Grass-G/jee-main-questions huggingface dataset:
Subject: Physics — Gravitation — Conservation of mechanical energy
Difficulty: Tough

Question: An asteroid is moving directly towards the centre of the earth.
when at a distance of 10R (R is theradius of the earth) from the earths
centre, it has a speed of 12 km/s. Neglecting the effect of
earth'satmosphere, what will be the speed of the asteroid when it hits
the surface of the earth (escape velocityfrom the earth is 11.2

1) 14
2) 12
3) 16
4) 18

Correct option: 3

Solution:
KE_i + PE_i = KE_f + PE_i
$\frac{1}{2}mu_{0}^{2} + \left( - \frac{GMm}{10R} \right) = \frac{1}{2}mv^{2} + \left( - \frac{GMm}{R} \right)$
$v^{2} = u_{0}^{2} + \frac{2GM}{R}\left\lbrack 1 - \frac{1}{10} \right\rbrack$
$v = \sqrt{u_{0}^{2} + \frac{9}{5}\frac{GM}{R}}$
$= \sqrt{12^{2} + \left( \frac{9}{5} \right)\frac{(11.2)^{2}}{2}}$
$= \sqrt{144 + 0.9(11.2)^{2}} = \sqrt{256.896}$
= 16.028 km/s $\simeq 16$
Subject: Physics —  — 
Difficulty: Easy

Question: A body is acted upon by a force which is inversely proportional to
the distance covered. The work done will be proportional to

1) s
2) s2
4) None

Correct option: 4

Solution:
[IMAGE] TOPIC: WPE
SUB TOPIC: WORK DONE BY VARIABLE FORCE
Subject: Physics — Projectile Motion — Velocity of projectile at an instant
Difficulty: Tough

Question: Two seconds after projection, a projectile is travelling in a
direction inclined at 30º to the horizontal after one more second, it is
travelling horizontally. Then the speed of projection(u) and angle of
projection ($\theta$) are given by

1) $u = 10\sqrt{3}m/s$and$\theta = 45^{\circ}$
2) $u = 10\sqrt{3}m/s$and$\theta = 60^{\circ}$
3) $u = 20\sqrt{3}m/s$and$\theta = 60^{\circ}$
4) $u = 20\sqrt{3}m/s$and$\theta = 45^{\circ}$

Correct option: 3

Solution:
$T = 6sec = \frac{2usin\theta}{g} \Rightarrow usin\theta = 30$
$\tan 30^{\circ} = \frac{{usin}\theta - gt}{{ucos}\theta} = \frac{30 - 20}{{ucos}\theta}$which
gives $ucos\theta = 10\sqrt{3}$
$$tan\theta =
Subject: Physics — Rigid Body Dynamics — Pure rolling
Difficulty: Moderate

Question: A body of mass m and radius R rolling horizontally without
slipping at a speed v climbs a ramp to a height[IMAGE].
The rolling body can be.

1) A sphere
2) A circular ring
3) A spherical shell
4) A circular disc

Correct option: 4

Solution:
From energy
$K = \frac{R}{\sqrt{2}}$
[IMAGE] Body is a disc
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 and absolute temperature respectively for ideal gas
then $\left( \gamma = \frac{C_{P}}{C_{v}} \right)$.

1) $UV^{\gamma} =$constant
2) ${UP}^{\gamma} =$constant
3) ${TU}^{\gamma - 1} =$constant
4) ${VU}^{\frac{1}{\gamma - 1}} =$constant

Correct option: 4

Solution:
$U\ \alpha\ T$
For adiabatic process ${TV}^{\gamma - 1} =$ constant
$\therefore{UV}^{\gamma - 1} =$constant
$\Rightarrow {VU}^{\frac{1}{\gamma - 1}} =$constant
Subject: Physics — Electronic devices — Diodes
Difficulty: Moderate

Question: The current (in A) flowing through battery in the given circuit
 ifjiFk


Answer: 2

Solution:
The diode D_1is reverse-biased and acts as an open switch as
shown in Fig. So, there is no currentthrough D_1and the
/kkjk izokfgr ugha gksxhA
However, D_2is forward biased and acts like a short-current or closed
switch. The current drawn isI = 12/(2 + 4) = 2A.;)fi, D_2 vxz
Subject: Physics — Mechanical Properties of Fluids — Buoyant force
Difficulty: Moderate

