Grass-G/jee-main-questions huggingface dataset:
Subject: Physics — Mechanical Properties of Fluids — Bernoulli's equation
Difficulty: Moderate

Question: In the figure shown cross-section area of holes are same and
very small in comparison to area of cross-section of water tank:
(tanks are the open from the top)

1) $x_{1}:x_{2}:x_{3} = \sqrt{5}:\sqrt{6}:\sqrt{3}$
2) $\ x_{1}:x_{2}:x_{3} = \sqrt{6}:\sqrt{5}:\sqrt{3}$
3) $x_{1}:x_{2}:x_{3} = \sqrt{3}:\sqrt{6}:\sqrt{5}$
4) $x_{1}:x_{2}:x_{3} = \sqrt{2}:\sqrt{5}:\sqrt{3}$

Correct option: 1

Solution:
For case -1
$\rho gh = 1/2\rho v_{1}^{2}$
$V_{1} = \sqrt{2gh}$
$t_{1} = \sqrt{(10h/g)}$
For case -2:
$\rho g2h + 2\rho gh = \frac{1}{2}2\rho v_{2}^{2}$
$V_{2} = \sqrt{4gh}$
$t_{2} = \sqrt{\frac{6h}{g}}$
For case-3:
$\rho g2h + 2\rho gh + 3\rho gh = \frac{1}{2}3\rho v_{3}^{2}$
$V_{3} = \sqrt{6gh}$
$t_{3} = \sqrt{\frac{2h}{g}}$
$x_{1} = v_{1}t_{1} = \sqrt{20}$ $x_{2} = \sqrt{24}$ $x_{3} = \sqrt{12}$
$x_{1}:x_{2}:x_{3} = \sqrt{20}:\sqrt{24}:\sqrt{12}$
$x_{1}:x_{2}:x_{3} = \sqrt{10}:\sqrt{12}:\sqrt{6}$
$x_{1}:x_{2}:x_{3} = \sqrt{5}:\sqrt{6}:\sqrt{3}$
Subject: Physics — Dual Nature of Matter — Einstein's photoelectric equation
Difficulty: Moderate

Question: Radiation, with wavelength 656.1 nm falls on a metal surface to
produce photoelectrons. The electrons are made to enter a uniform
magnetic field of[IMAGE]. If the radius of the largest
circular path followed by the electrons is[IMAGE], the
work function of the metal is close to

1) 1.8 eV
2) 0.8 eV
3) 1.6 eV
4) 1.1 eV

Correct option: 2

Solution:
Subject: Physics — Wave Optics — Diffraction
Difficulty: Moderate

Question: In a single slit diffraction of different light of wavelength
[IMAGE] by a slit of with t, the size of the central
maximum on a screen at a distance[IMAGE] is


Correct option: 3

Solution:
[IMAGE] Width of central Max. [IMAGE]
Subject: Physics — Mechanical Properties of Fluids — Hydrostatic pressure
Difficulty: Moderate

Question: Two immiscible liquids are poured in a U-tube having densities
[IMAGE]. If the ratio of heights (of the liquids
above their interface)
[IMAGE] is x, then find the value of x.
U-ufy nks vfeJ.kh; nzo ftuds ?kuRoρ_1= 1.0 × 10^3
kg/m^3rFkkρ_2= 3.0 × 10^3 kg/m^3gSA;fn (nksuks nzoks ds vUr i`"B ls) nzoksa dh vuqikr $\frac{h_{1}}{h_{2}}$dk eku
x


Answer: 3

Solution:
Subject: Physics — Error analysis — Maximum percentage error
Difficulty: Moderate

Question: A wire has a mass $0.3\ \pm \ 0.003$ g, radius
$0.5\ 0 \pm 0.005$cm and length $6 \pm 0.06$cm. What is the maximum
percentage error in the measurement of density?


