datavorous/entrance-exam-dataset huggingface dataset:
A bacterium divides every 35 minutes. If a culture containing 105 cells per ml is grown for 175 minutes, what will be the cell concentration per ml after 175 minutes?

Options:
A 32 X 105 cells
B 5 X 105 cells
C 35 X 105 cells
D 175 X 105 cells

The correct answer is:
32 X 105 cells

Solution:
The 'n', the time difference is 175/35 = 5 minutes, then, 2n = 25 = 32.
The culture initially had 105 cells/ml, therefore, 32 X 105 cells will be present after 175 minutes.

Topic: Biology, Cell Cycle and Cell Division, NEET
A bacterium divides every 35 minutes. If a culture containing 105 cells/ml is grown for 2 hours 55 minutes, what will be the cell concentration per ml?

Options:
A 32 Ã 105 cells
B 35Â ÃÂ 105 cells
C 64Â ÃÂ 105 cells
D 5Â ÃÂ 105 cells

The correct answer is:
32 Ã 105 cells

Solution:
The bacterium divides after every 35 minutes through simple mitotic division. The duration of cell division allowed is 2 hours 55 minutes (175 minutes). Therefore, the number of divisions are 17535=5
Since one bacterium on division produces two cells, the cell concentration after 175 minutes will be
=105×(2)5=32×105

Topic: Biology, Biological Classification, NEET
A bacterium having rDNA formed by using pBR322 and EcoRI restriction enzyme is resistant to:

Options:
A Penicillin and ampicillin
B Ampicillin and tetracycline
C Tetracycline only
D Ampicilin only

The correct answer is:
Ampicillin and tetracycline

Solution:
Plasmids are able to replicate in the host cell independent of the control of chromosomal DNA. Plasmids like pBR322 is used as a cloning vector in E. coli organisms. The pBR322 plasmid has resistance site for two antibiotics - tetracycline and ampillicin. If we ligate a foreign DNA at the site of tetracycline resistance the recombinant plasmids will not be tetracycline resistant. We can obtain the recombinants by growing them on ampicillin-resistant medium. The restriction site for EcoRI is neither on ampicillin resistance gene nor tetracycline resistant gene. Thus it will be resistant to both these antibiotics

Topic: Biology, Biotechnology - Principles and Processes, NEET
A ball A is thrown vertically upward with speed u. At the same instant, another ball B is released from rest at height h. At time t, the speed of A relative to B is (Assume both are in the air)

Options:
A u
B u-2gt
C u2-2gh
D u−gt

The correct answer is:
u-2gt

Solution:
From equation of motion, we have
v=u+gt
where u is initial velocity and t is time.
For ball A,â v A =uâgt
For ball B,â v B =0+gt
â´ Relative speed
Î v â = v A â â v B â = u - gt j ^ - + gt j ^ = uâ2gt j ^

Topic: Physics, Motion In One Dimension, NEET
A ball falling in a lake of 200 m shows a decrease of 0.1 % in its volume. The Bulk modulus of elasticity of the material of the ball is (take g=10 m s-2 )

Options:
A 109 N m-2
B 2×109 N m-2
C 3×109 N m-2
D 4×109 N m-2

The correct answer is:
2×109 N m-2

Solution:
We know that pressure difference between surface point and a point at depth
ΔP=hdg
⇒∆P=200×103×10
⇒∆P=2×106 N m-2
Bulk modulus K=-ΔPΔVV
K=-2×106-0.1100
K=2×109 N m-2

Topic: Physics, Mechanical Properties of Solids, NEET
A ball falls from a height of 5 m and strikes a lift which is moving in the upward direction with a velocity of 1 m s-1 , then the velocity with which the ball rebounds after collision will be

Options:
A 11 m s-1 downwards
B 12 m s-1 upwards
C 13 m s-1 upwards
D 12 m s-1 downwards

The correct answer is:
12 m s-1 upwards

Solution:
Velocity after a fall of 5 m is,
u1=2Ã10Ã5=10 m s-1
vÂ¯1-vÂ¯2=euÂ¯2-uÂ¯1
v1-1=1--10
v1=12 m s-1

Topic: Physics, Center of Mass Momentum and Collision, NEET
A ball having velocity v towards the right and having angular velocity clockwise approaches the wall. It collides elastically with wall and moves towards the left. Ground and wall are frictionless. Select the correct statement about the angular velocity of the ball after the collision.