Question: A small ball of mass m is released from rest from a height h
above the liquid surface. The density of ballis $\frac{\rho}{3}$:
(assume no splash)
 xsan dks fLFkjkoLFkk

1) ) If h ≥ H, then the ball will be able to enter the liquid of
higher density
 ?kuRo
2) If h ≥ $\frac{7H}{2}$, then the ball will be able to strike the
bottom of the vessel;fn h ≥ $\frac{7H}{2}$, gks rks xsan ik=k
3) If h = 2H, then the ball will cross the interface of two liquids
with kinetic energy = mgH and if h = 0 then ball will float on the
surface;fn h = 2H, gks rks xsan nksuks nzoks lrg ij xfrt = mgH h = 0 gks rks xasn lrg ij rSjxhA
4) All of the above

Correct option: 4

Solution:
[IMAGE] K.E. = mgH.
Subject: Physics — Newton's Laws of Motion — Newton's second law
Difficulty: Tough

Question: In the system shown in figure mA = 4m, mB = 3m and mC = 8m.
Friction is absent everywhere. String is light and inextensible. If the
system is released from rest find the acceleration of block B


Correct option: 2

Solution:
$T = m_{B}b$[Newtons II law for B in horizontal direction]
$T = \left\lbrack m_{A} + m_{e} \right\rbrack c$ I law for A
and C in horizontal direction]
$m_{A}g - T = m_{A}a$ [Newton's II law for A in vertical direction].
$\frac{m_{A}g - T}{m_{A}} = \frac{T}{m_{B}} + \frac{T}{m_{A} + m_{C}}$
$\frac{4mg - T}{4m} = \frac{T}{3m} + \frac{T}{12m} \Rightarrow T = \frac{3mg}{2} \Rightarrow b = \frac{g}{2}$
Subject: Physics — Ray Optics — Refraction (Lens)
Difficulty: Moderate

Question: If an object is placed at 20 cm in front of a half thin convex
lens of focal length 10 cm as shown in figure then taking object
position as shown the

1) x - coordinate of image is 20 cm
2) x - coordinate of image is 40 cm
3) y - coordinate of image is 0.2 cm
4) y - coordinate of image is 0.4 cm

Correct option: 3

Solution:
$\frac{1}{\text{v}} - \frac{1}{- 20} = \frac{1}{10} \Rightarrow \text{v} = 20$
Subject: Physics — Dual Nature of Matter — De Broglie wavelength
Difficulty: Moderate

Question: A particle moving with kinetic energy E has de Broglie
wavelength[IMAGE]. If energy [IMAGE] is
added to its energy, the wavelength becomes[IMAGE]. Value
of[IMAGE], is

1) 4E
2) 3E
3) 2E
4) E

Correct option: 2

Solution:
Subject: Physics —  — 
Difficulty: Tough

Question: A particle is released from a certain height H = 400 m. Due to the
wind the particle gathers the horizontal velocity component vx = ay
where [IMAGE] and y is the vertical displacement of the
particle from point of release, then the horizontal drift of the
particle when it strikes the ground is

1) 2.67 km
2) 8.67 km
3) 1.67 km
4) 5.1 km
 H=400m 
 vx = ay [IMAGE], 
 y,,

Correct option: 1

Solution:
Vy = [IMAGE]... (1)
Vx =[IMAGE]... (2)
[IMAGE] Dividing (1) by (2)
[IMAGE] or [IMAGE] dy = 2dx
Integrating
[IMAGE] ∴ x = 2.67 km
TOPIC:KINEMATIC
SUB TOPIC: 2 D
Subject: Physics —  — 

Question: A horizontal conductor is oriented north south and carries some
current. A positively charged particle located vertically above it and
having a velocity directed northward experiences an upward force. What
is the direction of the force if this charged particle were located to
the east of the conductor and had a velocity directed towards the
conductor.