Answer: 4

Solution:
$\frac{d\rho}{A} = \frac{dm}{m} + 2\frac{dr}{r} + \frac{dl}{l}$
$\%\frac{d\rho}{A} = \left( \frac{0.003}{0.3} + 2*\frac{0.005}{0.5} + \frac{0.06}{6} \right)*100 = 4\%$
Subject: Physics —  — 
Difficulty: Moderate

Question: A non-viscous fluid of density [IMAGE] is
flowing in a tube as shown in figure. Area of section

1) is double
that of section
2) . Centre of mass of section
3) Work done by gravitational force per unit volume from
section
4) Work done by elastic forces (pressure) per unit volume from
section

Correct option: 2

Solution:
Applying Bernoulli is equation from section-(1) and (2)
[IMAGE] and $P_{1} - P_{2} = \rho g(2h)$
Solving we get, [IMAGE] (3) Work done by gravitation force per unit volume = negative of the
increase in gravitational potential energy per unit volume =
$- (\rho gh)$
(4) Work done by elastic force per unit volume = decrease in
pressure = [IMAGE] = $2\rho gh$
TOPIC: Fluid dynamics
SUB TOPIC: Bernoulli's theorem
Subject: Physics — Collision — Perfectly inelastic collision
Difficulty: Moderate

Question: Two particles of equal mass m have respective initial
velocities[IMAGE] and [IMAGE]. They collide
completely inelastically. The energy lost in the process is


Correct option: 4

Solution:
Conserving momentum
[IMAGE] On solving
Subject: Physics — Capacitance — Distribution of charge on plates
Difficulty: Moderate

Question: The charge that will flow to earth when only switch S1 is closed
and right most plate is uncharged is

1) $- \left( \frac{Q}{2} \right)$
2) $\left( \frac{Q}{2} \right)$
3) Q
4) - Q

Correct option: 3

Solution:
At point D,$\overrightarrow{E}$must be zero because potential
of left most plate is zero.
So, x = 0
${\overrightarrow{E}}_{A} = 0$ (Conductor)
x = y + Q
y = - Q
Subject: Physics — Electrostatics — Gauss law
Difficulty: Moderate

Question: If finding the electric field using gauss law the
formula$|\overset{\rightarrow}{E}| = \frac{q_{enc}}{\varepsilon_{0}|A|}$is
applicable. In the formula $\varepsilon_{0}$ is permittivity of free
space, A is the area of Gaussian surface and q_enc is charge enclosed
by the Gaussiansurface. This equation can be used in which of the
following situation?

1) Only when the Gaussian surface is an equipotential surface.
2) Only when $|\overset{\rightarrow}{E}| =$ constant on the
surface.
3) Only when the Gaussian surface is an equipotential
surface and $\left| \overset{\rightarrow}{E} \right|$is constant of
the surface.
4) For any choice of Gaussian surface.

Correct option: 3

Solution:
Magnitude of electric field is constant & the surface is
equipotential
Subject: Physics — Wave Optics — YDSE
Difficulty: Moderate

Question: In YDSE arrangement, white light is used to illuminate the slits.
At point on the screen directly in front of slits, which of these
wavelengths is missing?
([IMAGE] = wavelength of light used)
(d = distance between slits)
(D = distance between slits and screen)
(Given d<<D)

2) $\frac{d^{2}}{6D}$
3) $\frac{d^{2}}{4D}$

Correct option: 1

Solution:
Wavelength missing is condition for destructive superposition
Subject: Physics — Ray Optics — Apparent shift
Difficulty: Moderate

Question: A vessel of depth 2h is half filled with a liquid of refractive
index[IMAGE] and the upper half with another liquid of
refractive index[IMAGE]. The liquids are immiscible. The
apparent depth of the inner surface of thebottom of vessel will be

1) $\frac{h}{3\sqrt{2}}$
2) $\frac{3}{4}h\sqrt{2}$
3) $\frac{h}{\sqrt{2}}$
4) $\frac{h}{2(\sqrt{2} + 1)}$

