datavorous/entrance-exam-dataset huggingface dataset:
A block of mass m = 1 kg, moving on a horizontal surface with speed vi = 2 m s–1 enters a rough patch ranging from x = 0.10 m to x = 2.01 m. The retarding force Fr on the block in this range is inversely proportional to x over this range, F= -k/x for 0.1 < x < 2.01 m = 0 for x < 0.1m and x > 2.01 m where k = 0.5 J. What is the final kinetic energy and speed vf of the block as it crosses this patch ?

Options:
A 0.5J , 1m/s
B 0.5J , 2m/s
C 1J, 0.5m/s
D 2J , 1m/s

The correct answer is:
0.5J , 1m/s

Solution:
Correct Option is : (A)
0.5J , 1m/s

Topic: Chemistry, Work Power Energy, NEET
A block of mass $M$ is attached to the lower end of a vertical spring. The spring is hung from a ceiling and has force constant value $k$. The mass is released from rest with the spring initially unstretched. The maximum extension produced in the length of the spring will be

Options:
A $M g / k$
B $2 \mathrm{Mg} / \mathrm{k}$
C $4 \mathrm{Mg} / \mathrm{k}$
D $M g / 2 k$

The correct answer is:
$2 \mathrm{Mg} / \mathrm{k}$

Solution:
Key Idea Use the law of conservation of energy. Let $x$ be the extension in the spring.
Applying conservation of energy
$$
\begin{array}{ll}
& \mathrm{mgx}-\frac{1}{2} \mathrm{kx}^2=0-0 \\
\Rightarrow \quad & \mathrm{x}=\frac{2 \mathrm{mg}}{\mathrm{k}}
\end{array}
$$

Topic: Physics, Work Power Energy, NEET, NEET 2009 (Screening)
A block of mass m is connected two springs of spring constant 2k and k, respectively, as shown in the vertical plane. At equilibrium, both springs are compressed by same length. If suddenly lower spring is cut, then acceleration of block, just after spring cut, is

Options:
A 2g downward
B g downward
C g upward
D None of these

The correct answer is:
2g downward

Solution:
2kx=mg+kx

⇒kx=mg

When the lower spring is cut the net downward force on the block is

Fnet=kx+mg=2mg

⇒a=2g (downward)

Topic: Physics, Oscillations, NEET
A block of mass m is kept on a platform. Platform starts moving upwards with an acceleration of g2. Find the work done by the normal force on the block in the first one second.

Options:
A 3 mg22
B zero
C 3mg28
D 3mg24

The correct answer is:
3mg28

Solution:
N-mg=mg2
N=3mg2
s=ut+12at2
=0+12g212
=g4
WN=3mg2Ãg4
=3mg28=3mg28

Topic: Physics, Work Power Energy, NEET
A block of mass m is placed on a smooth wedge of inclination Î¸ . The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block (g is acceleration due to gravity) will be

Options:
A mg cosÎ¸
B mg cos θ
C mg sin θ
D mg sinÎ¸

The correct answer is:
mg cosθ

Solution:
N=maâââsinâÎ¸+mgâcosÎ¸ â¦(i)
Also, mâgâsinâÎ¸=mâaââcosâÎ¸ â¦(ii)
From (1) and (2), a=gâtanâÎ¸
â´ââN=mg sin 2 Î¸ cosÎ¸ +mgâcosÎ¸âââââââââââââââââââââââââ[substitutingââtheââvalueââofâa] â´ââN= mg cosâÎ¸ ( sin 2 Î¸+ cos 2 Î¸)= mg cosâÎ¸ or,ââN= mg cosÎ¸

Topic: Physics, Laws of Motion, NEET
A block of mass m is placed over rough inclined plane having inclination 30Â°. The co-efficient of friction between the block and the inclined plane is 0.75. The contact force on the block is

Options:
A 34mg
B 2mg
C 54mg
D mg

The correct answer is:
mg

Solution:
mgsin⁡θ=32 mg=f
fl =μmgcos⁡θ
=34mg32=338mg
f<fr
f=mgsin⁡θ
N=mgcos⁡θ
fnet=mgsin2⁡θ+cos2⁡θ
=mg