1) North
2) South
3) Up
4) Down

Correct option: 1

Solution:
12 in first case according to [IMAGE] Magnetic field
will be towards west at position of charge particle for this current in
conductor should be towards south in second case magnetic field at
position of particular is upward so according to
[IMAGE] force will be towards north.
Subject: Physics — Simple Harmonic Motion — Time period of oscillation
Difficulty: Moderate

Question: A disc of mass is attached to a spring of
stiffness "k". During its motion, the disc rolls on the ground. When
released from some stretched position, the center of the disc will
execute simple harmonic motion with a time period of


Correct option: 3

Solution:
Subject: Physics —  — 
Difficulty: Moderate

Question: A particle is projected at an angle tan-1[IMAGE].
Find the height at which velocity of the particle is
[IMAGE] (m/sec). (g = 10 ms-2)

1) 1.8 m
2) 0.9 m
3) 0.7 m
4) 0.11 m
 tan-1 (m/sec) (g = 10
ms-2)

Correct option: 2

Solution:
tan θ
[IMAGE] Uy [IMAGE] 9 = 27 + 2(-10)Sy
Sy = 0.9
Subject: Physics —  — 
Difficulty: Moderate

Question: [IMAGE] = [IMAGE] X

2) 1
3) 2
4) - 1

Correct option: 1

Solution:
L.H.S. is dimensionless. So 
TOPIC: UNIT & DIMENSION
SUB TOPIC: DIMENSIONS
Subject: Physics — Motion in 1D — Non uniform acceleration
Difficulty: Moderate

Question: The velocity of an object moving rectilinearly is given as a
function of time by $v = 4t - 3t^{2}$, where v is in m/s and t is in
seconds. The average velocity of particle between t=0 to t=2 seconds is

1) 0
2) $- 2m/s$
3) $- 4m/s$
4) none of these

Correct option: 1

Solution:
$< v > = \frac{\int_{0}^{2}\mspace{2mu} vdt}{\int_{0}^{2}\mspace{2mu} dt} = \frac{\int_{0}^{2}\mspace{2mu}\left( 4t - 3t^{2} \right)dt}{2} = 0$
Subject: Physics — Work Power & Energy — Conservation of mechanical energy
Difficulty: Moderate

Question: A small block of mass m is released from rest from point A inside
a smooth hemisphere bowl of radius R, which is fixed on ground such that
OA is horizontal. The ratio (x) of magnitude of centripetal force and
normal reaction on the block at any point B varies with $\theta$ as


Correct option: 1

Solution:
$\frac{mv^{2}}{R} = N - mg\ sin\theta$
By energy conservation,
$mgR\ sin\theta = \frac{1}{2}{mv}^{2}$
$\frac{{mv}^{2}}{R} = 2mg\ sin\theta$
$N = 3mg\ sin\theta$
$Ratio = \frac{{mv}^{2}}{RN} = \frac{2}{3}$ (Constant)
$x = \frac{2}{3}$
Subject: Physics — Projectile Motion — Equation of trajectory
Difficulty: Moderate

Question: Projectiles are simultaneously projected from the same point
with speeds u and 2 u, at angles $\theta$ and $2\theta$ respectively
with horizontal. The locus of second projectile with respect to the
first is

1) $y = (\frac{2cos2\theta - cos\theta}{2sin2\theta - sin\theta})x$
2) $y = {(\frac{2sin2\theta - sin\theta}{2cos2\theta - cos\theta})}^{x}$
3) $y = (\frac{2cos2\theta + cos\theta}{2sin2\theta + sin\theta})x$
4) $y = (\frac{2sin2\theta + sin\theta}{2cos2\theta + cos\theta})x$

Correct option: 2

Solution:
For first projectile$y_{1} = usin\theta t - \frac{1}{2}gt^{2}$
$y_{2} = 2usin2\theta t - \frac{1}{2}{gt}^{2}$
$x_{1} = ucos\theta t$ $x_{2} = 2ucos2\theta t$
$y = y_{2} - y_{1} = t\lbrack 2usin2\theta - usin\theta\rbrack$
$x = x_{2} - x_{1} = t\lbrack 2ucos2\theta - ucos\theta\rbrack$
$\frac{y}{x} = \frac{2usin2\theta - usin\theta}{2ucos2\theta - ucos\theta}$
$y = x\lbrack\frac{2sin2\theta - sin\theta}{2cos2\theta - cos\theta}\rbrack$
Subject: Physics — Capacitance — Capacitor with dielectric
Difficulty: Tough