Options:
A It will be anti-clockwise
B It becomes zero
C Angular speed decreases
D It will be clockwise

The correct answer is:
It will be clockwise

Solution:
As collision is elastic, so after collision ball moves towards left with speed v. As walls and ground are smooth, there is no tangential torque on the balls. Only normal forces and mg force pass through the centre of the ball. So, their torques about its centre are zero. By τ=Iα, angular acceleration is zero, hence angular velocity does not change.

Topic: Physics, Center of Mass Momentum and Collision, NEET
A ball impinges directly on a similar ball at rest. The first ball is brought to rest by the impact. If half of the kinetic energy is lost by the impact, the value of the coefficient of restitution is

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

The correct answer is:
1 2

Solution:
Let u1 and v1 be the initial and final velocities for ball 1 and u2 and v2 be the initial and final velocities for ball 2. Here, u2=0 and v1=0.
Ki=12mu12+12mu22=12mu12
K f = 1 2 mv 1 2 + 1 2 mv 2 2 = 1 2 mv 2 2
Loss of K.E = 1 2 mu 1 2 - 1 2 mv 2 2
According to question,
1 2 1 2 mu 1 2 = 1 2 mu 1 2 - 1 2 mv 2 2
(as half of K.E. is lost by the impact)
∴ u 1 2 = 2 v 2 2 or v 2 = u 1 2
∴ e ⁡ = v 2 - v 1 u 1 - u 2 = v 2 u 1 = 1 2

Topic: Physics, Work Power Energy, NEET
A ball is dropped from a certain height on a horizontal floor. The coefficient of restitution between the ball and the floor is 1 2 . The displacement-time graph of the ball will be

Options:
A
B ​
C
D ​

The correct answer is:

Solution:
The ball will stop after a long time. The final displacement of the ball will be equal to the height. The motion is first accelerated, then retarded, then accelerated and so on. Hence, the correct graph is

Topic: Physics, Center of Mass Momentum and Collision, NEET
A ball is dropped from a height h on the ground. If the coefficient of restitution is e, the height to which the ball goes up after it rebounds for the nth time is

Options:
A he2n
B hen
C e2nh
D he2n

The correct answer is:
he2n

Solution:
After each collision speed becomes e times
â´ Speed after 1st collision is Î½1=eÎ½0
â´ Maximum height after 1st collision is,
h1=Î½122g=e2Î½022g=e2h
Similarly maximum height after 2nd collision is,
h2=e2h1=e4h
maximum height after 3rd collision is,
h3=e2h2=e6h
â´ The height to which the ball goes up after it rebounds for the nth time is hn=e2nh

Topic: Physics, Work Power Energy, NEET
A ball is dropped from a height of 20 m above the surface of the water in a lake. The refractive index of water is 4 3 . A fish inside the lake, in the line of fall of ball is looking at the ball. At an instant, when the ball is 12.8 m above the water surface, the fish sees speed of ball as

Options:
A 9 m s -1
B 1 2 m s -1
C 1 6 m s -1
D 21.33 m s -1

The correct answer is:
1 6 m s -1

Solution:
Ï = 2 gh = 2 Ã 1 0 Ã 7
= 1 2 ms -1
In this case when eye is inside water,

x app. = Î¼ x
â´ d x app. dt = Î¼ Â· d x dt
or, Ï app. Î¼ Ï = 4 3 Ã 1 2 = 1 6 m s -1

Topic: Physics, Ray Optics, NEET
A ball is dropped from a high rise platform at $\mathrm{t}=0$ starting from rest. After $6 \mathrm{~s}$ another ball is thrown downwards from the same platform with a speed $\mathrm{v}$. The two balls meet at $\mathrm{t}=18 \mathrm{~s}$. What is the value of $\mathrm{v}$ ? (take $g=10 \mathrm{~ms}^{-2}$ )