Topic: Physics, Laws of Motion, NEET
A block of mass M is resting on a smooth horizontal plane. One end of a uniform rope of mass M4 is fixed to the block, which pulled it in the horizontal direction by applying a force F at the other ends. The tension in the middle of the rope is

Options:
A F2
B F5
C 910F
D F

The correct answer is:
910F

Solution:
According to the question,
Acceleration of the system (the block and the rope),
a=FM+M4=4F5M
Let the tension at the mid-point of the rope is T, FBD of the system
Applying Newton's 2nd law,
T=M+M8a=9M8Ã4F5M=910F

Topic: Physics, Laws of Motion, NEET
A block of mass m is stationary with respect to a wedge of mass M moving with uniform speed v on a horizontal surface. Find the work done by friction force on the block in t second.

Options:
A zero
B -mgvt2sin⁡2θ
C -mgvt2
D -mgvt2sin2⁡θ

The correct answer is:
-mgvt2sin⁡2θ

Solution:
As the block is at rest with respect to wedge,
fr=mgsin⁡θ
Work done by friction
W=FScos⁡θ
W=-mgsin⁡θvtcos⁡θ
W=-mgsin⁡2θ2vt

Topic: Physics, Laws of Motion, NEET
A block of mass m is stationary with respect to the wedge of mass M moving with uniform speed v on a horizontal surface. The work done by the friction force on the block as a function of time is

Options:
A zero
B âmgvt2 sin2Î¸
C –mgvt2
D –mgvt4 sin2θ

The correct answer is:
âmgvt2 sin2Î¸

Solution:
fr=mg sinÎ¸
Work done by fr=â mg sinÎ¸ cosÎ¸ vt
= â mg sin2 Î¸2(v Ãt)

Topic: Physics, Work Power Energy, NEET
A block of mass m is stationary with respect to the wedge of mass M moving with uniform speed v on horizontal surface. Work done by friction force on the block in t seconds is

Options:
A zero
B âmgvt2 sinâ¡2Î¸
C âÂ mgvt2
D âmgvt2 sinâ¡Î¸

The correct answer is:
âmgvt2 sinâ¡2Î¸

Solution:
Since block doesn't move with respect towedge, so friction force should balance the component of weight along the inclined plane.
f=mg sinâ¡Î¸
Component of displacement along the inclined plane in time t is given by vtcosÎ¸
Work done by f=â mg sinâ¡Î¸ cosâ¡Î¸ vt
= â mg sinâ¡2Î¸2v Ãt

Topic: Physics, Work Power Energy, NEET
A block of metal weighing 2 kg is resting on a frictionless plane. It is struck by a jet releasing water at a rate of 1 kg s-1 and at a speed of 5 m s-1. The initial acceleration of the block will be

Options:
A 2.5m s-2
B 5.0m s-2
C 10 m s-2
D None of the above

The correct answer is:
2.5m s-2

Solution:
Force on the block F=u dmdt=5×1=5 N
∴ Acceleration of block a=Fm=52=2.5 m s-2

Topic: Physics, Laws of Motion, NEET
A block of the mass of 1 kg is moving on the x-axis. A force F acting on the block is shown. The velocity of the block at time t=2 s is -3 m s-1. What is the speed of the block at time t=4 s?

Options:
A 8 m s-1
B 2 m s-1
C 3 m s-1
D 5 m s-1

The correct answer is:
3 m s-1

Solution:
Given, the mass of block m=1 kg
The velocity of the block at time t=2  s is -3 m s-1
Impulse during t=2 s to t=4 s .
From the graph,
I=Area of ΔABC+Area of ΔCDE
=123-2×5+12(4-3)(-5)
=52-52=0
We know that,
Impulse = Change in momentum
0=mvf-mvi
⇒ mvf-mvi=0
⇒ mvf=mvi
vf=vi=-3 m s-1 or |3 m s-1|

Topic: Physics, Laws of Motion, NEET
A block of wood floats in freshwater with two-third of its volume submerged. In oil, the block floats with one-fourth of its volume submerged. The density of oil is

Options:
A 2666.7 kg m-3
B 5333.3 kg m-3
C 1333.3 kg m-3
D 3333.3 kg m-3

The correct answer is:
2666.7 kg m-3

Solution:
Ïw 23 V=ÏbV
Ïoil 14 V=ÏbV
â´ 23 Ïw=14 Ïoil
Ïoil=83 Ïw
=80003=2666.7

Topic: Physics, Mechanical Properties of Fluids, NEET
A block performs simple harmonic motion with equilibrium point x=0. Graph of acceleration of the block as a function of time is shown. Which of the following statement is correct about the block?