Correct option: 2

Solution:
$d = \frac{h}{\sqrt{2}} + \frac{h}{2\sqrt{2}}$
Subject: Physics — EM wave — Relation $\overset{\rightarrow}{E}$ =
Difficulty: Moderate

Question: A plane electromagnetic wave of frequency 25 GHz is propagating
in vacuum along the z-direction. At aparticular point in space and time,
the magnetic field is given
by$\overset{\rightarrow}{B} = 5 \times 10^{- 8}\overset{\hat{}}{j}T$.The
corresponding electric field E is (Speed of light
c = 3 × 10^8ms^-1)

1) $1.66 \times 10^{- 16}\overset{\hat{}}{i}V/m$
2) $15\overset{\hat{}}{i} \vee /m$
3) $- 15\overset{\hat{}}{i} \vee /m$
4) $- 1.66 \times 10^{- 16}\overset{\hat{}}{i} \vee /m$

Answer: 2

Solution:
$\frac{E}{B} = c$
$\overset{\rightarrow}{E}$ = $\overset{\rightarrow}{B}$ ×
$\overset{\rightarrow}{c}$
= $15\overset{\hat{}}{i} \vee /m$
Subject: Physics —  — 

Question: Diagrams show readings of a micrometer screw gauge. Figure (i)
shows the zero error reading when the screw gauge is closed, figure (ii)
the reading when the screw gauge is being used to measure the diameter
of a ball-bearing. What is the diameter of the ball-bearing ? There are
50 divisions on circular scale.

1) 5.29 mm
2) 5.26 mm
3) 5.32 m
4) 5.28 mm

Correct option: 2

Solution:
Zero error is 3 division
LC = [IMAGE] = 0.01 mm
zero error = 0.03 mm
Reading = 5.0 + 29 × 0.01
d = 5.29 - 0.03 = 5.26 mm
SECTION - B [INTEGER ANSWER section contains
Subject: Physics — Kinematics — Acceleration displacement graph
Difficulty: Moderate

Question: The acceleration-displacement graph of a particle moving in a
straight line is as shown in figure. Initial velocity of particle is
zero. What is the velocity (in m /s) of the particle when displacement
of the particle is s=16 m.


Answer: 8

Solution:
$v\ dv = a\ ds$
$\therefore\int_{0}^{v}vdv = \int_{0}^{12m}ads$
$\therefore\frac{v^{2}}{2} = area\ under\ a - s\ graph\ froms = 0\ to\ s = 16m$
$= 2 + 12 + 6 + 12$
$= 32m/s$
or$v = \sqrt{64}m/s = 8m/s$
Subject: Physics — Nuclear Physics — Radioactivity
Difficulty: Moderate

Question: The activity of a radioactive sample falls from
[IMAGE] to [IMAGE] in 30 minutes. Its half
life is close to

1) 62 min
2) 66 min
3) 72 min
4) 52 min

Correct option: 1

Solution:
$\Rightarrow \text{\quad}\frac{ln2}{ln(7/5)} = \frac{t_{\frac{1}{2}}}{(30\min)} \Rightarrow \text{\quad}(2.06004)30 = t_{\frac{1}{2}} = 61.8\text{min}$
Subject: Physics — Error analysis — Percentage error
Difficulty: Easy

Question: A physical quantity A is related to fourobservation a, b, c and
d as follows$A = \frac{a^{2}b^{3}}{c\sqrt{d}}$
The percentage errors of measurement in a, b, c and d are 1, 3, 2 and 2 respectively. Whatis the percentage error in the quantity A ?

