datavorous/entrance-exam-dataset huggingface dataset:
A body is executing SHM. When the displacement of the body from the mean position is 4 cm and 5 cm, the values of the corresponding velocity of the body are 10 cm s-1 and 8 cm s-1, respectively. The time period of the body is

Options:
A 2π s
B Ï2 s
C Ï s
D 3Ï2 s

The correct answer is:
Ï s

Solution:
10=Ïa2-16 ....(1)
8=Ïa2-25 .....(2)
542=a2-16a2-25
or 2516=a2-16a2-25
9a2=625-256
a2=3699=41
from equation (i)
10=Ï41-16=ÏÃ5
or Ï=2
2ÏT=2
or T=Ï sec

Topic: Physics, Oscillations, NEET
A body is executing simple harmonic motion with frequency n, the frequency of its potential energy is

Options:
A n
B 2n
C 3n
D 4n

The correct answer is:
2n

Solution:
Given,
frequency of SHM=n
As we know time period is the time in which particle complete its one revolution.

Now we know that in one time period particle passes through mean position and extreme position two times.

So, In one time period PE become maximum & minimum two times. It means frequency of PE is 2n.

Topic: Physics, Oscillations, NEET, NEET 2021
A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to

Options:
A ${{t}^{1/2}}$
B t
C ${{t}^{3/2}}$
D ${{t}^{2}}$

The correct answer is:
t

Solution:
Correct Option is : (B)
t

Topic: Chemistry, Work Power Energy, NEET
A body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time t is proportional to

Options:
A t34
B t32
C t14
D t12

The correct answer is:
t32

Solution:
Power = Work Time⁡ = Force × distance Time =Force × velocity
∴Force×velocity=constant (K)
or (m⁡a)(at⁡)=K⁡
or a=(K⁡m⁡t⁡)1/2
∵s⁡=12at⁡2
∴s⁡=12(K⁡m⁡t⁡)1/2t⁡2=12(K⁡m⁡)1/2t3/2
or s is proportional to t32.

Topic: Physics, Work Power Energy, NEET
A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time t is proportional to

Options:
A ${{t}^{1/2}}$
B t
C ${{t}^{3/2}}$
D ${{t}^{2}}$

The correct answer is:
${{t}^{3/2}}$

Solution:
Correct Option is : (C)
${{t}^{3/2}}$

Topic: Chemistry, Work Power Energy, NEET
A body is moving with velocity $30 \mathrm{~m} / \mathrm{s}$ towards east. After $10 \mathrm{~s}$ its velocity becomes $40 \mathrm{~m} / \mathrm{s}$ towards north. The average acceleration of the body is

Options:
A $7 \mathrm{~m} / \mathrm{s}^2$
B $\sqrt{7} \mathrm{~m} / \mathrm{s}^2$
C $5 \mathrm{~m} / \mathrm{s}^2$
D $1 \mathrm{~m} / \mathrm{s}^2$

The correct answer is:
$5 \mathrm{~m} / \mathrm{s}^2$

Solution:
$$
\begin{aligned}
& \text { Average acceleration }=\frac{\text { Change in velocity }}{\text { Total time }} \\
& \qquad \begin{aligned}
a & =\frac{\left|\overrightarrow{\mathbf{v}}_f-\overrightarrow{\mathbf{v}}_i\right|}{\Delta t} \\
& =\frac{\sqrt{30^2+40^2}}{10}=\frac{\sqrt{900+1600}}{10} \\
& =5 \mathrm{~ms}^{-2}
\end{aligned}
\end{aligned}
$$

Topic: Physics, Motion In One Dimension, NEET, NEET 2011 (Screening)
A body is orbiting very close to the earth's surface with kinetic energy K. The energy required to completely escape from it is

Options:
A K
B 2K
C K2
D 3K2

The correct answer is:
K

Solution:
For a satellite in a closed orbit, we know that T.E.=P.E.2=-K.E.
Hence, in order to make the body escape, we should make its total energy 0, which can be achieved by providing additional energy equal to the kinetic energy of the body itself as from the abiove relation we can clearly see that T.E.=-K.E..
So, the energy given to the body so as to completely escape from its orbit is equal to its kinetic energy K.

