac{2\pi}{\lambda} \times D(\frac{4d^{2}}{2D}) = \frac{8\pi}{3}$
[IMAGE] Now using phase diagram
$A_{R} = A\sqrt{3}andIR = 3Io$
Subject: Physics —  — 
Difficulty: Moderate

Question: A stone is dropped from a certain height and can reach the ground
in 5s. But in its fall, the stone is stopped after, 3s of the fall for a
moment and is dropped again at once. Now the stone reaches the ground in
total time of seconds. Find the value of t?
 5s 
 3s 
 
 t


Answer: 07

Solution:
h = [IMAGE] × 10 × 52 = 125 m
h1 = [IMAGE] × 10 × 32 = 45
80 = [IMAGE] × 10 × t2
t = 4 sec. Total time = 3 + 4 = 7 sec
TOPIC:KINEMATIC
SUB TOPIC: MOTION UNDER GRAVITY
Subject: Physics — Current Electricity — Potentiometer
Difficulty: Moderate

Question: In a meter bridge experiment S is a standard resistance. R is a
resistance wire. It is found thatbalancing length is $\mathcal{l}$ = 25
cm. If R is replaced by a wire of half length and half diameter that of
R ofsame material, then the balancing distance is$\mathcal{l}'$ (in cm).
What is the value of $\mathcal{l}^{'}/10$.


Answer: 4

Solution:
$\frac{x}{R} = \frac{75}{25} = 3$
$R = \frac{\rho\mathcal{l}}{A} = \frac{4\rho\mathcal{l}}{\pi d^{2}}$
$R^{'} = \frac{4\rho\left( \frac{\mathcal{l}}{2} \right)}{\pi\left( \frac{d}{2} \right)^{2}} = 2R$
 then $\frac{X}{R^{'}} = \left( \frac{100 - \mathcal{l}}{\mathcal{l}} \right)$
$\frac{100 - \mathcal{l}}{\mathcal{l}} = \frac{X}{2R} = \frac{3}{2}$
$\mathcal{l =}40.00cm$
Subject: Physics — Wave Optics — YDSE
Difficulty: Easy

Question: In a Young's double slit experiment, the separation between the
slits is 0.15 mm. In the experiment, asource of light of wavelength 589
mm is used and the interference pattern is observed on a screen kept1.5
m away. The separation between the successive bright fringes on the
screen is

1) 4.9 mm
2) 3.9 mm
3) 5.9 mm
4) 6.9 mm

Correct option: 3

Solution:
Subject: Physics — Kinetic Theory of Gases — Gases in equilibrium
Difficulty: Tough

Question: A ring shaped tube contains two ideal gases with equal masses
and atomic mass numbers M_1=32 and M_2=28. The gases are separated by
one fixed partition X and another movable conducting partition Y which
can move freely without friction inside the ring. The angle is α, in
equilibrium as shown in thefigure (in degrees). Find the value of
[IMAGE],d vkdkj dh Vîqc esa nks vkn'kZ xSl ftudk æO;eku leku rFkk
ijek.kq æ M_1=32 rFkk M_2=28
foHkktd X ls vyx&vyx dh tkrh gS rFkk nqljk foHkktd,d xfreku pkydfoHkktd
X tks fcuk ?k"kZ.k xfr dj ldrk gSA fp=kuqlkj
dks.k α ¼fMxzh esa½ rks [IMAGE]


Answer: 4

Solution:
At equilibrium, pressure and temperature are same. rFkk rki leku gSA
 Also, let cross-sectional area of tube be A and radius of ring be r.
ekuk nkc
Subject: Physics — EMI — Induced current in a loop
Difficulty: Moderate

Question: The magnetic field is increasing at a constant rate$\alpha$
through a circular loop of radius aandthe axis of magnetic field
coincides with axis of loop as shown in figure. Theresistance per unit
length of loop wire is$\rho$. Choose correct option(s).

1) Current in loop PQRS is $\frac{a\alpha}{2\rho}$ anticlockwise.
2) Current in the loop PQRS is $\frac{a\alpha}{2\rho}$ clock wise.
3) Current in wire PR is $\frac{\pi a\alpha}{\rho}$.
4) Current in wire PR is$\frac{\pi a\alpha}{2\rho}$.