1) 12
2) 7
3) 5
4) 14

Correct option: 4

Solution:
$\frac{dA}{A} = 2\frac{da}{a} + 3\frac{db}{b} + \frac{dc}{c} + \frac{1}{2}\frac{dd}{d}$
$\%\frac{dA}{A} = 2*1 + 3*3 + 1*2 + \frac{1}{2}*2 = 14\%.$
Subject: Physics — Current Electricity — Potentiometer
Difficulty: Moderate

Question: The internal resistance of cell E2

1) $7.5\ \Omega$
2) $8\ \Omega$
3) $10\ \Omega$
4) $2\ \Omega$

Correct option: 1

Solution:
[IMAGE] V_AB = 2
$i^{'} = \frac{0.5}{r + 5}$
$0.5 - i^{'}r$
$= \frac{V_{AB}}{50} \times 5$
$0.5 - \frac{0.5}{r + 5}r = \frac{2}{10}$
$0.5 - 0.2 = \frac{0.5r}{r + 5}$
0.3 r +1.5=0.5 r
0.2r = 1.5
r = 7.5 $\Omega$
Subject: Physics —  — 
Difficulty: Easy

Question: 100 m 
 XY.: (Y) )


Correct option: 3

Solution:
Range=[IMAGE] Rmax =[IMAGE] The speed becomes minimum at the highest point.
H =[IMAGE] = 100[IMAGE] =25 m.
Hence, ([IMAGE] ) are the coordinates.
TOPIC:KINEMATIC
SUB TOPIC: 2 D
Subject: Physics — Heat — Heat transfer by conduction
Difficulty: Moderate

Question: Three metal rods of the same material an identicalin all
respects are joined as shown in the figure. The temperatures at the ends
are maintained asindicated. Assuming no loss of heat from thecurved
surfaces of the rods, the temperature at thejunction X inºCis 10n. Find
n.


Answer: 6

Solution:
Let T be the temperature at the junction.
Let L and A be the length and area of cross-sectionof each rod
respectively.
[IMAGE] $\therefore$ Heat current from Y to X is 
$H_{1} = \frac{KA\left( 90^{\circ}C - T \right)}{L}$
$H_{2} = \frac{KA\left( 90^{\circ}C - T \right)}{L}$
Heat current from X to W is
$H_{3} = \frac{KA\left( T - 0^{\circ}C \right)}{L}$
 At the junction X 
$H_{1} + H_{2} = H_{3}$
$\therefore 90^{\circ}C - T + 90^{\circ}C - T = T$
or $3T = 180^{\circ}C\ or\ T = 60^{\circ}C$
Subject: Physics — Motion in 1D — Non uniform acceleration
Difficulty: Tough

Question: A point moves in a straight line under the retardation av^2,
where is a positive constant and v is speed. If the initial speed
is u, the distance covered in seconds is

1) $\ a\ u\ t\ $
2) $\frac{1}{a}ln(aut)$
3) $\frac{1}{a}ln(1 + aut)$
4) $\ a\ In\ (aut)$

Correct option: 3

Solution:
The retardation is given by
$\frac{dv}{dt} = - av^{2}$
integrating between proper limits
$\Rightarrow - \int_{u}^{v}\mspace{2mu}\frac{dv}{v^{2}} = \int_{0}^{t}\mspace{2mu} adt$or$\frac{1}{v} = at + \frac{1}{u}$
$\Rightarrow \frac{dt}{dx} = at + \frac{1}{u} \Rightarrow dx = \frac{udt}{1 + aut}$
integrating between proper
limits$\Rightarrow \int_{0}^{s}\mspace{2mu} dx = \int_{0}^{t}\mspace{2mu}\frac{udt}{1 + aut} \Rightarrow S = \frac{1}{a}\ ln\ (1 + \ aut\ )$
Subject: Physics — Wave Optics — Limit of resolution
Difficulty: Moderate

Question: The aperture diameter of a telescope is 5 m. The separation
between the moon and the earth is [IMAGE]. With light of
wavelength of[IMAGE], the minimum separation between
objects on the surface ofmoon, so that they are just resolved, is close
to

1) 600 m
2) 20 m
3) 200 m
4) 60 m

Correct option: 4

Solution:
[IMAGE] Distance [IMAGE] Distance [IMAGE] Answer from options =60 m
(Minimum distance)
Subject: Physics —  — 