Options:
A Displacement from equilibrium is maximum at t=4 s
B Speed is maximum at t=4 s
C Speed is minimum at t=4 s
D Speed is maximum at t=3 s

The correct answer is:
Speed is maximum at t=4 s

Solution:
From the figure given in the question, the speed of block will be maximum at t=0, 2 and 4. So, the block can execute the simple harmonic motion. At extreme points at time t=1 s and t=3 s, the direction of acceleration is changed. So, at t=4 s the speed of the block is maximum.

Topic: Physics, Oscillations, NEET
A block rests on a fixed rough inclined plane and a horizontal force F is applied to it, as shown in the figure. Which of the following statements are correct?

(1) Normal reaction on the block is Fsin⁡θ+mgcos⁡θ

(2) The frictional force is zero when Fcos⁡θ=mgsin⁡θ

(3) The value of limiting friction is μ(mgsin⁡θ+Fcos⁡θ)

(4) The value of limiting friction is μ(mgsin⁡θ-Fcos⁡θ)

Options:
A 1, 2
B 3, 4
C 2, 4
D 2, 3

The correct answer is:
1, 2

Solution:
As the object is at rest, therefore from FBD,
Normal Reaction N=mgcos⁡θ+Fsin⁡θ .....(i)
Limiting Friction fL=μN=μmgcos⁡θ+Fsin⁡θ ......(ii)
Also, if mgsin⁡θ=Fcos⁡θ
then the object has no tendency of motion; hence friction is zero in this case.

Topic: Physics, Laws of Motion, NEET
A block rests on a rough inclined plane making an angle of 30Â° with the horizontal. The coefficient of static friction between the block and the plane is 0.8. If the frictional force on the block is 10 N, the mass of the block is Take g=10 m s-2

Options:
A 1 kg
B 2 kg
C 3 kg
D 4 kg

The correct answer is:
2 kg

Solution:
As the block is at the state of rest,
â´R=mgcosâ¡Î¸ , which is balanced.
and f1= force of friction =10 N (given)
Hence, f1=mgsinâ¡Î¸
10=mÃ10Ãsinâ¡30o
m=1010Ãsinâ¡30o=2 kg

Topic: Physics, Laws of Motion, NEET
A block whose mass is 1 kg is fastened to a spring. The spring has a spring constant of 50 N m–1. The block is pulled to a distance x = 10 cm from its equilibrium position at x = 0 on a frictionless surface from rest at t = 0. Calculate the kinetic energy of the block when it is 5 cm away from the mean position.

Options:
A 0.19J
B 0.31J
C 0.38J
D 0.11J

The correct answer is:
0.19J

Solution:
Correct Option is : (A)
0.19J

Topic: Chemistry, Oscillations, NEET
A block whose mass is 1 kg is fastened to a spring. The spring has a spring constant of 50 N m–1. The block is pulled to a distance x = 10 cm from its equilibrium position at x = 0 on a frictionless surface from rest at t = 0. Calculate the total energy of the block when it is 5 cm away from the mean position.

Options:
A 0.19J
B 0.31J
C 0.25J
D 0.11J

The correct answer is:
0.25J

Solution:
Correct Option is : (C)
0.25J

Topic: Chemistry, Oscillations, NEET
A block whose mass is 1 kg is fastened to a spring. The spring has a spring constant of 50 N m–1. The block is pulled to a distance x = 10 cm from its equilibrium position at x = 0 on a frictionless surface from rest at t = 0. Calculate the potential energy of the block when it is 5 cm away from the mean position.