Topic: Physics, Gravitation, NEET
A body is performing simple harmonic motion of amplitude A and time period T. The figure shows position-time graph of the body. At any time t, acceleration of the body is f, then which of the following graphs is/are appropriate?

Options:
A
B
C
D

The correct answer is:

Solution:
Since initial phase is π 2 as per the given graph.
Using the equation of position of SHM x=Asin( ωt+ π 2 )
⇒x=Acosωt
⇒ acceleration f= d 2 x d t 2 =−A ω 2 cosωt
Which is represented in option third.

Topic: Physics, Oscillations, NEET
A body is projected horizontally with speed 20 m s-1. The speed of the body after 5 s is nearly

Options:
A 54 m s-1
B 35 m s-1
C 20 m s-1
D 70 m s-1

The correct answer is:
54 m s-1

Solution:
vx=u=20 m s-1
vy=vu+gt
=0+10×5=50 m s-1
v=ux2+uy2
=202+502 =54 m s-1

Topic: Physics, Motion In Two Dimensions, NEET
A body is projected up a smooth inclined plane with a velocity v0 from point A as shown in the figure. The angle of inclination is 45Â° and top B of the plane is connected to a well of diameter 40 m. If the body just manages to cross the well, the value of v0 is (Length of the inclined plane is 202 m, and g=10 m s-2)

Options:
A 20 m s-1
B 202 m s-1
C 40 m s-1
D 402 m s-1

The correct answer is:
202 m s-1

Solution:
Let v 2 be the velocity at point B at it cross the well. thus the range of the projection motion.
R=u2sin 2θg=v22sin2θg=40 m
also as the inclined plane is at an angle 45 ο thus θ for projection motion is also 45 ο
thus v 2 2 sin 2 θ g = 40 = v 2 2 sin 90 ο g = v 2 g
v 2 2 = 40 g
v 2 2 = 400
v 2 = 20
now given initial velocity is v 0 so apply the equation of motion
v 2 = u 2 + 2 as
20 2 = v 0 2 + 2 g sin θ . 20 2
v 0 2 = 400 + 2.20. 10
v0=202 m s-1

Topic: Physics, Motion In Two Dimensions, NEET
A body is projected up a smooth inclined plane with velocity v from the point A as shown in the figure. The angle of inclination is 45Â° and the top is connected to a well of diameter 40 m. If the body just manages to cross the well, what is the value of v? Length of the inclined plane is 202 m.

Options:
A 40 m s-1
B 402 m s-1
C 20 m s-1
D 202 m s-1

The correct answer is:
202 m s-1

Solution:
vâ²2g=40 âµÎ¸=45Â°
vâ²2=40 g but vâ²2=v2-2al
Or 40g=v2-2g sin45Â°.202
Or 40g=v2-40g
Or v2=80g
Or v=202 m s-1

Topic: Physics, Motion In Two Dimensions, NEET
A body is projected up with a velocity equal to 34th of the escape velocity from the surface of the earth. The height it reaches from the center of the earth is

Options:
A 10R/9
B 16R/7
C 9R/8
D 10R/3

The correct answer is:
16R/7

Solution:
v=34ve
K.E.=12mv2=12m34ve2=932mve2=932m2GMR
916GMmR-GMmR=-GMmr
r=16R7

Topic: Physics, Gravitation, NEET
A body is projected up with a velocity equal to 9GM8R, where M is the mass of the earth and R is the radius of the earth. The maximum distance it reaches from the centre of the earth is

Options:
A 10R9
B 16R7
C 10R3
D 9R8

The correct answer is:
16R7

Solution:
Let us assume that the maximum distance of the body from the centre of the earth is h.