Correct option: 1

Solution:
Magnitude of Induced emf = $A\frac{dB}{dt}$
$\varepsilon = \pi a^{2}\alpha$
$I_{PQRS\ } = \frac{a\alpha}{2\rho}$
I_PR=0
Subject: Physics — Collision — Inelastic collision
Difficulty: Moderate

Question: A ball of mass m is suspended from a fixed point by an
inextensible string. A small particle of same mass m, moving downward at
an angle 37º with the vertical, collides head on with v_0 velocity.
Coefficient of restitution is 4/5.

1) Velocity of ball after impact is [IMAGE]
2) Velocity of ball after impact is
$\left( \frac{25}{34} \right)\text{v}_{\text{0}}$
3) Impulse produced by tension is m v_0[IMAGE]
4) Impulse produced by tension is [IMAGE]

Correct option: 3

Solution:
Say V_1 is velocity attained by ball after collision
and $\frac{4}{5} = \frac{\frac{3V_{1}}{5} + V}{V_{0}}$
$V_{1} = \left( \frac{27}{34} \right)v_{0}:V = \frac{11}{34}v_{0}$
Impulse produced by tension $= mv_{0}cos37^{\circ} + mvcos37^{\circ}$
Subject: Physics — Electronic devices — Diodes
Difficulty: Moderate

Question: In the figure, potential difference (in V) between A and B is
2n. Find n:
 fp=k


Answer: 5

Solution:
Diode is in forward bias, so it will behave as simple wire so,
 voLFkk vr;g lk
[IMAGE] So vc $V_{ab} = \frac{30}{5 + 10} \times 5 = 10V$
Subject: Physics — Kinetic theory of gases — Root mean squared speed
Difficulty: Moderate

Question: Two moles of hydrogen are mixed with n moles of helium. The root
mean square speed of gas molecules in the mixture is $\sqrt{2}$ times
the speed of sound in the mixture. Then n is.
gkbMªkstu ds nks eksyksa n eksyksa ds lkFk

1) 3
2) 2
3) 1.5
4) 2.5

Correct option: 2

Solution:
$V_{rms} = \sqrt{\frac{3RT}{M_{mix}}}$
$V_{sound\ } = \sqrt{\frac{\gamma RT}{M_{mix}}}$
$V_{rms} = \sqrt{2}V_{sound\ }$
$\sqrt{\frac{3RT}{M_{mix}}} = \sqrt{2}\sqrt{\frac{\gamma RT}{M_{mix}}}$
$\gamma_{mix} = \frac{3}{2}$
$\gamma_{mix} = \frac{n_{1}C_{P_{1}} + n_{2}C_{P_{2}}}{n_{1}C_{v_{1}} + n_{2}C_{v_{2}}}$
$\frac{3}{2} = \frac{2 \times \frac{7R}{2} + n \times \frac{5R}{2}}{2 \times \frac{5R}{2} + n \times \frac{3R}{2}} \Rightarrow \frac{3}{2} = \frac{14 + 5n}{10 + 3n} \Rightarrow 30 + 9n = 28 + 10n \Rightarrow n = 2$
Subject: Physics —  — 

Question: A reflecting surface is represented by the equation
[IMAGE] A ray travelling horizontally becomes vertical after reflection. The
co-ordinates of points where this ray incident is


Correct option: 4

Solution:
The slope of curve at such points is [IMAGE] or [IMAGE] TOPIC: Ray Optics
SUB TOPIC:Reflection by mirror
LEVEL:Moderate
Subject: Physics — Current Electricity — Meter bridge
Difficulty: Moderate

Question: In a meter bridge experiment S is a standard resistance. R is a
resistance wire. It is found that balancing length is
$\mathcal{l =}25cm$. If R is replaced by a wire of half length and half
diameter that of R of same material, then the balancing distance
$\mathcal{l}'$ (in cm) will now be_______.

1) 40.00
2) 41.00
3) 42.50
4) 40.50

Correct option: 1

Solution:
$\frac{X}{R} = \frac{75}{25} = 3$
$R = \frac{\rho\mathcal{l}}{A} = \frac{4\rho\mathcal{l}}{\pi d^{2}}$
$R^{'} = \frac{4\rho(\frac{\mathcal{l}}{2})}{\pi{(\frac{d}{2})}^{2}} = 2R$
Then $\frac{X}{R^{'}} = (\frac{100 - \mathcal{l}}{\mathcal{l}})$
$\frac{100 - \mathcal{l}}{\mathcal{l}} = \frac{X}{2R} = \frac{3}{2}$
$\mathcal{l =}40.00cm$
Subject: Physics —  — 