Question: In the figure shown a parallel plate capacitor has a dielectric
of width d/2 and dielectric constant K = 2.The other dimensions of the
dielectric are same as that of the plates. The plates P_1 and P_2 of
thecapacitor have area each. The energy of the capacitor is:
fp=k lekUrj iê la/kkfj=k esa ijkoS|qr dh pkSM+kbZ d/2
rFkk ijkoS|qrkad K = 2 gSAijkoS|qr

1) $\frac{\epsilon_{0}AV^{2}}{3d}$
2) $\frac{2\epsilon_{0}AV^{2}}{d}$
3) $\frac{3}{2}\frac{\epsilon_{0}AV^{2}}{d}$
4) $\frac{2\epsilon_{0}AV^{2}}{3d}$

Correct option: 4

Solution:
$U = \frac{1}{2}C_{eq}v^{2}$
$C_{1} = \frac{k\varepsilon_{0}A}{d/2} = \frac{2\varepsilon_{0}A}{(d/2)}$
and $C_{2} = \frac{\varepsilon_{0}A}{d/2}$
$C_{eq} = \frac{C_{1}C_{2}}{C_{1} + C_{2}};C_{eq} = \frac{\left( 2\frac{\varepsilon_{0}A}{d/2} \right) \times \frac{\varepsilon_{0}}{d/2}}{3\frac{\varepsilon_{0}}{d/2}}$
$= \frac{4}{3}\frac{\varepsilon_{0}A}{d}\ or\ C_{eq} = \frac{\varepsilon_{0}A}{d - \frac{d}{2} + \frac{d}{4}} = \frac{4\varepsilon_{0}A}{3d}$
$U = \frac{1}{2}\left( \frac{4}{3}\frac{\varepsilon_{0}A}{d} \right)V^{2} = \frac{2}{3}\left( \frac{\varepsilon_{0}}{d} \right){AV}^{2}$
Subject: Physics — Capacitance — Capacitor with dielectric
Difficulty: Moderate

Question: In the figure shown a parallel plate capacitor has a dielectric
of width d/2 and dielectric constant K = 2.The other dimensions of the
dielectric are same as that of the plates. The plates P_1 and P_2 of
thecapacitor have area each. The energy of the capacitor is

1) $\frac{\epsilon_{0}AV^{2}}{3d}$
2) $\frac{2\epsilon_{0}AV^{2}}{d}$
3) $\frac{3}{2}\frac{\epsilon_{0}AV^{2}}{d}$
4) $\frac{2\epsilon_{0}AV^{2}}{3d}$

Correct option: 4

Solution:
$U = \frac{1}{2}C_{eq}v^{2}$
$C_{1} = \frac{k\varepsilon_{0}A}{d/2} = \frac{2\varepsilon_{0}A}{(d/2)}$
and $C_{2} = \frac{\varepsilon_{0}A}{d/2}$
$C_{eq} = \frac{C_{1}C_{2}}{C_{1} + C_{2}};$
$C_{eq} = \frac{\left( 2\frac{\varepsilon_{0}A}{d/2} \right) \times \frac{\varepsilon_{0}}{d/2}}{3\frac{\varepsilon_{0}}{d/2}}$
$= \frac{4}{3}\frac{\varepsilon_{0}A}{d}\ or\ C_{eq}$
$= \frac{\varepsilon_{0}A}{d - \frac{d}{2} + \frac{d}{4}} = \frac{4\varepsilon_{0}A}{3d}$
$U = \frac{1}{2}\left( \frac{4}{3}\frac{\varepsilon_{0}A}{d} \right)V^{2}$
$= \frac{2}{3}\left( \frac{\varepsilon_{0}}{d} \right){AV}^{2}$
Subject: Physics — Rigid Body Dynamics — Rolling over a movable platform
Difficulty: Moderate

Question: A sphere of mass m is given some angular velocity about a
horizontal axis through its centre, andgently placed on a plank of mass
m. The coefficient of friction between the two is $\mu$. The plank rests
on a smooth horizontal surface. The initial acceleration of the sphere
relative to the plank will be $x(\mu g)$. Find the value of x.