Question: A circular loop is placed with its plane parallel to a long
current carrying wire as shown. Consider the following statements.
[IMAGE] (i) Circular loop is rotated about y-axis
(ii) Circular loop is rotated about x-axis
(iii) Circular loop is rotated about z-axis
(iv) Circular loop is rotated about the current carrying wire.
Then emf induced in the loop is zero in

1) (i) and (ii) only
2) (iii) and (iv) only
3) (i), (ii) and (iii) only
4) in all four cases

Correct option: 2

Solution:
[IMAGE] Due to current carrying wire [IMAGE] is towards negative
z-axis. Also the area vector of the square loop is along positive
z-axis.
∴ If square loop is rotated about x-axis or y-axis the angle
⇒ Flux through the square loop changes
⇒ emf is induced in the loop in case (i) & (ii)
Subject: Physics — Current Electricity — KVL
Difficulty: Moderate

Question: In the figure, potential difference between A and B is

1) 15 V
2) Zero
3) 10 V
4) 5 V

Correct option: 3

Solution:
Diode is in forward bias, so it will behave as simple wire so,
[IMAGE] So, $V_{ab} = \frac{30}{5 + 10} \times 5 = 10V$
Subject: Physics —  — 

Question: The sinusoidal voltage of an a.c. source is given as
V = 100 sin (100πt) cos (100πt). Then

1) Peak value of voltage is 100 V and frequency is 100 Hz.
2) Peak value of voltage is 50 V and frequency is 50 Hz.
3) Peak value of voltage is 100 V and frequency is 200 Hz.
4) Peak value of voltage is 50 V and frequency is 100 Hz.

Correct option: 4

Solution:
The above relation can be written as
V = 50 (2 sin (100 πt) cos (100 πt))
or V = 50 sin (200πt)
∴ V_o = 50 V
f = 100 Hz
Subject: Physics — Mechanical Properties of Fluids — Capillary tube
Difficulty: Moderate

Question: A vertical capillary tube with inner radius 0.5 mm, is submerged
into water so that the length of its part above the water surface is h =
25 mm. Radius of curvature of meniscuses formed is
$\frac{\alpha}{10}mm$. Find $\alpha$. (surface tension of water = 0.075
N/m. Density of water 1000kgm^-3, angle of contact = 0°)


Answer: 6

Solution:
$h_{0} = \frac{2{Scos}\theta}{r_{0}\rho g} = \frac{2 \times 7.5 \times 10^{- 2}}{5 \times 10^{- 4} \times 10^{3} \times 10}$
$= \frac{15 \times 10^{- 2}}{5} = 3 \times 10^{- 2}m = 30mm$
$h_{0}r_{0} = hr$
$\Rightarrow r = \frac{h_{0}}{h}\ r_{0} = \left( \frac{30}{25} \right)(0.5) = 0.6mm$
Subject: Physics — Newton's Laws of Motion — Friction
Difficulty: Tough

Question: A weight W is to be moved from the bottom to the top of an
inclined plane of inclination $\theta$ to the horizontal. If a smaller
force is to be applied to drag it along the plane in comparison to lift
it verticallyup, the coefficient of friction should be such that

1) $\mu > tan(\frac{\pi}{4} - \frac{\theta}{2})$
2) $\mu < tan(\frac{\pi}{4} - \frac{\theta}{2})$
3) $\mu < tan\theta$
4) $\mu > tan\frac{\pi}{4}$