Options:
A $74 \mathrm{~ms}^{-2}$
B $55 \mathrm{~ms}^{-1}$
C $40 \mathrm{~ms}^{-1}$
D $60 \mathrm{~ms}^{-1}$

The correct answer is:
$74 \mathrm{~ms}^{-2}$

Solution:
. For first ball, $\mathrm{u}=0$
$$
\therefore \quad \mathrm{s}_1=\frac{1}{2} \mathrm{gt}_1^2=\frac{1}{2} \times \mathrm{g}(18)^2
$$
For second ball, initial velocity $=\mathrm{v}$
$$
\begin{aligned}
& \therefore \mathrm{s}_2=\mathrm{vt}_2+\frac{1}{2} \mathrm{gt}^2 \\
& \Rightarrow \quad \mathrm{t}_2=18-6=12 \mathrm{~s} \\
& \Rightarrow \quad \mathrm{s}_2=\mathrm{v} \times 12+\frac{1}{2} \mathrm{~g}(12)^2
\end{aligned}
$$
Here, $\quad \mathrm{s}_1=\mathrm{s}_2$
$$
\begin{aligned}
\frac{1}{2} g(18)^2 & =12 \mathrm{v}+\frac{1}{2} g(12)^2 \\
\Rightarrow \quad v & =74 \mathrm{~ms}^{-1}
\end{aligned}
$$

Topic: Physics, Motion In One Dimension, NEET, NEET 2010 (Screening)
A ball is dropped from height H on to a horizontal surface. If the coefficient of restitution is e then the total time after which it comes to rest is

Options:
A 2Hg1- e1+ e
B 2Hg1+e1-e
C 2Hg1+e21- e2
D 2Hg1- e21+e2

The correct answer is:
2Hg1+e1-e

Solution:
The time it takes for the ball to fall from a height is given by

t=2Hg

The height up to which the ball bounces after the nth collision from the floor is

Hn=e2nH

t=2Hg+22H1g+22H2g+…∞
t=2Hg1+2e+2e2+ ……..∞
t=2Hg1+e1-e

Topic: Physics, Center of Mass Momentum and Collision, NEET
A ball is dropped on a floor and bounces back to a height somewhat less than the original height. The curve which depicts its motion correctly about y and t is

Options:
A
B
C
D

The correct answer is:

Solution:
When a ball is dropped on a floor then
y=12 gt2 â¦i
So, the graph between yand t is a parabola. Here y decrease as time decrease. When the ball is bounces back then
y=ut-12gt2 â¦ii
Eq. (ii) is also the form of a general equation of parabola so the graph between y and t will be a parabola. Here y increases when the time increases. Hence, the required graph between y and t is shown in the figure.

Topic: Physics, Motion In One Dimension, NEET
A ball is dropped vertically from a height d above the ground. It hits the ground and bounces up vertically to a height d2, Neglecting subsequent motion and air resistance, its velocity v varies with height h above the ground as

Options:
A
B
C
D

The correct answer is:

Solution:
(i) For uniformly accelerated/deaccelerated motion, v2=u2±2gh
i.e., v - h graph will be a parabola (because the equation is quadratic).
(ii) Initially, velocity is downwards (-ve) and then after the collision it reverses its direction with lesser magnitude. i.e., velocity is upwards (+ve). Graph (a) satisfies both these conditions.
Note that time t=0 corresponds to the point on the graph where h = d.

Topic: Physics, Motion In One Dimension, NEET
A ball is projected from a certain point on the surface of a planet at a certain angle with the horizontal surface. The horizontal and vertical displacement x and y vary with time t (in seconds) as x=103t and y=10t-t2. The maximum height attained by the ball is

Options:
A 100 m

B 75 m

C 50 m

D 25 m

The correct answer is:
25 m

Solution:
dy dt = v y = 10 - 2t
at maximum height v y = 0
t = 5 s
So maximum height attained ⇒ y at = 5 = 10 t - t 2
= 50 - 25
= 25 m

Topic: Physics, Motion In Two Dimensions, NEET
A ball is projected from a certain point on the surface of a planet at a certain angle with the horizontal surface. The horizontal and vertical displacement x and y vary with time t (in seconds)
as x=103t and y=10t-t2. The maximum height attained by the ball is