Applying the conservation of mechanical energy, we get

12m9GM8R2-GMmR=-GMmh
-716GMmR=-GMmh

⇒h=16R7

Topic: Physics, Gravitation, NEET
A body is projected with an initial velocity of (8i^+6j^) m s-1. The horizontal range is (g=10 m s-2)

Options:
A 9.6 m

B 14 m

C 50 m
D 100 m

The correct answer is:
9.6 m

Solution:
velocity = 8 i ^ + 6 j ^
v = 10 m s - 1
and angle tan Î¸ = 6 8 = 3 4
Range =u2sin2Î¸g=u22sinÎ¸cosÎ¸g
= 10 2 2 3 5 4 5 10
= 48 5 = 9.6 m

Topic: Physics, Motion In Two Dimensions, NEET
A body is released from a height and falls freely towards the earth. Exactly 1 s later another body is released. What is the distance between the two bodies 2 s after the release of the second body, if g=9.8 m s-2?

Options:
A 24.5 m

B 25.6 m

C 12.3 m

D 30.5 m

The correct answer is:
24.5 m

Solution:
According to the given problem 2nd body falls for 2 s so that
h2=12g22 ...(i)
while 1st has fallen for 2+1=3 s; so
h1=12g32 ...(ii)
â´ The separation between two bodies 2 s after the release of the 2nd body,
d=h1-h2=12g32-22=4.9Ã5=24.5 m

Topic: Physics, Motion In Two Dimensions, NEET
A body is thrown from a point with speed 50 m s-1 at an angle 37Â° with horizontal. When it has moved a horizontal distance of 80 m then its distance from point of projection is

Options:
A 40Â m
B 402Â m
C 405 m
D none

The correct answer is:
405 m

Solution:
As y=xtanθ-gx22u2cos2θ
y=80×34-10×80×80×252×50×50×16=40 m
∴ distance from point of projection
=(80)2+(40)2 m=405 m

Topic: Physics, Motion In Two Dimensions, NEET
A body is thrown up in a lift with a velocity 5 m s-1 relative to the lift and the time of flight is found to be 0.8 s. The acceleration with which the lift is moving up is (g=10 m s-2)

Options:
A 1.5 m s-2
B 2 m s-2
C 2.5 m s-2
D 3 m s-2

The correct answer is:
2.5 m s-2

Solution:
Acceleration of body w. r. t. lift=-(g +a)
∴ Time of flight of body, t = 2ug+a
⇒a=2u-gtt=(2×5)-(10×0.8)0.8 = 2.5 m s-2

Topic: Physics, Motion In One Dimension, NEET
A body is thrown up with a speed u, at an angle of projection θ. If the speed of the projectile becomes u2 on reaching the maximum height, then the maximum vertical height attained by the projectile is

Options:
A u24g
B u23g
C u22g
D u2g

The correct answer is:
u24g

Solution:
Speed of projectile at maximum height
=u2 .....(i)
Also, we know that speed of a projectile at maximum height =ucos⁡θ
∴ ucos⁡θ=u2 ⇒ cos⁡θ=12 .....(ii)
⇒ θ=45o
The maximum height is given by the formula
H=u2sin2⁡θ2g=u22g122=u24g

Topic: Physics, Motion In Two Dimensions, NEET
A body is thrown vertically up with certain intial velocity, the potential and kinetic energies of the body are equal at a point P in its path. If the same body is thrown with double the velocity upwards, the ratio of potential and kinetic energies of the body when it crosses the same point, is

Options:
A 1Â :1
B 1Â :4
C 1 :7
D 1Â :8

The correct answer is:
1 :7

Solution:
Potential energy + Kinetic energy = Translation energy
Translation energy = 2×Potential energy
T.E1=4T.E ∵V1=2V
PE1KE1=PE8PE-PE=17

Topic: Physics, Work Power Energy, NEET
A body is thrown vertically upwards. Which one of the following graphs correctly represents the velocity v vs time t?