Options:
A 0.0351J
B 0.0625J
C 0.15J
D 0.11J

The correct answer is:
0.0625J

Solution:
Correct Option is : (B)
0.0625J

Topic: Chemistry, Oscillations, NEET
A boat covers 24 km upstream and 36 km downstream in 6 hours while it covers 36 km upstream and 24 km downstream in 612 hours. The velocity of the current is:

Options:
A 1 km hr-1

B 1.5 km hr-1

C 2 km hr-1

D 2.5 km hr-1

The correct answer is:
2 km hr-1

Solution:
Let rate of upstream = x kmph and rate of downstream = y kmph.
Then, 24x+36y=6….(i) and 36x+24y=132….(ii)
Solving (i) & (ii) y=12 km hr-1 and x=8 km hr-1
∴Speed upstream = 8 kmph, Speed downstream = 12 kmph.
Hence, rate of current =1212-8 kmph = 2 kmph.
and rate of Boat =1212+8=10 kmph

Topic: Physics, Motion In One Dimension, NEET
A boat is sent across a river with a velocity of 8âkm hrâ1â. If the resultant velocity of the boat is 10âkmâhrâ1,âthen the velocity of the river is

Options:
A 12.8 km h−1
B 6 km hâ1
C 8 km h−1
D 10 km hâ1

The correct answer is:
6 km hâ1

Solution:
Let the velocity of the river = v → R

∴|v→boat|=8 km h−1

v → bR = v → b − v → R
v → b = v → bR + v → R
∵ v → bR ⊥ v → R

v b = v bR 2 + v R 2

v b 2 = v bR 2 + v R 2

(10) 2 = (8) 2 + v R 2

vR=6km h-1

Topic: Physics, Motion In Two Dimensions, NEET
A boat which has a speed of 5 km h-1 in still water crosses a river of width 1 km along the shortest possible path in 15 minutes. The velocity of the river water (in km h-1) is

Options:
A 3
B 4
C 21
D 1

The correct answer is:
3

Solution:
Speed along the shortest path =vâR
âââ|vâR|=distancetime=1(1560)ââkmââhâ1âââ|vâR|=4ââkmââhâ1
â´â Speed of water, vâwaterâ=â52â42â=â3âkmââhâ1

Topic: Physics, Motion In Two Dimensions, NEET
A bob hangs from a rigid support by an inextensible string of length l. It is released from rest when string makes an angle 60o with vertical. The speed of the bob at the lowest position is

Options:
A gl
B 3gl
C 2gl
D 5gl

The correct answer is:
gl

Solution:
This mean h=l1-cos⁡60o
⇒h=l×12
Velocity of bob v=2gh
⇒v = 2gl2
⇒v = gl

Topic: Physics, Work Power Energy, NEET
A bob is whirled in a horizontal circle by means of a string at an initial speed of 10 rpm. If the tension in the string is quadrupled while keeping the radius constant, the new speed is:

Options:
A 20 rpm
B 40 rpm
C 5 rpm
D 10 rpm

The correct answer is:
20 rpm

Solution:
In horizontal circular motion,
$T=m \omega^2 r$
For constant $m$ and $r, T \propto \omega^2$
$T=4 T$ (Given)
$\Rightarrow \omega^{\prime}=2 \omega$
$=20 \mathrm{rpm}$

Topic: Physics, Laws of Motion, NEET, NEET 2024 (Re-NEET)
A bob is whirled in a horizontal plane by means of a string with an initial speed of $\omega$ rpm. The tension in the string is $T$. If speed becomes $2 \omega$ while keeping the same radius, the tension in the string becomes:

Options:
A 4T
B T/4
C $\sqrt{2} T$
D T

The correct answer is:
4T

Solution:
$T=m \ell \omega^2$
$\begin{aligned} & T^{\prime}=m \ell(2 \omega)^2 \\ & T^{\prime}=4 T\end{aligned}$

Topic: Physics, Motion In Two Dimensions, NEET, NEET 2024
A bob of mass M is suspended by a massless string of length L. The horizontal velocity v at position A is just sufficient to make it reach point B. The angle Î¸ at which the speed of the bob is half of that at A satisfies

Options:
A θ = π 4
B π 4 < θ < π 2
C Ï 2 < Î¸ < 3 Ï 4
D 3 Ï 4 < Î¸ < Ï

The correct answer is:
3 π 4 < θ < π

Solution:
Suppose at point P velocity becomes half. We also know minimum velocity required to complete the circle is

vA=5gL

Using conservation of energy from point A to P

12mvA2=12mvP2+mgh
12mv2=12mv22+mgh
38mv2=mgh

38Ã5gL=ghh=158L

cosÎ¸=L-hL=1-hL=-78

Thus option d is right answer.