Answer: 2

Solution:
$a_{sphere\ \ } = \frac{\mu mg}{m} = \mu g( + \widehat{i})$
$a_{plank\ \ } = \frac{\mu mg}{m} = \mu g\left( - \widehat{i} \right)$
${\ a}_{relative}\ \ = 2\mu g$
Subject: Physics —  — 

Question: has four choices

1) ,
2) ,
3) ,
4) out of which ONLY ONE is correct]

Correct option: 1

Solution:
Subject: Physics — Mechanical Properties of Fluids — Fluid in motion
Difficulty: Moderate

Question: A container is filled with a liquid of density ρ. Container is
accelerating on a horizontal surface with acceleration
[IMAGE] towards right hand side and liquid is at
rest with respect to container as shown in figure. If there is a point A
in liquid then, which of the following is correct: (assume atmospheric
pressure is zero and AB is vertical and AC is horizontal line),d ik=$\rho$?kuRo ds nzo ls Hkjk gSA ik=k {kSfrt lrg
ij$\frac{3g}{4}$Roj.k vksj Rofjr gS rFkk nzo fp=kkuqlkj ik=k ds
fodYi gS P_atm = 0 o AB

1) Pressure at A is[IMAGE] Aij nkc[IMAGE] gksxkA
2) Pressure at A is [IMAGE] Aij nkc[IMAGE] gksxkA
4) All of the above

Correct option: 4

Solution:
For point A from top surface PA=[IMAGE] and from
right wall P_A=[IMAGE] lrg ls fcUnq A ds fy, PA = [IMAGE] rFkk nka;h nhokj
lsP_A=[IMAGE] and rFkk
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:Fluid
SUB TOPIC:Bouyancy
LEVEL: Moderate
Subject: Physics — Rigid Body Dynamics — Moment of inertia
Difficulty: Easy

Question: A disc has mass 30 kg and radius 1m. The moment of inertia in
kgm^2about a tangent to its rim in itsplane is 7.5n. Find the value on
n.


Answer: 5

Solution:
Moment of inertia of a disc about its diameter is
$I_{d} = \frac{1}{4}{MR}^{2}$
Now, according to perpendicular axis theorem moment of inertia of disc
about a tangent passingthrough rim and in the plane of disc is
$$= = 37.5$$
Subject: Physics — Work Power & Energy — Potential energy
Difficulty: Hard

Question: The potential energy of a particle of mass 1 kg in a
conservative field is given as $U = (3x^{2}y^{2} + 6x)J$, where x and y
are measured in meter. Initially particle is at (1,1) & at rest then

1) Initial acceleration of particle is $6\sqrt{5}{ms}^{2}.$
2) If particle is left free it does not move in a straight line.
3) Work done to slowly bring the particle to origin is - 9 J.
4) All of the above

Correct option: 4

Solution:
$\overset{\rightarrow}{F} = - \frac{\partial u}{\partial x}\overset{\hat{}}{i} - \frac{\partial u}{\partial y}\overset{\hat{}}{j}$
$\overset{\rightarrow}{F} = - (6xy^{2} + 6)\overset{\hat{}}{i} - (6x^{2}y)\overset{\hat{}}{j}$
At (1,1)
$\overset{\rightarrow}{F} = - 12\overset{\hat{}}{i} - 6\overset{\hat{}}{j}$
$|\overset{\rightarrow}{a}| = \frac{|\overset{\rightarrow}{F}|}{m}$ =
$\sqrt{12^{2} + 6^{2}} = 6\sqrt{5}ms^{- 2}$
Also,
$w_{ext} = \Delta u = u_{f} - u_{i} = 0 - \left( 3.1^{2}.1^{2} + 6.(1) \right) = - 9J$
Subject: Physics — Thermodynamics — Adiabatic process
Difficulty: Moderate

Question: A monoatomic ideal gas follows the process:
TV^3/2 = constant
The molar specific heat for this process is: (R = gas constant).