Options:
A
B
C
D

The correct answer is:

Solution:
v=u-gt
y=c+mx
m=-g = slope
c=+u
Velocity becomes negative after some time and the negative slope is constant.

Topic: Physics, Motion In Two Dimensions, NEET
A body is thrown with the velocity 20 m s-1 at an angle of 60Â° with the horizontal. Find the time gap between the two positions of the body where the velocity of the body makes an angle of 30Â° with horizontal

Options:
A 1.15 s

B 0.95 s
C 1 s
D 1.5 s

The correct answer is:
1.15 s

Solution:
tanα= V y V x
⇒ 1 3 = 20 o sin60−10 t 1 20.cos60
⇒ t 1 = 2 3
−1 3 = 20sin60−10 t 2 20cos 60 o
t 2 = 4 3
⇒Δt=23=1.15 s

Topic: Physics, Motion In Two Dimensions, NEET
A body is thrown with the velocity 20 m s-1 at an angle of 60Â° with the horizontal. Find the time gap between the two positions of the body where the velocity of the body makes an angle of 30Â° with horizontal.

Options:
A 1.15 s

B 0.95 s
C 1 s
D 1.5 s

The correct answer is:
1.15 s

Solution:
tanα= V y V x
⇒ 1 3 = 20 o sin60−10 t 1 20.cos60
⇒ t 1 = 2 3
−1 3 = 20sin60−10 t 2 20cos 60 o
t 2 = 4 3
⇒Δt=23=1.15 s

Topic: Physics, Motion In Two Dimensions, NEET
A body is vibrating in simple harmonic motion with an amplitude of 0.06 m and frequency of 15 Hz. The maximum velocity and acceleration of the body is

Options:
A 9.80 m s-1 and 9.03 × 102 m s-2
B 8.90 m s-1 and 8.21 × 102 m s-2

C 6.82 m s-1 and 7.62 × 102 m s-2

D 5.65 m s-1 and 5.32 × 102 m s-2

The correct answer is:
5.65 m s-1 and 5.32 × 102 m s-2

Solution:
Amplitude, A = 0.06 m
Frequency, f = 15 Hz
Maximum velocity of the body,
v m = ω A
= 2 π f A
= 2 × 3.14 × 1 5 × 0.06
= 5.65 m s-1

Maximum acceleration of the body.
a m = ω 2 A
= 2 π f 2 A = 4 π 2 f 2 A
= 4 × 3.14 2 × 1 5 2 × 0.06
=5.32×102 m s-2

Topic: Physics, Oscillations, NEET
A body moves along a circular path of radius 10 m and the coefficient of friction is 0.5. What should be its angular speed in rad s-1, if it is not to slip from the surface? (g=9.8 m s-2)

Options:
A 5
B 10
C 0.1
D 0.7

The correct answer is:
0.7

Solution:
For body to move on circular path. Frictional force provides the necessary centripetal force,
ie, frictional force = centripetal force
or Î¼mg=mv02r=mrÏ2
or Î¼g=rÏ2
â´ 0.5Ã9.8=10 Ï2
or Ï=0.7 rad s-1

Topic: Physics, Laws of Motion, NEET
A body moves along a circular path of the radius 5 m. The coefficient of friction between the surface of the path and the body is 0.5. The angular velocity in rad s-1, with which the body should move so that it does not have to leave the path is (Take g=10 m s-2)

Options:
A 4
B 3
C 2
D 1

The correct answer is:
1

Solution:
Here, Î¼ = 0.5
r = 5 m, g = 10 m s-2
The friction force provides the centripetal force
â´ mv 2 r = Î¼ mg or v 2 = Î¼ rg
or v = Î¼ rg = 0.5 5 m 1 0 m s - 2
= 5 m s - 1
As v = r Ï or Ï = v r = 5 m s - 1 5 m = 1 rad s - 1

Topic: Physics, Laws of Motion, NEET
A body moves along an uneven horizontal road with a constant speed at all points. The normal reaction of the road on the body is

Options:
A maximum at A
B maximum at B
C minimum at C
D same at A,B and C

The correct answer is:
maximum at A

Solution:
∴ NA=mg+mν2RA
NB=mg-mν2RB
NC=mg+mν2RC
From the curve,
RA<RC
∴ NA>NC
NA is maximum.