Topic: Physics, Work Power Energy, NEET
A bob of pendulum of mass 50 g is suspended by a string with the roof of an elevator. If the lift is falling with a uniform acceleration of 5 m s-2, the tension in the string is g=10 m s-2

Options:
A 0.5 N
B 0.225Â N
C 0.25 N
D 0.025Â N

The correct answer is:
0.25 N

Solution:
Here, mass of bob, m=50 g
=0.05 kg
Since, the lift is falling downward with a uniform acceleration of 5 m s-2 , hence tension in the string,
T=mg-a
=0.0510-5
=0.05×5
=0.25 N

Topic: Physics, Laws of Motion, NEET
A body A of mass M while falling vertically downwards under gravity breaks into two parts; a body B of mass M3 and a body C of mass 2M3. The centre of mass of bodies B and C taken together shifts compared to that of a body A towards

Options:
A depends on height of breaking
B does not shift
C body C
D body B

The correct answer is:
does not shift

Solution:
The centre of mass of the two parts B and C of the body will not shift. It is because, after breaking into two parts, no external force acts on the system. It still moves under the effect of gravity. Hence, the centre of mass of the two parts B and C will follow the original path.

Topic: Physics, Center of Mass Momentum and Collision, NEET
A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by F=$-i+2j+3k$ N where $i,j,k~$are unit vectors along the x-, y- and z-axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis ?

Options:
A 10J
B 11J
C 12J
D 13J

The correct answer is:
12J

Solution:
Correct Option is : (C)
12J

Topic: Chemistry, Work Power Energy, NEET
A body constrained to move along Z-axis of a coordinate system is subjected to a constant force F given by Fâ¡=-iâ¡^+2jâ¡^+3k^Nâ¡ where i^, j^ and k^ are the unit vector along the X, Y and Z-axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the Z-axis?

Options:
A 12 J

B 13 J

C 10 J

D 9 J

The correct answer is:
12 J

Solution:
Force,F⁡=-i^+2j^+3k^ N
Displacement, s⁡=4k^ m=0i^+0j^+4k^ m⁡
Work done by the force is given by
W =F.s
=-i^+2j^+3k^·0i^+0j^+4k^
= (– 1 × 0) + (2 × 0) + (3 × 4)
= 0 + 12 ∵ i ^ · j ^ = j ^ · j ^ = k ^ · k ^ = 1 and i ^ · j ^ = j ^ · k ^ = k ^ · i ^ = 0
= 12 J

Topic: Physics, Work Power Energy, NEET
A body constrained to move in the y direction is subjected to a force given by F→=(-2i^+15j^+6k^) N. What is the work done by this force in moving the body through a distance 10 m along the y axis?

Options:
A 20Â J
B 150 J
C 160Â J
D 190Â J

The correct answer is:
150 J

Solution:
Work done along the y-axis is due to force along y-axis which will cause displacement along the y-axis.
Therefore, F=15 N and displacement is 10 m along y-axis.

⇒W=15×10=150 J

Topic: Physics, Work Power Energy, NEET
A body constrained to move in y-direction is subjected to a force given by Fâ¡â=-2iâ¡^+15jâ¡^+6k^ Nâ¡. The work done by this force in moving the body distance of 10 m along the y-axis is

Options:
A 20 J
B 150 J
C 160 J
D 190 J

The correct answer is:
150 J

Solution:
F⁡→=-2i⁡^+15j⁡^+6k^N⁡
S⁡→=0i⁡^+10j⁡^+0k^m⁡
∴ Work done
F⁡→·S⁡→=-2i⁡^+15j⁡^+6k^·0i⁡^+10j⁡^+0k^=150J

Topic: Physics, Work Power Energy, NEET
A body cools from 50 Â°C to 40 Â°C in 5 min. If the temperature of the surroundings is 20 Â°C, the temperature of the body after the next 5 min would be