Options:
A 100 m
B 75 m
C 50 m
D 25 m

The correct answer is:
25 m

Solution:
dy dt = v y = 10 - 2t
at maximum height v y = 0
t = 5 sec
So maximum height attained ⇒ y at = 5 = 10 t - t 2
= 50 - 25
= 25 m

Topic: Physics, Motion In Two Dimensions, NEET
A ball is projected from point $A$ with velocity $20 \mathrm{~m} \mathrm{~s}^{-1}$ at an angle $60^{\circ}$ to the horizontal direction. At the highest point $B$ of the path (as shown in figure), the velocity $v \mathrm{~m} \mathrm{~s}^{-1}$ of the ball will be

Options:
A Zero
B $10$
C $20$
D $10 \sqrt{3}$

The correct answer is:
$10$

Solution:
At the highest point of a projectile, vertical component of velocity becomes zero. Therefore at point $B, v=20 \cos 60^{\circ}=20 \times \frac{1}{2}=10 \mathrm{~m} / \mathrm{s}$

Topic: Physics, Motion In Two Dimensions, NEET, NEET 2023 (Manipur)
A ball is projected from the bottom of an inclined plane of inclination 30o , with a velocity of 30 m s-1, at an angle of 30o with the inclined plane. If g=10 ms-2, then the range of the ball on given inclined plane is

Options:
A 12 m
B 60 m
C 120 m
D 600 m

The correct answer is:
60 m

Solution:
Range on incline plane,
R = 2 u 2 sin Î± - Î² cos Î± g cos 2 Î²
Here, Î² = 30 Î¿ and Î± = 30 Î¿ + 30 Î¿ = 60 Î¿
R= 2 ( 30 ) 2 sin( 30Â° )cos( 60Â° ) 10 cos 2 â( 30Â° )
= 2 ( 30 ) 2 ( 1 2 )( 1 2 ) 10( 3 4 )
= 60 m

Topic: Physics, Motion In Two Dimensions, NEET
A ball is projected upwards from the top of the tower with a velocity 50 m s-1 making an angle 30Â° with the horizontal. The height of tower is 70 m . After how many seconds the ball will strike the ground?

Options:
A 3 s
B 5 s
C 7 s
D 9 s

The correct answer is:
7 s

Solution:
YÂ¯=vÂ¯yt+12aÂ¯yt2
70=-50sinâ¡30Â°t+12Ã10Ãt2
70=-25t+5t2
or t2-5t-14=0
t=7 s ( -ve time is not possible)

Topic: Physics, Motion In Two Dimensions, NEET
A ball is projected with a velocity, 10 m s-1, at an angle of 60° with the vertical direction. Its speed at the highest point of its trajectory will be

Options:
A 53 m s-1
B 5 m s-1
C 10 m s-1
D Zero

The correct answer is:
53 m s-1

Solution:
The speed of a projectile at highest point is equal to the component of velocity of projection along the horizontal direction. The angle given with the vertical is 60° therefore, the angle with the horizontal will be 30°.

Therefore,

V=ucosθ ⇒V=10cos30°=10×32

⇒V=53 m s-1

Topic: Physics, Motion In Two Dimensions, NEET, NEET 2022 (Phase 1)
A ball is released from the top of a tower of height h metres. It takes T seconds to reach the ground. What is the position of the ball in T 3 seconds?

Options:
A

17h 18 metre from the ground

B

8h 9 metre from the ground

C

7h 9 metre from the ground

D

h 9 metre from the ground

The correct answer is:

8h 9 metre from the ground

Solution:
⇒h=12gT2

or T= 2h g

Now, S= 1 2 g ( T 3 ) 2 = g 18 × 2h g = h 9

Height above the ground =h− h 9 = 8h 9

Topic: Physics, Motion In One Dimension, NEET
A ball is released from the top of a tower. The ratio of work done by the force of gravity in 1st second, 2nd second and 3rd second of the motion of the ball is

Options:
A 1 : 2 : 3

B 1 : 4 : 16
C 1 : 3 : 5
D 1 : 9 : 25

The correct answer is:
1 : 3 : 5

Solution:
Initial velocity of ball is zero ie., u=0
â´ Displacement of ball in tth second
s=gt-12g=gt-12
sât-12
or s1:s2:s3=1-12:2-12:3-12
=1:3:5
Now, W=mgs
Wâs
â´W1:W2:W3=1:3:5