Topic: Physics, Laws of Motion, NEET
A body moves from a position r1=2i^-3j^-4k^ m to a position, r2=(3i^-4j^+5k^) m under the influence of a constant force F=4i^+j^+6k^ N. The work done by the force is

Options:
A 57 J
B 58 J
C 59 J
D 60 J

The correct answer is:
57 J

Solution:
Given,r1=2i^-3j^-4k^
And r2=3i^-4j^+5k^
Now, r2-r1=i^-j^+9k^
And F=4i^+j^+6k^
∴ work done = F.r
W=4i^+j^+6k^.(i^-j^+9k^)
=4-1+54=57 J

Topic: Physics, Work Power Energy, NEET
A body moves from a position r1=2i^-3j^-4k^ to a position, r 2 = 3 i ^ - 4 j ^ + 5 k ^ m under the influence of a constant force F=4i^+j^+6k^ N. The work done by the force is

Options:
A 57 J

B 58 J
C 59 J
D 60 J

The correct answer is:
57 J

Solution:
Given,r1=2i^-3j^-4k^
And r2=3i^-4j^+5k^
Now, r2-r1=i^-j^+9k^
And F=4i^+j^+6k^
∴ work done = F.r
W=4i^+j^+6k^.i^-j^+9k^
=4-1+54=57 J

Topic: Physics, Work Power Energy, NEET
A body moves with speed V1 for distance L and then with speed V2 for distance 2L. The average speed for the motion is

Options:
A V1+V22
B 3V1V2V1+2V2
C 3V1V22V1+V2
D 3V1V2V1+V2

The correct answer is:
3V1V22V1+V2

Solution:
Average speed =Total distanceTotal time

=3LLV1+2LV2

=3V1V2V2+2V1

Topic: Physics, Motion In One Dimension, NEET
A body moving in a straight line with constant acceleration covers distances a and b in successive equal time interval of t. The acceleration of the body is

Options:
A a+bt2
B b-at2
C 2b-a2t2
D b-2at2

The correct answer is:
b-at2

Solution:
a=ut+12 At2
b=(u+At)×t+12At2
⇒ A=(b-a)t2

Topic: Physics, Motion In One Dimension, NEET
A body moving with uniform acceleration describes 40 m in the first 5 s and 70 m in the next 5 s. Its initial velocity will be

Options:
A 4 m s-1
B 2.5 m s-1
C 5 m s-1
D 11 m s-1

The correct answer is:
5 m s-1

Solution:
s1=40 m, s2=70 m

∆t1=5 s, ∆t2=5 s

40=u∆t1+12a∆t12=5u+12a×25

⇒2u+5a=16

110=u∆t1+∆t2+12a∆t1+∆t22

110=10u+12a×100

⇒u+5a=11

u=5 m s-1

Topic: Physics, Motion In One Dimension, NEET
A body of mass 1 g and carrying a charge 10-8 C passes from two points P and Q . P and Q are at electric potentials 600 V and 0 V respectively. The velocity of the body at Q is 20 cm s-1 . Its velocity in m s-1 at P is

Options:
A 0.028
B 0.056
C 0.56
D 5.6

The correct answer is:
0.028

Solution:
Potentials, VP= 600 V VQ= 0 V
Conserving mechanical energy,
KP+ UP= KQ+ UQ
12mvP2+qvR=12mvQ2+qvQ
12mvP2-vQ2=qVQ-VP
1210-3vP2-0.22=10-80-600
vP2-0.04=-12Ã10-3
vP2=0.04-0.012
vP=0.028 m s-1

Topic: Physics, Electrostatics, NEET
A body of mass 1 kg is rotated in a horizontal circle of radius 1 m and moves with velocity 2 m s-1 . The work done in 10 revolutions is

Options:
A 40 J
B 20 J
C 4 × 2π 10 J
D zero

The correct answer is:
zero

Solution:
There is no change in the P.E. as well as in the K.E. Therefore, work done is zero.