Question: In the circuit shown, equivalent resistance between A and B is
nR,then value of n is


Answer: 02

Solution:
Subject: Physics — Collision — Projectile
Difficulty: Moderate

Question: A particle of mass m is projected with a speed u from the ground
at an angle [IMAGE] horizontal (x-axis). When it has
reached its maximum height, it collides completely inelastically with
another particle of the same mass and velocity[IMAGE].
The horizontal distance covered by the combined mass before reaching the
ground is


Correct option: 1

Solution:
[IMAGE] so horizontal range after collision[IMAGE]
Subject: Physics — Wave Optics — YDSE
Difficulty: Moderate

Question: The distance between two slits in a Young's double slit
experiment is 3 mm. The distance of the screen from the slits is 1 m.
Microwaves of wavelength 1 mm are incident on the plane of the slits
normally. The distance (in cm) of the first maxima on the screen from
the central maxima is $\frac{100}{\sqrt{n}}$. Find n.


Answer: 8

Solution:
At first maxima [IMAGE]
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 liquid surface inside the tube is
hemispherical in shape. The tube is now pushed down so that the height
of the 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 be hemispherical.

Correct option: 3

Solution:
The angle of contact at the free liquid surface inside the
capillary tube will change such that the vertical component of the
surface tension forces just balance the weight of the liquid column.
Subject: Physics — Collision — Perfectly inelastic collision
Difficulty: Moderate

Question: A bullet of mass m hits the hanging block of mass M at its centre
completely inelastically with a velocity v and gets embedded in it as
shown, then choose the correct option

1) The tension in the string just before and after the collision changes
by$\frac{m^{2}v^{2}}{(m + M)^{\mathcal{l}}}$
2) The tension in the string just before and after the collision changes
by
$$\left\lbrack g + \right\rbrack m$$
3) Tension in the string just after the collision
becomes$m\left\lbrack g + \frac{mv^{2}}{(m + M)^{\mathcal{l}}} \right\rbrack$
4) Tension in the string will not change after the collision

Correct option: 2

Solution:
As T_1 = Mg
$T_{2} = (M + m)g + \frac{m^{2}v^{2}}{(M + m)^{\mathcal{l}}}$
$\therefore T_{2} - T_{1} = m\left\lbrack g + \frac{mv^{2}}{(m + M)^{\mathcal{l}}} \right\rbrack$
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 — Thermodynamics — Thermodynamic processes
Difficulty: Moderate

Question: Select the correct alternative for an ideal gas

1) The change in internal energy in a constant pressure process from
temperature $T_{1}\ to\ T_{2}\ $is
equal to$\ nC_{v}\left( T_{2} - T_{1} \right)$ where $C_{v}\ $is the
molar specific heat at constant volume and n the number of moles of
the gas.
2) The change in internal energy of the gas and the work done by the
gas are equal in magnitude in an adiabatic process.
3) The internal energy does not change in an isothermal process and No
heat is added or removed in an adiabatic process.
4) All of the above

Correct option: 4

Solution:
$\Delta U = n\frac{f}{2}R\left( T_{2} - T_{1} \right) = nC_{v}\left( T_{2} - T_{1} \right)$
 For adiabatic process $Q_{12} = 0$
$Q_{12} = \Delta U + W_{12}$
$W_{12} = - \Delta U$
For Isothermal process Temperature is constant.
So, change internal energy is zero.
Subject: Physics — Gravitation — Conservation of momentum
Difficulty: Moderate

Question: A body A of mass m is moving in a circular orbit of radius R
about a planet. Another body B of mass [IMAGE] collides
with A with a velocity which is half [IMAGE] the
instantaneous velocity [IMAGE] of A. The collision
iscompletely inelastic. Then, the combined body

1) Escapes from the planet's Gravitational field
2) Continues to move in a circular orbit
3) Falls vertically downwards towards the planet
4) Starts moving in a elliptical orbit around the planet

Correct option: 4

Solution:
Conserving momentum
[IMAGE] thus the combined mass will go on to an
elliptical path
Subject: Physics —  — 

Question: Find the resultant focal length (in cm ) for following system
where the common radius of curvature is 15 cm. Glass has refractive
index of 1.5 and water has refractive index.