Correct option: 2

Solution:
Force applied to drag$F_{1} = Wsin\theta + \mu Wcos\theta$
Force applied to lift $F_{2} = W$
As, $F_{1} < F_{2} \Rightarrow Wsin\theta + \mu Wcos\theta < W$
$\Rightarrow \mu < \frac{1 - sin\theta}{cos\theta} = \frac{\sin^{2}\frac{\theta}{2} + \cos^{2}\frac{\theta}{2} - 2sin\frac{\theta}{2}cos\frac{\theta}{2}}{\cos^{2}\frac{\theta}{2} - \sin^{2}\frac{\theta}{2}}$
$= \frac{{(cos\frac{\theta}{2} - sin\frac{\theta}{2})}^{2}}{(\cos^{2}\frac{\theta}{2} - \sin^{2}\frac{\theta}{2})} = \frac{cos\frac{\theta}{2} - sin\frac{\theta}{2}}{cos\frac{\theta}{2} + sin\frac{\theta}{2}} = \frac{1 - tan\frac{\theta}{2}}{1 + tan\frac{\theta}{2}}$
$\Rightarrow \mu < tan(\frac{\pi}{4} - \frac{\theta}{2})$
Subject: Physics —  — 

Question: A convex lens is placed between a fixed object and a fixed
screen. The distance between the object and the screen is 150 cm. The
real images of the object are formed on the screen for two successive
positions of the lens separated by a distance of 50 cm. The focal length
of the lens is

1) 25 cm
2) 33.3 cm
3) 45 cm
4) 66.6 cm


Solution:
B
According to displacement method
[IMAGE] = 33.3 cm
∴ (B)
Subject: Physics — Rigid Body Dynamics — Rolling
Difficulty: Moderate

Question: A uniform sphere of mass 20 kg and radius 10 cm is placed on a
rough horizontal surface. The coefficient of friction between the sphere
and the surface is 0.5. A horizontal force of magnitude 14 N is applied
on the sphere as shown. If the friction force (in N) acting on the
sphere is F the value of F is (g = 10 ms^-2)


Answer: 4

Solution:
$f_{e} = \mu mg = 100N$
$F - f = ma$
$fR = I\alpha \Rightarrow f = \frac{2}{5}ma$
$\therefore f = \frac{2}{5} \times 20 \times 0.5 = 4N$
Subject: Physics — Magnetics — Magnetic field due to current carrying conductors
Difficulty: Moderate

Question: A very long wire ABDMNDC is shown in figure carrying current I.
AB and BC parts are straight, long and at right angle. At D wire forms a
circular turn DMND of radius R. AB, BC parts are tangential to circular
turn at N and D. Magnetic field at the centre of circle is

1) $\frac{\mu_{0}I}{2\pi R}(\pi + \frac{1}{\sqrt{2}})$
2) $\frac{\mu_{0}I}{2R}$
3) $\frac{\mu_{0}I}{2\pi R}(\pi - \frac{1}{\sqrt{2}})$
4) $\frac{\mu_{0}I}{2\pi R}(\pi + 1)$

Correct option: 1

Solution:
$\frac{\mu_{0}i}{4\pi R}\lbrack sin90^{\circ} - sin45^{\circ}\rbrack \otimes + \frac{\mu_{0}I}{2R} \odot + \frac{\mu_{0}i}{4\pi R}(sin45^{\circ} + sin90^{\circ}) \odot$
$= \frac{- \mu_{0}i}{4\pi R}\lbrack 1 - \frac{1}{\sqrt{2}}\rbrack + \frac{\mu_{0}i}{2R} + \frac{\mu_{0}i}{4\pi R}\lbrack\frac{1}{\sqrt{2}} + 1\rbrack \odot$
$= \frac{\mu_{0}i}{4\pi R}\lbrack - 1 + \frac{1}{\sqrt{2}} + 2\pi + \frac{1}{\sqrt{2}} + 1\rbrack \odot$
$= \frac{\mu_{0}i}{4\pi R}\lbrack\sqrt{2} + 2\pi\rbrack \odot$
$= \frac{\mu_{0}i}{2\pi R}\lbrack\frac{1}{\sqrt{2}} + \pi\rbrack \odot$
Subject: Physics —  — 

Question: A uniform cylinder floats with its axis vertical with
[IMAGE] parts of its length in the upper liquid and
[IMAGE] part of its length in lower liquid. If the
densities of upper and lower liquids are [IMAGE] then the
density of the material of the cylinder is


Correct option: 1

Solution:
Let [IMAGE] be the be the length of the cylinder then
 is the density of the cylinder.
TOPIC: Hydrostatics
SUB TOPIC:Bouyancy
LEVEL: Moderate
Subject: Physics — Simple Harmonic Motion — kinematics of SHM
Difficulty: Moderate

Question: A 20 gm particle is subjected to two simple harmonic motions
x_1 = 2 sin 10 t,
$x_{2} = 4sin\left( 10t + \frac{\pi}{3} \right)$where x_1& x_2 are in
metre& t is in sec.