Topic: Physics, Work Power Energy, NEET
A body of mass 1 kg begins to move from rest under the action of a time dependent force F→=2t i^+3t2 j^N, where i^and j^ are unit vectors along x and y axis. What power will be developed by the force at the time t?

Options:
A 2t2+3t3 W
B 2t2+4t4 W
C 2t3+3t4 W
D 2t3+3t5 W

The correct answer is:
2t3+3t5 W

Solution:
a→=2ti^+3t2j^
V → = 2t 2 2 i ^ + 3 3 t 3 j ^
F→=2ti^+3t2j^
P=F→.V→=2t3+3t5

Topic: Physics, Work Power Energy, NEET, NEET 2016 (Phase 1)
A body of mass 1 kg is connected to a string and is undergoing a complete vertical circular motion of radius 1 m. If it is barely able to complete the vertical circle, then the difference in its kinetic energy at the top and bottom of the circle is (Take g=10 ms-2)

Options:
A 10 J
B 20 J
C 30 J
D 50 J

The correct answer is:
20 J

Solution:
Difference in KE= 1 2 m[ ( 5gr ) 2 − ( gr ) 2 ]
=2mgr =20 J

Topic: Physics, Laws of Motion, NEET
A body of mass $1 \mathrm{~kg}$ is thrown upwards with a velocity $20 \mathrm{~ms}^{-1}$. It momentarily comes to rest after attaining a height of $18 \mathrm{~m}$. How much energy is lost due to air friction ? $\left(\mathrm{g}=10 \mathrm{~ms}^{-2}\right)$

Options:
A $20 \mathrm{~J}$
B $30 \mathrm{~J}$
C $40 \mathrm{~J}$
D $10 \mathrm{~J}$

The correct answer is:
$20 \mathrm{~J}$

Solution:
Key Idea The energy lost due to air friction is equal to difference of initial kinetic energy and final potential energy.
Initially body posses only kinetic energy and after attaining a height the kinetic energy is zero.
Therefore, loss of energy $=\mathrm{KE}-\mathrm{PE}$
$$
\begin{aligned}
& =\frac{1}{2} \mathrm{mv}^2-\mathrm{mgh} \\
& =\frac{1}{2} \times 1 \times 400-1 \times 18 \times 10 \\
& =200-180=20 \mathrm{~J}
\end{aligned}
$$

Topic: Physics, Work Power Energy, NEET, NEET 2009 (Screening)
A body of mass 1 kg tied to a string is just able to move in a vertical circular path of the radius 1 m. The difference between the kinetic energies at its highest and lowest positions of the body is

Options:
A 45 J
B 20 J

C 10 J

D 30 J

The correct answer is:
20 J

Solution:
Kinetic energy is the energy possessed by a body of mass m due to its velocity v.
KE=12mv2
At the highest point velocity, vA=rg
∴ KE=12mvA2
At the lowest point velocity, vB=5 rg
KE=12mvB2
Difference n KE=12mvB2-vA2=12m5 rg-rg
=2mrg=2×1×1×10=20 J

Topic: Physics, Work Power Energy, NEET
A body of mass 2âkg is thrown up vertically with kinetic energy of 490âJ . If g=9.8 m s-2, the height at which the kinetic energy of the body becomes half of the original value, is