Topic: Physics, Work Power Energy, NEET
A ball is suspended by a thread from the ceiling of a tram car. The brakes are applied and the speed of the car changes uniformly from 36 Km h-1 to zero in 5 s. The angle by which the ball deviates from the vertical is g=10 m s-2

Options:
A tan-1⁡13
B sin-1⁡15
C tan-1⁡15
D cot-1⁡13

The correct answer is:
tan-1⁡15

Solution:
u=36 km h-1=10 m s-1

v=u+at , tan⁡θ=ag

v=10-a×5 (given, v=0)

a = 2

tan⁡θ=210

Topic: Physics, Motion In One Dimension, NEET
A ball is thrown upwards, it takes 4 s to reach back to the ground. Find its initial velocity

Options:
A 30 m/s
B 10 m/s
C 40 m/s
D 20 m/s

The correct answer is:
20 m/s

Solution:
The ball traverses distance upwards and then downwards.
Hence, time taken is going from A to B is t 2 = 4 2 = 2 s
Also at the maximum height B , speed is zero, therefore
v=u-g tmax

where v=0, tmax=2s, g = 10â m/s 2
â´ u=g tmax = 10Ã2=20âm/s

Topic: Physics, Motion In One Dimension, NEET
A ball is thrown vertically downward with a velocity of 20 m s-1 from the top of a tower. It hits the ground after some time with a velocity of 80 m s-1. The height of the tower is: g=10 m s-2

Options:
A 340 m
B 320 m
C 300 m
D 360 m

The correct answer is:
300 m

Solution:
We know that,

v2=u2+2gh
⇒802=202+2×10h
⇒h=300 m

Topic: Physics, Motion In Two Dimensions, NEET, NEET 2020 (Phase 1)
A ball is thrown vertically downwards from a height of 20 m with an initial velocity v0. It collides with the ground, loses 50 percent of its energy in collision and rebounds to the same height. The initial velocity v0 is: (Take g=10 m s-2)

Options:
A 20 m s-1
B 28 m s-1
C 10 m s-1
D 14 m s-1

The correct answer is:
20 m s-1

Solution:
kinetic energy of the ball on reaching ground
K E 1 =mgh+ 1 2 m v 0 2
after collision K E 1 2 =mgh
mgh+ 1 2 m v 0 2 2 =mgh
v 0 = 2gh =20 m   s −1

Topic: Physics, Work Power Energy, NEET, NEET 2015 (Phase 2)
A ball is thrown vertically downwards from a height of 20 m with an initial velocity v0. It collides with the ground, loses 50 percent of its energy in collision and rebounds to the same height. The initial velocity v0 is: (Take g=10 m s-2)

Options:
A 20 m s-1
B 28 m s-1
C 10 m s-1
D 14 m s-1

The correct answer is:
20 m s-1

Solution:
kinetic energy of the ball on reaching ground
K E 1 =mgh+ 1 2 m v 0 2
after collision K E 1 2 =mgh
mgh+ 1 2 m v 0 2 2 =mgh
v 0 = 2gh =20âmââ s â1

Topic: Physics, Work Power Energy, NEET
A ball is thrown vertically. It has a speed of $10 \mathrm{~m} / \mathrm{sec}$ when it has reached one half of its maximum height. How high does the ball rise? Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$.

Options:
A $10 \mathrm{~m}$
B $5 \mathrm{~m}$
C $15 \mathrm{~m}$
D $20 \mathrm{~m}$

The correct answer is:
$10 \mathrm{~m}$

Solution:
Using equation of motion
$$
V^2=u^2-2 g h
$$
After reading max height velocity because zero.
$\begin{aligned} 0 & =(10)^2-2 \times 10 \times \frac{1}{2} \\ \Rightarrow \quad h & =\frac{200}{20}=10 \mathrm{~m}\end{aligned}$