1) $\frac{5R}{2}$
2) $\frac{5R}{6}$
3) $\frac{3R}{2}$
4) R

Correct option: 2

Solution:
TV^3/2 = constant
$\frac{(P \vee )V^{3/2}}{nR} =$constant
${PV}^{\frac{5}{2}} =$constant 
${PV}^{x} =$constant
$x = \frac{5}{2}$
$C = C_{v} + \frac{R}{1 - x}$
For polytropic process,
$C = \frac{3}{2}R + \frac{R}{1 - \frac{5}{2}} = \frac{5R}{6}$
Subject: Physics — Work Power & Energy — Potential energy
Difficulty: Moderate

Question: A thin rod of length 1 m is kept vertical and is hinged from the
lower end Its linear mass density varies with the distance (x)
from the end $H^{'}$ as $\lambda = (12x)kg/m$, where x is distance along
rod from point H. The potential energy (in Joules) of the rod is 5 N.
Find N (at H the potential energy is zero)
$\left( g = 10m/s^{2} \right)$.


Answer: 8

Solution:
Lets take an element of dx width at a distance x from the lowest
end.
$dU = (dm)gx,U = 12g\int_{x = 0}^{x - 1}\mspace{2mu} x^{2}dx = 40J$
Subject: Physics —  — 

Question: A ball is given velocity
[IMAGE] as shown. If the ratio of centripetal acceleration totangential
acceleration at the point where the ball leaves the circular path is
[IMAGE] then n is: [Neglect size of the ball].

1) and [IMAGE].....
2) [IMAGE] Now TOPIC: Circular motion
SUB TOPIC: Vertical circle
LEVEL: Moderate

Answer: 8

Solution:
and [IMAGE].....(2)
[IMAGE] Now TOPIC: Circular motion
SUB TOPIC: Vertical circle
LEVEL: Moderate
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

1) 627°C
2) 527°C
3) 427°C
4) 327°C

Correct option: 4

Solution:
Here, [IMAGE] As changes are adiabatic,
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
[IMAGE]. Here α is an integer. Find α. (surface
tension of water = 0.075 N/m. Density of water
[IMAGE], angle of contact = 0°)
0.5 mm vkUrfjd f=,d ikuh esa bl çdkj Mwch
gqbZ gS fd ikuh dh lrg ds blds Hkkx dh yEckbZ h = 25 mm gSA;gk¡α,d iw.kkZad gSAα dk eku Kkr dhft,A (ikuh dk i`"B ruko = 0.075
N/m, ty


Answer: 6

Solution:
Subject: Physics —  — 
Difficulty: Moderate

Question: If the block in the shown arrangement is acted upon by a constant
force F on a frictionless surface for t ≥ 0, its maximum speed will be

4) [IMAGE], t ≥ 0 
F,

Correct option: 1

Solution:
For vmax, [IMAGE] ⇒ Fnet = 0
⇒ Kx = F
x =[IMAGE] W = ∆KE
[IMAGE] + Fx = [IMAGE] mv2 - 0
vmax =[IMAGE] TOPIC: WPE
SUB TOPIC: WORK ENERGY THEOREM
Subject: Physics — Wave Optics — YDSE
Difficulty: Moderate

Question: In the arrangement shown in figure, wavelength of light used
is[IMAGE]. The distance between slits
[IMAGE] is [IMAGE] The distance between
[IMAGE] If the ratio of maximum to minimum intensity
received on screen P is K. Then K is


Answer: 9

Solution:
If intensity of light incident on slits [IMAGE] is [IMAGE] The intensity through [IMAGE] will be
Subject: Physics — Kinetic Theory of Gases — Heat capacity
Difficulty: Moderate

Question: Two moles of an ideal gas with [IMAGE] are mixed
with 3 moles of another ideal gas with [IMAGE] Thevalue
of [IMAGE] for the mixture is
eksy=

1) 1.42
2) 1.47
3) 1.50
4) 1.45

Correct option: 1

Solution:
[IMAGE] On rearranging we getiqu ij
Subject: Physics — Wave Optics — Interference
Difficulty: Moderate