Options:
A 50Â m
B 25Â m
C 12.5 m
D 19.6Â m

The correct answer is:
12.5 m

Solution:
Initial kinetic energy is 490 J
Let h be the height at which kinetic energy becomes half of the initial value
KE 2 = 490 2 =245âJ
Gain in Potential energy at height h is mgh
This is equal change in kinetic energy
ÎKE=â490âJâ245âJ=245âJâ
mgh=245 J
h= 245 2Ã9.8 =12.5âm

Topic: Physics, Work Power Energy, NEET
A body of mass 2 kg moving on a horizontal surface with an initial velocity of 4 m s-1 comes to rest after 2 s . If one wants to keep this body moving on the same surface with a velocity of 4 m s-1 the force required is

Options:
A 8 N
B 4 N
C Zero
D 2 N

The correct answer is:
4 N

Solution:
u=4m/s, v=0, t=2 sec
v=u+at ⇒ 0=4+2a ⇒ a=-2m/s2
∴ Retarding force=ma=2×2=4 N
This force opposes the motion. If the same amount of force is applied in forward direction, then the body will move with constant velocity.

Topic: Physics, Laws of Motion, NEET
A body of mass 3 kg acted upon by a constant force is displaced by S metre, given by relation S=13t2 , where t is in second. Work done by the force in 2 seconds is :

Options:
A 8/3 J
B 19/5Â J
C 5/19Â J
D 3/8Â J

The correct answer is:
8/3 J

Solution:
S=t23
v=dSdt
dSdt=23t=v;
W=âK=Kt=2-Kt=0
W=12mvt=22-vt=02=12â3169-0= 8/3 J

Topic: Physics, Work Power Energy, NEET
A body of mass 3 kg is moving under the action of a force such that its displacement is given by the relation s=13t2, where t is time in second. The work done by the force in 2 s is

Options:
A 83 J
B 195 J
C 519 J
D 38 J

The correct answer is:
83 J

Solution:
s=13t2
v=dsdt=23t,a=d2sdt2=23
F=ma=3×23=2 N
W= 2×13t2
At t=2 s,
W=2×13×2×2=83 J

Topic: Physics, Work Power Energy, NEET
A body of mass $3 \mathrm{~kg}$ is under a constant force which causes a displacement $s$ in metres in it, given by the relation $s=\frac{1}{3} t^2$, where $t$ is in seconds. Work done by the force in 2 seconds is:

Options:
A $\frac{19}{5} \mathrm{~J}$
B $\frac{5}{19} \mathrm{~J}$
C $\frac{3}{8} \mathrm{~J}$
D $\frac{8}{3} \mathrm{~J}$

The correct answer is:
$\frac{8}{3} \mathrm{~J}$

Solution:
It is given that:
$s=\frac{t^2}{3}$
Differentiating w.r.t. $t$ we have
$U=\frac{d s}{d t}=\frac{2 t}{3}$
Again differentiating w.r.t. $t$ we have
$a=\frac{d^2 s}{d t^2}=\frac{2}{3}$
Now $\mathrm{W}=\int \mathrm{F} d s=\int \mathrm{mads}$
$\begin{aligned}
& =\int m \times \frac{2}{3} \times \frac{2 t}{3} d t \\
& =\frac{4}{9} m \int t d t \\
W & =\frac{4}{9} m \times\left[\frac{t^2}{2}\right]_0^2 \\
& =\frac{4}{9} \times 3 \times 2=\frac{8}{3} \mathrm{~J}
\end{aligned}$

Topic: Physics, Work Power Energy, NEET, NEET 2006
A body of mass 3 kg moving with a velocity 2i^+3j^+3k^ m s-1collides with another body of mass 4 kg moving with a velocity 3i^+2j^-3k^ m s-1. The two bodies stick together after collision. The velocity of the composite body is

Options:
A 174i^+6j^-3k^
B 1718i^+17j^-3k^
C 17 6i^+4j^-6k^
D 17 9i^+8j^-6k^

The correct answer is:
1718i^+17j^-3k^

Solution:
Vâcm=m1Vâ1+m2Vâ2m1+m2
V→cm=3 2i^+3j^+3k^ +4 ( 3i^+2j^-3k^ )3+4
=18i^+17j^-3k^7