Topic: Physics, Motion In One Dimension, NEET, NEET 2005
A ball moving on a horizontal frictionless plane hits an identical ball at rest with a velocity of 50 cm s-1. If the collision is elastic, calculate the speed imparted to the target ball if the speed of the striking ball after the collision is 30 cm s-1

Options:
A 20 cm s-1
B 30 cm s-1
C 40 cm s-1
D 50 cm s-1

The correct answer is:
40 cm s-1

Solution:
As the collision is perfectly elastic
Hence 12mu12+12mu22=12mv12+12mv22
âu12+u22=v12+v22
0.52+0=0.32+v22
v2=0.52-0.32
v2=40 cm s-1

Topic: Physics, Work Power Energy, NEET
A ball moving with a certain velocity hits another identical ball at rest. If the plane is frictionless and collision is an elastic oblique collision, the angle between the directions in which the balls move after the collision, will be

Options:
A 30°
B 60°
C 90°
D 120°

The correct answer is:
90°

Solution:
This is an example of elastic oblique collision. When a moving body collides obliquely with another identical body in rest, then during elastic collision, the angle of divergence will be 90°

Topic: Physics, Center of Mass Momentum and Collision, NEET
A ball moving with velocity $2 \mathrm{~ms}^{-1}$ collides head on with another stationary ball of double the mass. If the coefficient of restitution is 0.5 , then their velocities (in $\mathrm{ms}^{-1}$ ) after collision will be

Options:
A 0,1
B 1,1
C $1,0.5$
D 0,2

The correct answer is:
0,1

Solution:
If two bodies collide head on with coefficient of restitution
$$
\mathrm{e}=\frac{\mathrm{v}_2-\mathrm{v}_1}{\mathrm{u}_1-\mathrm{u}_2}
$$
From the law of conservation of linear momentum
$$
\begin{array}{r}
\mathrm{m}_1 \mathrm{u}_1+\mathrm{m}_2 \mathrm{u}_2=\mathrm{m}_1 \mathrm{v}_1+\mathrm{m}_2 \mathrm{v}_2 \\
\Rightarrow \quad \mathrm{v}_1=\left[\frac{\mathrm{m}_1-\mathrm{em}_2}{\mathrm{~m}_1+\mathrm{m}_2}\right] \mathrm{u}_1+\left[\frac{(1+\mathrm{e}) \mathrm{m}_2}{\mathrm{~m}_1+\mathrm{m}_2}\right] \mathrm{u}_2
\end{array}
$$
Substituting $\mathrm{u}_1=2 \mathrm{~ms}^{-1}, \mathrm{u}_2=0, \mathrm{~m}_1=\mathrm{m}$ and
$$
\mathrm{m}_2=2 \mathrm{~m}, \mathrm{e}=0.5
$$
we get
$$
\begin{array}{lc}
\text { we get } & \mathrm{v}_1=\left[\frac{\mathrm{m}-\mathrm{m}}{\mathrm{m}+2 \mathrm{~m}}\right] \times 2 \\
\Rightarrow & \mathrm{v}_1=0
\end{array}
$$
Similarly,
$$
\begin{aligned}
\mathrm{v}_2 & =\left[\frac{(1+\mathrm{e}) \mathrm{m}_1}{\mathrm{~m}_1+\mathrm{m}_2}\right] \mathrm{u}_1+\left[\frac{\mathrm{m}_2-\mathrm{em}_1}{\mathrm{~m}_1+\mathrm{m}_2}\right] \mathrm{u}_2 \\
& =\left[\frac{1.5 \times \mathrm{m}}{3 \mathrm{~m}}\right] \times 2 \\
& =1 \mathrm{~ms}^{-1}
\end{aligned}
$$

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2010 (Screening)
A ball of mass 0.2 kg rests on a vertical post of height 5 m. A bullet of mass 0.01 kg, travelling with a velocity V m s-1 in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of 20 m and the bullet at a distance of 100 m from the foot of the post. The velocity V of the bullet is

Options:
A 250 m s-1

B 350 m s-1
C 400 m s-1
D 500 m s-1

The correct answer is:
500 m s-1

Solution:
Time of flight tf=2hg=2×510=1 s
Horizontal range (R) = horizontal velocity × time of flight
∴ Horizontal velocities of the bullet and of the ball after the collision respectively are
vbullet=1001=100 m s-1
vball=201=20 m s-1
From the conservation of momentum, Total initial momentum = total final momentum
⇒mbullet×V=mbullet×vbullet+mball×vball
⇒0.01 V=0.01 ×100+0.2×20
⇒V=500 m s-1