1) The displacement of the particle at t = 0 will be $2\sqrt{3}m$
2) Maximum speed of the particle will be m/s. $20\sqrt{7}m$
3) Magnitude of maximum acceleration of the particle will be
4) All of the above

Correct option: 4

Solution:
At t = 0
Displacement $x = x_{1} + x_{2} = 4sin\frac{\pi}{2} = 2\sqrt{3}m$
Resulting Amplitude
$A = \sqrt{2^{2} + 4^{2} + 2(2)(4)cos\pi/3} = \sqrt{4 + 16 + 8} = \sqrt{28} = 2\sqrt{7}m$
Maximum speed $= A\omega = 20\sqrt{7}m/s$
Maximum acceleration $= A\omega^{2} = 200\sqrt{7}m/s^{2}$
Subject: Physics — Circular motion — Tangential acceleration
Difficulty: Moderate

Question: Kinetic energy of a particle of mass 2 kg moving along a circle
varies with time t as $K = 3t^{2}$, where t in second and K is in
Joules. Then rate at which work is being done on particle at t=1 s in
watt is

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

Correct option: 1

Solution:
$K = \frac{1}{2}{mv}^{2} = 3t^{2}$
$v = \sqrt{3}t$
$a_{t} = dv/dt = \sqrt{3}$
$= ma_{t}v$
$= 2.\sqrt{3} \cdot \sqrt{3} \times 1 = 6watt$
Subject: Physics — Thermodynamics — Thermodynamic process
Difficulty: Tough

Question: A system undergoes three quasi-static processes sequentially as
indicated in the given figure. $1 \rightarrow 2$is an isobaric process,
$2 \rightarrow 3$ is a polytrophic process with
$\gamma = \frac{4}{3}$and $3 \rightarrow 1$ is a process in which PV=
constant.
$P_{2} = P_{1} = 4 \times 10^{5}Nm^{- 2},P_{3} = 1 \times 10^{5}Nm^{- 2}$and$\ V_{1} = 1m^{3}$.
The heat transfer of the cycle is $\Delta Q,$ change in internal
energy is $\Delta U$ and the work done is $\Delta W$. for polytropic
process $PV^{\gamma} =$ constant.
(Take $\mathcal{l}$n2 = 0.693). Then,d s fp=kuqlkj rhu vnZ~/k&fLFkj çØeks ls xqtkjk A
 vkarfjd


Correct option: 2

Solution:
For the process izØe ds fy, 3 → 1
[IMAGE] For the processizØe ds fy, 2→ 3
[IMAGE] For the processizØe ds fy, 1 → 2
[IMAGE] For the processizØe ds fy, 2 → 3
[IMAGE] For the process 3 → 1
[IMAGE] For cyclic process pØh
$= 1.08 \times 10^{5}J$
Subject: Physics — Rigid Body Dynamics — Conservation of angular momentum
Difficulty: Moderate

Question: A rod AB is free to rotate on a smooth horizontal plane about one
of its end A which is hinged. Initially the rod is at rest and D and E
are fixed points located in horizontal plane in such a way that DA is
perpendicular to rod and AE is along the length of the rod. A particle P
hits the rod perpendicularly at point C some distance from point A and
gets imbedded in it, then

1) Angular momentum of rod plus particle system about point C must
increase after collision
2) Angular momentum of rod plus particle
system about point D is conserved during collision.
3) Angular momentum of rod plus particle system about point D is not
conserved during collision.
4) Angular momentum of rod plus particle system about point E is
conserved