Topic: Physics, Center of Mass Momentum and Collision, NEET
A body of mass 4 kg is accelerated upon by a constant force, travels a distance of 5 m in the first second and a distance of 2 m in the third second. The force acting on the body is

Options:
A 2 N

B 4 N

C 6 N

D 8 N

The correct answer is:
6 N

Solution:
Distance travelled by the body in nth second is given by
Sn=u+a2(2n-1)
5=u+a22Ã1-1
5=u+a2 â¦..i
2=u+a22Ã3-1
2=u+52aâ¦.ii
Solving Eqs. (i) and (ii), we get
a=-64 m s-2
ie, body is decelerating
Given, mass=4 kg
â´F=mÃa=4Ã64=6 N

Topic: Physics, Laws of Motion, NEET
A body of mass 4m is lying in x - y plane at rest. It suddenly explodes into three pieces. Two pieces, each of mass m move perpendicular to each other with equal speeds υ. The total kinetic energy generated due to explosion is:

Options:
A mυ2
B 32mυ2
C 2 mυ2
D 4 mυ2

The correct answer is:
32mυ2

Solution:
The total momentum of the system will be conserved before and after explosion.

Then, pi=pf
O=mvi^+mvj^+2mv→
v→=-v2i^-v2j^
v→=v2
Total kinetic energy generated due to explosion is KE=v2mv2+v2mv2+v22mv22=mv2+mv22=32mv2

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2014
A body of mass 5 kg is thrown vertically up with a kinetic energy of 490 J. The height at which the kinetic energy of the body becomes half of the original value is : (take g as 9.8 m s-2)

Options:
A 12.5 m
B 10 m
C 2.5 m
D 5 m

The correct answer is:
5 m

Solution:
According to the law of conservation of energy
12mu2=12mv2+mgh
490=245+5×9.8×h
h=24549=5 m

Topic: Physics, Work Power Energy, NEET
A body of mass 5 kg starts from the origin with an initial velocity u=30i^+40j^ m s-1. If a constant force -6i^-5j^ N acts on the body, the time in which the y component of the velocity becomes zero is

Options:
A 5 s
B 20 s
C 40 s
D 80 s

The correct answer is:
40 s

Solution:
From Newton's second law of motion,

Fy=â³Pyâ³t
â³t=â³PyFy
=0-5Ã40-5=40 s

Topic: Physics, Laws of Motion, NEET
A body of mass 0.5 kg travels in a straight line with velocity v =a x 3/2 where a = 5 m–1/2 s–1 . What is the work done by the net force during its displacement from x = 0 to x = 2 m ?

Options:
A 50 J
B 60J
C 70J
D 80J

The correct answer is:
50 J

Solution:
Correct Option is : (A)
50 J

Topic: Chemistry, Work Power Energy, NEET
A body of mass 0.5 kg travels in a straight line with velocity, v = ax3/2, where a = 5m –1/2/s. What is the work done by the net force during its displacement from x = 0 to x = 2 m?

Options:
A 50 J
B 10 J
C 20 J
D 30 J

The correct answer is:
50 J

Solution:
Mass of the body (m) = 0.5 kg
Velocity of the body (v) = ax3/2
where, a = 5m–1/2/s
Velocity of the body at x = 0,
v1 = 5 × 0 = 0
Velocity of the body at c = 2 m
v2 = 5 × (2)3/2m/s
According to work-energy theorem
Work done = Change in K.E.
= 1 2 mv 2 2 - 1 2 mv 1 2
= 1 2 mâ¡ v 2 2 - v 1 2 = 1 2 × 0.5 5 × 2 3 / 2 2 - 0 2
= 1 2 × 0.5 × 2 5 × 8
= 50 J

Topic: Physics, Work Power Energy, NEET