Options:
A 36 Â°C
B 35 Â°C
C 33.33 Â°C
D 30Â°C

The correct answer is:
33.33 Â°C

Solution:
θ2-θ1Δt=Kθ2+θ12-θ0
50-405 min=K50+402-20 .......(1)
40-θ5 min=K40+θ2-20 .......(2)
2÷1
40-θ50-40=40+θ2-2050+402-20
40-θ10=θ225
200-5θ=θ
θ=2006=33.33 °C

Topic: Physics, Thermal Properties of Matter, NEET
A body cools from 50 Â°C to 40 Â°C in 5 minutes. Its temperature comes down to 33.33 Â°C in the next 5 minutes. The temperature of the surroundings is

Options:
A 15 °C
B 20 °C
C 25 Â°C
D 10 °C

The correct answer is:
20 °C

Solution:
Using Newton's law of cooling,
log Î¸2-Î¸0Î¸1-Î¸0=-Kt
log 40-Î¸050-Î¸0= -KÃ5 .....(i)
log 33.33-Î¸040-Î¸0 = -KÃ5 .....(ii)
From Eqs.(i) and (ii),
40-Î¸050-Î¸0 = 33.33-Î¸040-Î¸0
On solving, we get
Î¸0=19.95Â°C â20 Â°C

Topic: Physics, Thermal Properties of Matter, NEET
A body cools from 50 Â°C to 40 Â°C in 5 minutes. Its temperature comes down to 33.33 Â°C in the next 5 minutes. The temperature of the surroundings is

Options:
A 15 °C
B 20 Â°C
C 25 Â°C
D 10 °C

The correct answer is:
20 Â°C

Solution:
Let Î¸0 = temperature of surrounding,

Using Newton's law of cooling,
log Î¸2-Î¸0Î¸1-Î¸0=-Kt
log 40-Î¸050-Î¸0=-KÃ5 .....(i)
log 33.33-Î¸040-Î¸0 =-KÃ5 .....(ii)
From Eqs.(i) and (ii),
40-Î¸050-Î¸0 = 33.33-Î¸040-Î¸0
On solving, we get
Î¸0=19.95Â°C â20 Â°C

Topic: Physics, Thermal Properties of Matter, NEET
A body cools from a temperature 3T to 2T in 10 minutes. The room temperature is T. Assume that Newton's law of cooling is applicable. The temperature of the body at the end of next 10 minutes will be

Options:
A 74T
B 32T
C 43T
D T

The correct answer is:
32T

Solution:
T1-T2∆t=KT1+T22-T0
3T-2T10=K2.5T-T
⇒T10=K1.5T
K=115
Now, 2T-x10=K2T+x2-T
Solving x=3T2

Topic: Physics, Thermal Properties of Matter, NEET, NEET 2016 (Phase 2)
A body covers 25th of the total distance with speed v1 and 35th with v2 . The average speed of the body will be

Options:
A 5v1v23v1+2v2
B v1+v22
C 12v1v2
D 2v1v2v1+v2

The correct answer is:
5v1v23v1+2v2

Solution:
Let the total distance is x, then we have
t1=25 xv1 and t2=35 xv2
Since, average speed, vav=total distancetotal time=xt1+t2
=x2x5v1+3x5v2=5v1v23v1+2v2
∴ vav=5v1v23v1+2v2

Topic: Physics, Motion In One Dimension, NEET
A body falls freely for 10 s. Its average velocity during this journey is Take g=10 m s-2

Options:
A 100 m s-1
B 10Â m s-1Â
C 50Â m s-1Â
D 5Â m s-1

The correct answer is:
50Â m s-1Â

Solution:
Average velocity
=Total DisplacementTime interval
In 10 s, the distance covered by the body
s=12gt2
â´ s=1210Ã102=500 m
Î½av=50010=50 m s-1

Topic: Physics, Motion In One Dimension, NEET
A body falls towards Earth in air. Will its total mechanical energy be conserved during the fall?

Options:
A Yes, because the resistance of air does not affect the body's total mechanical energy.
B No, because the resistance of air is a non-conservative force leading to a decrease in mechanical energy.
C Yes, because the gravitational force will balance out the resistance of air during the fall.
D No, because the force of gravity will be greater than the resistance of air, leading to a loss in mechanical energy.

The correct answer is:
No, because the resistance of air is a non-conservative force leading to a decrease in mechanical energy.

Solution:
Correct Option is : (B)
No, because the resistance of air is a non-conservative force leading to a decrease in mechanical energy.