Topic: Physics, Laws of Motion, NEET
A ball of mass 0.15 kg is dropped from a height 10 m strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is g=10 m s-2 nearly :

Options:
A 0 kg m s-1
B 2.1 kg m s-1
C 4.2 kg m s-1
D 1.4 kg m s-1

The correct answer is:
4.2 kg m s-1

Solution:
m=0.15 kg; h=10 m

v=2gh=2×9.8×10=14 m s-1

As we know,

Impulse=change in momentum

=pf-pi=0.15×14--0.15×14

=2×0.15×14

=4.2 kg m s-1

( To raise back to same height, rebounds velocity is same as the hitting velocity)

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2021
A ball of mass 1 kg moving with a velocity of 0.4 m s-1 collides with another stationary ball. After the collision, the first ball moves with a velocity of 0.3 m s-1 in a direction making an angle of 90o with its initial direction. The momentum of the second ball after the collision will be (in kg m s-1)

Options:
A 0.1
B 0.3
C 0.5
D 0.7

The correct answer is:
0.5

Solution:
In collision, final momentum = Initial momentum
p→f=p→i
p→2+10.3j^=10.4i^
p→2=0.4i^-0.3j^
p→2=0.5kg⁡-m/s

Topic: Physics, Center of Mass Momentum and Collision, NEET
A ball of mass $2 \mathrm{~kg}$ and another of mass $4 \mathrm{~kg}$ are dropped together from a 60 feet tall building. After a fall of 30 feet each towards earth, their respective kinetic energies will be in the ratio of:

Options:
A $\sqrt{2}+1$
B $1: 4$
C $1: 2$
D $1: \sqrt{2}$

The correct answer is:
$1: 2$

Solution:
Let $E_1$ and $E_2$ be the K.E. of two bodies.
Then $\quad E_1=\frac{1}{2} m_1 v_1^2$
and $E_2=\frac{1}{2} m_2 v_2{ }^2$
$$
\therefore \quad \frac{E_1}{E_2}=\frac{\frac{1}{2} m_1 v_1^2}{\frac{1}{2} m_2 v^2}=\frac{m_1 v_1^2}{m_2 v_2^2}
$$
The initial velocity of both the bodies are zero.
$$
\begin{aligned}
& \therefore \quad v^2=2 g h \text { is same for both } \\
& \text { bodies } \\
& \therefore \quad \frac{E_1}{E_2}=\frac{m_1}{m_2}=\frac{2}{4}=\frac{1}{2}
\end{aligned}
$$

Topic: Physics, Work Power Energy, NEET, NEET 2004
A ball of mass 2 kg and another of mass 4 kg are dropped together from a 60 ft tall building. After a fall of 30 ft each towards earth, their respective kinetic energies will be in the ratio of

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

The correct answer is:
1:2

Solution:
Since travelling distance is same for both balls, their velocities at the end of 30 feet journey will be same
â´K.E1K.E2=m1v02m2v02=m1m2=24=12

Topic: Physics, Motion In One Dimension, NEET
A ball of mass 0.2 kg moves with a velocity of 20 m s-1 and stops in 0.1 s, then the force on the ball is

Options:
A 40 N
B 20 N
C 4 N
D 2 N

The correct answer is:
40 N

Solution:
F=ma=mΔvΔt=0.2×200.1=40 N

Topic: Physics, Laws of Motion, NEET
A ball of mass 150 g is hit by a force, which acts on it for 0.1 s thereby creating an acceleration 20âmââsâ2. The impulse of force is :

Options:
A 0.5 N s
B 0.1 N s
C 0.3 N s
D 1.2 N s

The correct answer is:
0.3 N s

Solution:
Mass = 150gm = 150×10-3 kg
F=ma=150×10-3×20
F=3 N
Impulse = F∆t=3×0.1=0.3 N s

Topic: Physics, Laws of Motion, NEET