Question: In a double-slit experiment, at a certain point on the screen
the path difference between the two interfering waves is
a wavelength. The ratio of the intensity of light at that point to that
at the centre of a bright fringe is

1) 0.568
2) 0.760
3) 0.853
4) 0.672

Correct option: 3

Solution:
Subject: Physics —  — 

Question: A positively charged particle is moved in a straight line to
the right from position x = -4 m to position x = -2 m by an external
agent. The electric potential at these positions in space are V(x =
-4m) = -4 volts and V(x = -2m) = -2 volts. Which statement is true
about the work done by the external agent moving the charge between
these positions and the electric field component parallel to the
direction of motion that the moving charged particle experiences? Assume
the particle is moved at constant speed.

1) The external agent does negative work and the electric field is
directed to the right.
2) The external agent does positive work and the electric field is
directed to the right.
3) The external agent does negative work and the electric field is
directed to the left.
4) The external agent does positive work and the electric field is
directed to the left.

Correct option: 4

Solution:
wext = q (vs - vi) = q (-4 - (-2)) = -2q = -ve
E goes from higher potential to lower potential.
Subject: Physics — Gravitation — Conservation of mechanical energy
Difficulty: Moderate

Question: A rocket is fired vertically from the surface of the earthwith a
speed v. How far from the earth does the rocketgo before returning to
the earth (where [IMAGE] is the radius of the earth and
g is acceleration due to gravity)


Correct option: 3

Solution:
Let the rocket reached a height h from thesurface of earth.
Total energy at the surface of the earth is [IMAGE] Where m and [IMAGE] are the masses of rocket andearth
respectively.
At highest point, the velocity of the rocketbecomes zero.
Total energy at the highest point is
[IMAGE] According to law of conservation of energy,
Subject: Physics —  — 
Difficulty: Easy

Question: One unit of power of a motor is 150 Watts in some system of units.
If the unit of force is doubled and unit of velocity tripled, then one
new unit of power of motor is ( in W )
 1 150,


Answer: 25

Solution:
P = f v
nu = constant
n = no. attached, u = size of unit
P' =[IMAGE] so, size of power will become 6 times
TOPIC: UNIT & DIMENSION
SUB TOPIC: MAGNITUDE
Subject: Physics — Wave Optics — Diffraction
Difficulty: Easy

Question: The first diffraction minima due to a single slit diffraction is
at [IMAGE] for a light of
wavelength[IMAGE]. The width of the slit is


Correct option: 2

Solution:
Subject: Physics —  — 

Question: Two tuning forks are producing 6 beats/s. The frequency of
first tuning fork is 256 Hz. Now, first tuning fork is loaded with wax.
As a result the beat frequency becomes 4 beats/s. Now some portion of
the wax is removed from first tuning fork. As a result the beat
frequency becomes 0 beats/s. Then the frequency of first tuning fork
after waxing (the condition when beat frequency is 4 beats/s) will be

1) 250 Hz
2) 262 Hz
3) 254 Hz
4) 246 Hz

Correct option: 4

Solution:
Since beat frequency is 6 Hz, therefore, frequency of second
fork will be either 262 Hz or 250 Hz. Since on loading with wax beat
frequency decreases, frequency of second fork should be 250 Hz. So after
waxing, the frequency of first tuning fork should be either 252 or 246
Hz. Since after removing a portion of wax from first fork beat frequency
becomes zero. Therefore frequency of first fork after waxing should be
246 Hz.
Subject: Physics — Mechanical Properties of Fluids — Surface tension
Difficulty: Moderate

Question: In a cylinder-piston arrangement, air is under a pressure P_1.
A soap bubble of radius r lies inside the cylinder, soap bubble has
surface tension T. The radius of soap bubble is to be reduced to half,
The new pressure P_2 to which air should be compressed isothermally.
(Assume r is very small as compared to height of cylinder),d csyu&fiLVu mifLFkr gS rFkk;g P_1 nkc ij gSA blesa,d lkcqu dk cqycqyk ftldh mifLFkr gSA;fn
lkcqu ds cqycqys dh f= djuh gks rks] csyu 
eku r csyu


Correct option: 3

Solution:
Isothermal process. ¼lerkih; çØe