Answer: 40

Solution:
Subject: Physics — Capacitance — Dielectric strength
Difficulty: Moderate

Question: A fiber sheet with relative permittivity 5 is inserted between
two parallel metal plates 0.25 cm apart. Apotential difference of 2500 v
is applied between the plates. If dielectric strength of air is 3 ×
10^6 v/mthan.,d Qkbcj dh ifV~Vdk ftl fo|qr'khyrk 5 gS 0-25 lseh-
fo|qr foHko 2500 v o ok;q

1) Air will not breakdown
ok;q
2) Air will breakdown
ok;q
3) Potential gradient strength in fiber sheet exceeds dielectric
strength of air
Qkbcj ifV~Vdk
4) All the above are correct
lHkh fodYi lgh gSA

Correct option: 2

Solution:
As D;ksafd V = V_1 + V_2
2500 = 0.0002 E_1 + 0.0023 E_2
2E_1 + 23 E_2 = 25 x 10^6
Also rFkk; E_1 = 5E_2
Solving gy E_2 = 0 757 x 10^6 V/m
E_1 = 3.788 x 10^6 V/m
As E_1> 3.6 x 10^6 Hence Air will breakdown
Subject: Physics — Electrostatics — Electric field due to uniformly charged sphere
Difficulty: Tough

Question: Consider a sphere of radius R which carries a uniform charge
density$\rho$. If a sphere of radius $\frac{R}{2}$is carved out of it,
as shown, the ratio
$|\frac{{\overset{\rightarrow}{E}}_{A}}{{\overset{\rightarrow}{E}}_{B}}|$
of magnitude of electric field
${\overset{\rightarrow}{E}}_{A}$and${\overset{\rightarrow}{E}}_{B}$,
respectively, at points A and B due to the remaining portion is

1) $\frac{18}{34}$
2) $\frac{21}{34}$
3) $\frac{17}{54}$
4) $\frac{18}{54}$

Correct option: 1

Solution:
For a solid sphere
$E = \frac{\rho r}{3\varepsilon_{0}}$
$E_{A} = \frac{- \rho R}{2(3\varepsilon_{0})}$
$|E_{A}| = \frac{\rho R}{6\varepsilon_{0}}$
Electric field at point$B = E_{B} = E_{1A} + E_{2A}$
$E_{1A}$Electric Field Due to solid sphere of radius R at point
B$= \frac{\rho R}{3\varepsilon_{0}}$
$E_{2A}$Electric Field Due to solid sphere of radius $R/2$ (which having
charge density $- \rho$)
$= - \frac{\rho R}{54\varepsilon_{0}}$
$E_{B} = E_{1A} + E_{2A} = \frac{\rho R}{3\varepsilon_{0}} - \frac{\rho R}{54\varepsilon_{0}} = \frac{17\rho R}{54\varepsilon_{0}}$
$\frac{|E_{A}|}{|E_{B}|} = \frac{9}{17}$
Subject: Physics — Centre of Mass — Center of mass at rest in the horizontal direction
Difficulty: Moderate

Question: A bead can slide on a smooth straight wire and a particle of mass
m attached to the bead by a lightstring of length L. The particle is
held in contact with the wire and with the string taut and is then let
fall.
If the bead has mass 2 m then when the string makes an angle $\theta$
with the wire, the bead will haveslipped a distance

1) $L(1 - cos\theta)$
2) $\ (L/2)\ (1 - cos\theta)$
3) $(L/3)(1 - cos\theta)$
4) $(L/6)(1 - cos\theta)$

Correct option: 3

Solution:
Since the centre of mass will not move
2mx=my$\Rightarrow$y=2x
x + y = $L - L\cos\theta = L\left( 1 - \cos\theta \right)$
$3x = L(1 - cos\theta)$
⇒ $x = \frac{L(1 - cos\theta)}{3}$
Subject: Physics — Magnetics — Cross field
Difficulty: Tough

Question: A charged particle of mass moving under
the influence of uniform electric field $E\overset{\rightarrow}{i}$and a
uniform magnetic field $B\overset{\rightarrow}{k}$follows a trajectory
from point P to Q as shown in figure. The velocitiesat P and Q are
respectively,
$v\overset{\rightarrow}{i}$and$- 2v\overset{\rightarrow}{j}$. Then which
of the following statements (A, B, C, D) are the correct? (Trajectory
shown is schematic and not to scale)

1) $E = \frac{3}{4}(\frac{mv^{2}}{qa})$
2) Rate of work done by the electric field at P
is$\frac{3}{4}\left( \frac{{mv}^{3}}{a} \right)$and at Q is zero.
3) The difference between the magnitude of angular momentum of the
particle at P and Q is 2mav
4) . All of the above