Topic: Chemistry, Work Power Energy, NEET
A body floats in a liquid contained in a beaker. If the whole system falls under gravity, then the upthrust on the body due to liquid is

Options:
A Zero
B Equal to the weight of the immersed part of the body
C Equal to the weight of the body in liquid
D Equal to the weight of the body in air

The correct answer is:
Zero

Solution:
In freefall the effective acceleration due to gravity is zero and hence the upthrust is zero

Topic: Physics, Mechanical Properties of Fluids, NEET
A body has equal amount of rotational kinetic energy and translational kinetic energy while rolling without slipping on a horizontal surface. Given body is an example of

Options:
A disc
B sphere
C ring
D cylinder

The correct answer is:
ring

Solution:
Translational Kinetic Energy = 1 2 m v 2
Rotational Kinetic Energy = 1 2 I Ï 2
Translational Kinetic Energy = Rotational Kinetic Energy
12mv2=12IÏ2

Put v=rÏ, you will get I=M R 2 which is inertia of ring.

Topic: Physics, Rotational Motion, NEET
A body having a temperature of 27 Â°C is kept in a room having a temperature of 27 Â°C. Does the body emit any radiation in this case when the room temperature is the same as body temperature?

Options:
A Yes
B No
C May emit, depends on the material of body
D Can't say anything

The correct answer is:
Yes

Solution:
All bodies emit radiation at all temperatures, governed by the Stefan's law.

Topic: Physics, Thermal Properties of Matter, NEET
A body initially at rest and sliding along a frictionless track from a height h (as shown in the figure) just completes a vertical circle of diameter AB=D. The height h is equal to,

Options:
A 75D
B D
C 32D
D 54D

The correct answer is:
54D

Solution:
To just complete a vertical circle, speed at A should be
vA=5gR
Using energy conservation
mgh=12mvA2
h=12vA2g=125ggD2 R=D2
h=5D4

Topic: Physics, Work Power Energy, NEET, NEET 2018
A body is displaced from 0, 0 to 1 m, 1 m along the path x=y by a force Fâ=x2j^+yi^ N . The work done by this force will be

Options:
A 43 J
B 56 J
C 32 J
D 75 J

The correct answer is:
56 J

Solution:
W=â«0,01,1Fâ.dâx
Here, dâs=dxi^+ dyj^ + dzk^
â´W=â«0,01,1x2dy+ydx
= â«0,01,1x2dy+x.dx
(As, x=y )
â´W=y33+x220,01,1=56 J

Topic: Physics, Work Power Energy, NEET
A body is displaced from (0,0) to (1 m,1 m) along the path x=y by a force Fâ=(yi^+x2j^) N, then the work done by the force is

Options:
A 43J
B 56J
C 32J
D 75J

The correct answer is:
56J

Solution:
W=∫(0,0)(1,1)F→.d→x
Here d→s=dxi^+ dyj^ + dzk^
∴W=∫(0,0)(1,1)(x2dy+ydx)
= ∫(0,0)(1,1)(x2dy+x.dx)
(As x=y)
∴W=y33+x22(0,0)(1,1)=56J

Topic: Physics, Work Power Energy, NEET
A body is displaced from 0 m, 0 m to 1 m, 1 m along the path x=y by a force F→=x2j^+yi^ N. The work done by this force will be

Options:
A 43Â J
B 56 J
C 32Â J
D 75Â J

The correct answer is:
56 J

Solution:
As force is a function of r, therefore
W=∫0,01,1F→.d→x
Here ds→=dxi^+dyj^+dzk^

∴ W=∫0,01,1x2dy+ydx=∫0,01,1y2dy+xdx (as x=y )
∴ W=y33+x220,01,1=56 J

Topic: Physics, Work Power Energy, NEET
A body is dropped from a certain height above the surface of the earth. When it loses U amount of its potential energy, it acquires a velocity v. The mass of the body is

Options:
A 2Uv2
B 2vU2
C 2vU
D U22v

The correct answer is:
2Uv2

Solution:
Using the conservation of mechanical energy

Ki+Pi=Kf+Pf

0+P=12mv2+P-U

12mv2=U

m=2Uv2

Topic: Physics, Work Power Energy, NEET