Correct option: 2

Solution:
The force applied by hinge on rod may be towards point D or
away from D or zero depending on distance of point C from A. Reaction at
hinge has torque about E therefore, angular momentum is not conserved.
Subject: Physics — Electrostatics — Electic field
Difficulty: Moderate

Question: Three charged particles A, B and C with charges -4q, 2q and
-2q are present on the circumference of acircle of radius d. The
charged particles A, C and centre O of the circle formed an equilateral
triangle asshown in figure. Electric field at O along x-direction is

1) $\frac{\sqrt{3}q}{\pi\varepsilon_{0}d^{2}}$
2) $\frac{\sqrt{3}q}{4\pi\varepsilon_{0}d^{2}}$
3) $\frac{3\sqrt{3}q}{4\pi\varepsilon_{0}d^{2}}$
4) $\frac{2\sqrt{3}q}{\pi\varepsilon_{0}d^{2}}$

Correct option: 1

Solution:
$= \frac{q\sqrt{3}}{\pi\varepsilon_{0}d^{2}}$
Subject: Physics —  — 
Difficulty: Moderate

Question: In a certain series RLC circuit it is found that at resonance,
the inductive reactance is equal to the resistance. It is also found
that rms current in this circuit at resonance was
[IMAGE]. If now angular frequency of source is doubled
without changing any other parameter, what will be new rms current in
ampere in circuit?


Answer: 2

Solution:
At resonance:
[IMAGE] Atangular frequency of source is double of resonance frequency:
[IMAGE] TOPIC:Alternating current
SUB TOPIC:RLC circuit
Subject: Physics — Thermodynamics — Adiabatic process
Difficulty: Moderate

Question: An ideal gas whose internal energy U is at absolute zero
temperature is equal to zero undergoes areversible adiabatic
compression. If U, P, V, T represents the internal energy, pressure,
volume andabsolute temperature respectively for ideal gas then,d ije ij mRØe.kh;
fy, U, P, V, T Øe'k vkUrfjd rFkk ije rki


Correct option: 4

Solution:
[IMAGE] For adiabatic process [IMAGE] constant
Subject: Physics — Newton's Laws of Motion — Newton's second law
Difficulty: Moderate

Question: Two forces of 6 N and 3 N are acting on the two blocks of 2 kg
and 1 kg kept on frictionless floor. The force exerted on 2 kg block by
1 kg block is 2 n then find the value of n.


Answer: 2

Solution:
[IMAGE] Both blocks are constrained to move with same acceleration.
$6 - N = 2a\ \ \ \ \ \ \ $[Newtons II law for 2kg block]
$N - 3 = 1a$ [Newton's II law for 1 kg block]
This gives a= 1m/s^2
Therefore, N =4 N
Subject: Physics — Work Power & Energy — Work done by variable force
Difficulty: Tough

Question: A force F= 4t acts on a particle of mass 1kg. If the particle
starts from rest the work done by the force during the first 2 second
will be 8n J, find n.

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

Correct option: 1

Solution:
$\frac{dv}{dt} = 4t$ $\int_{0}^{v}{dv} = 4\int_{0}^{t}{t\ dt}$
$V = 2t^{2}$
$ds = 2t^{2}dt$
$W = \int_{}^{}{Fds\ cos0^{\circ}}$
$= \int_{0}^{2}\mspace{2mu} 4t \cdot 2t^{2}dt$
$= 8\left\lbrack \frac{t^{4}}{4} \right\rbrack_{0}^{2}$ $= 32J$
Subject: Physics —  — 

Question: An object is moving along x-axis. It is definitely moving
towards origin if


Correct option: 3

Solution:
When x > 0, the particle will be moving towards origin if
[IMAGE] ⇒ [IMAGE] When x < 0, the particle will be moving towards origin if
[IMAGE] ⇒ [IMAGE] Hence in both cases [IMAGE]