Correct option: 4

Solution:
(A) by work energy theorem
$0 + {qE}_{0}2a = \frac{3}{2}{mv}^{2}$
$$E = 
(B) Rate of work done at A= power of electric force
$= {qE}_{0}V$
$= \frac{3}{4}\frac{{mv}^{3}}{a}$
(C) at$Q,\frac{dW}{dt} = 0$for both forces
(D)
$\Delta\overset{\rightarrow}{L} = ( - m2v2a\overset{\hat{}}{k}) - ( - mva\overset{\hat{}}{k})$
$|\Delta\overset{\rightarrow}{L}| = 3mva$
Subject: Physics — Kinematics — Uniform accelerated motion
Difficulty: Moderate

Question: Consider a particle initially moving with a velocity 7 m/s
starts decelerating at a constant rate of$\ 2m/s^{2}$.
The distance travelled in the fourth second is$\frac{x}{10}m$. Find
the value of x.


Answer: 5

Solution:
Here particle is decelerating so velocity of particle becomes
 after t=3.5 sec. the particle has a turning point at t=3.5 sec.
V=U+at
$0 = 7 - 2t$
$t = 3.5\sec$
$y_{2} = 7 \times 3.5 - \frac{1}{2} \times 2 \times (3.5)^{2} = \frac{49}{4}m$
$y_{2} - x_{1} = \frac{49}{4} - 12 = \frac{1}{4}m$.
Due to symmetry displacement of the particle at t=3 and t=4sec. is same.
so total distance traveled in the fourth sec
is$= \frac{1}{4} + \frac{1}{4} = \frac{1}{2}m$
Subject: Physics — Rigid Body Dynamics — Torque
Difficulty: Moderate

Question: The rod is uniform & weigh 500 N. Find magnitude of tension so
that tension in both strings (S_1, S_2) aresame (strings ideal).

1) 1000 N
2) 1500 N
3) 2000 N
4) 500 N

Correct option: 2

Solution:
[IMAGE] Taking torque about W.
- T_1 (0.4)+T_2 (0.3)+500(0.2)=0
T = 1000
Subject: Physics —  — 

Question: A Young's double slit experiment is conducted in water
[IMAGE] as shown in the figure, and a glass plate of
thickness t and refractive index [IMAGE] is placed in the
path of[IMAGE]. The magnitude of the phase difference at
O is[IMAGE]. Find n. (Assume that
 is the wavelength of light in air). O is
symmetrical w.r.t. [IMAGE] and [IMAGE].


Correct option: 3

Solution:
Here path difference will be:
[IMAGE] TOPIC:Waves Optics
SUB TOPIC:Optical path difference
LEVEL:Moderate
Subject: Physics — Mechanical properties of liquid — Buoyancy
Difficulty: Moderate

Question: For a fluid which is flowing steadily, the level in the vertical
tubes is best represented by


Correct option: 1

Solution:
From continuity equation, velocity at cross-section (1) is more
than that at cross-section (2).
Hence $P_{1} < P_{2}$
Hence option (1) is correct.
Subject: Physics —  — 

Question: The P-V diagram of a cyclic thermodynamic process for an
ideal gas is as shown in figure. Its P-T curve will be



Solution:
A
Process A → B is isochoric process, therefore, options (C) and (D)
cannot be correct. Again B → C is an adiabatic process, therefore, B → C
should be described by equati­on [IMAGE] constant.
As γ is always greater than
Subject: Physics — Thermodynamics — Thermodynamic cycle
Difficulty: Moderate

Question: $\ $Figure below shows two paths that may be taken by a gas to go
from a state A to a state C.
[IMAGE] In process AB, 400J of heat is added to the system and in process BC,
100J of heat is added to the system.The heat absorbed by the system in
the process AC will be

1) 500 J
2) 460 J
3) 300 J
4) 380 J

Correct option: 2

Solution:
For a complete cycle
$Q_{cycle\ } = W_{cycle\ }$
$+ 400 + 100 + Q_{C \rightarrow A} = \frac{1}{2}\left( 2 \times 10^{- 3} \right)\left( 4 \times 10^{4} \right)$
$\Rightarrow \ \ \ \ Q_{C \rightarrow A} = - 460J$
$\Rightarrow \ \ \ \ Q_{A \rightarrow C} = + 460J$