datavorous/entrance-exam-dataset huggingface dataset:
A bomb travelling in a parabolic path under gravity explodes in mid-air. The centre of mass of fragments will move

Options:
A vertically upwards and then downwards
B vertically downwards
C in an irregular path
D in the parabolic path as the unexploded bomb would have travelled.

The correct answer is:
in the parabolic path as the unexploded bomb would have travelled.

Solution:
As there is no net external force in the horizontal direction, hence the motion of the centre of mass of fragments should have been as before.

Topic: Physics, Center of Mass Momentum and Collision, NEET
A bottle containing two immiscible liquids is given to you. They may be separated by using a

Options:
A Fractionating column
B Separating funnel
C Vacuum distillation
D Steam distillation

The correct answer is:
Separating funnel

Solution:
Being immiscible liquids, they form different layers hence can be separated with the help of separating funnel.

Topic: Chemistry, General Organic Chemistry, NEET
A bottle of cold drink contains 200 mL liquid in which CO2 is 0.1 molar. Suppose CO2 behaves like an ideals gas, the volume of the dissolved CO2 at STP is

Options:
A 0.224 litre
B 0.448 litre
C 22.4 litre
D 2.24 litre

The correct answer is:
0.448 litre

Solution:
Number of mole of CO2 in 200 mL solution
=200×10-3L×0.1molL=0.0200 mol
Since the CO2 behaves like an ideal gas therefore, volume of 0.0200 mol of CO2 at STP
=0.0200 mol ×22.4LCO21mol=0.448 L.

Topic: Chemistry, Some Basic Concepts of Chemistry, NEET
A bottle of dry ammonia (NH3) and a bottle of dry hydrogen chloride connected through a long tube are opened simultaneously at both ends, the white ammonium chloride ring first formed will be

Options:
A at the centre of the tube
B near the hydrogen chloride bottle
C near the ammonia bottle
D throughout the length of the tube

The correct answer is:
near the hydrogen chloride bottle

Solution:
Ammonia being lighter gas (mol. mass =17 g mol-1 ) diffuses faster than dry hydrogen chloride gas (mol. mass =36.5 g mol-1 ). It means white fumes will be formed near HCl bottle.

Topic: Chemistry, States of Matter, NEET
A box of mass 5 kg is pulled by a cord, up along a frictionless plane inclined at $30^{\circ}$ with the horizontal. The tension in the cord is 30 N . The acceleration of the box is (Take $g=10 \mathrm{~m} \mathrm{~s}^{-2}$ )

Options:
A $2 \mathrm{~m} \mathrm{~s}^{-2}$
B Zero
C $0.1 \mathrm{~m} \mathrm{~s}^{-2}$
D $1 \mathrm{~m} \mathrm{~s}^{-2}$

The correct answer is:
$1 \mathrm{~m} \mathrm{~s}^{-2}$

Solution:
$T-M g \sin 30=M a$
$30-5 \times 10 \times \frac{1}{2}=5 a$
$a=1 \mathrm{~m} / \mathrm{s}^2$

Topic: Physics, Laws of Motion, NEET, NEET 2024 (Re-NEET)
A boy and a man carry a uniform rod of length L horizontally in such a way that the boy gets 14th of the load. If the boy is at one end of the rod, the distance of the man from the other end is

Options:
A L3
B L4
C 2L3
D 3L4

The correct answer is:
L3

Solution:
Let normal reaction on the rod due to boy is N1 and due to man is N2
Given, N1=W4
∴ N2=W-W4=3W4
Let x be the distance of the man from the other end.

For rotational equilibrium net torque must be zero.
⇒N1×L2=N2×L2-x
⇒ W4×L2=3W4×L2-x
⇒ x=L3

Topic: Physics, Rotational Motion, NEET
A boy and a man carry a uniform rod of length L, horizontally in such a way that the boy gets 14th of the load. If the boy is at one end of the rod, the distance of the man from the other end is

Options:
A L3
B L4
C 2L3
D 3L4

The correct answer is:
L3

Solution:
Weight of the rod=w
Reaction of boy RB=w4
Reaction of man RM=3w4
As the rod is in rotational equilibrium
∴ ∑τ=0
RB×L2-RM×x=0
⇒ w4×L2-3w4×x=0
⇒ x=L6
∴ Distance from other end, y=L2-x
⇒ y=L2-L6=2L6=L3

Topic: Physics, Rotational Motion, NEET
A boy is pushing a box on a horizontal floor from a position of rest to rest, while moving along a straight line. Consider the three phases of motion. The floor is rough with a small friction coefficient.

(i) Initially a constant hard push on the box to get it moving and attain a maximum velocity.

(ii) Mild push to keep the box moving with constant velocity.

(iii) To pull back the box to bring it to stop with the same retardation.

Which of the following graphs is CORRECT?

Options:
A
B
C
D

The correct answer is:

Solution:
The velocity of the box is always only in one direction. It is just increased and decreased during the motion, without changing its direction. Hence, velocity cannot be negative.
For the same reason, the direction of the friction also does not change and the friction is constant and always acts at the limiting value.
Acceleration is zero in the middle part of the motion as it is given that the velocity becomes constant. Hence, in the acceleration-time graph, the middle portion should be zero.
The worker first pushed the block harder to get it started, then he exerts some less force to sustain the motion and lastly, he reverses the direction of force to make the box retard. Also, he only exerts forces of constant magnitudes.

Topic: Physics, Laws of Motion, NEET
A boy is standing on an open truck. The truck is moving with acceleration 2 m s-2 on a horizontal road. When the speed of the truck is 10 m s-1 and reaches to an electric pole, boy projected a ball with velocity 10 m s-1 in a vertically upward direction relative to himself take g=10 m s-2. Neglect the height of boy and truck. The distance of the ball from the pole where ball land is

Options:
A 20 m

B 10 m

C 30 m

D 40 m

The correct answer is:
20 m

Solution:
Take x-axis along the motion of truck and y-axis vertically upward
vâball/boy=10i^ m s-1
vâboy/ground=10j^ m s-1
vâball/ground=10i^+10j^
The distance of the ball from pole
=u2sinâ¡2Î¸g=1022Ãsinâ¡90Â°g
=20 m

Topic: Physics, Motion In Two Dimensions, NEET
A boy is trying to start a fire by focusing sunlight on a piece of paper using an equiconvex lens of focal length $10 \mathrm{~cm}$. The diameter of the sun is $1.39 \times 10^9 \mathrm{~m}$ and its mean distance from the earth is $1.5 \times 10^{11} \mathrm{~m}$. What is the diameter of the sun's image on the paper?

Options:
A $9.2 \times 10^{-4} \mathrm{~m}$
B $6.5 \times 10^{-4} \mathrm{~m}$
C $6.5 \times 10^{-5} \mathrm{~m}$
D $12.4 \times 10^{-4} \mathrm{~m}$

The correct answer is:
$9.2 \times 10^{-4} \mathrm{~m}$

Solution:
From the relation
$\frac{I}{O}=\frac{v}{u}$
Here, $O=1.39 \times 10^9, v=0.1 \mathrm{~m}$,
$\begin{aligned}
u & =1.5 \times 10^{11} \mathrm{~m} \\
\therefore I & =\frac{0.1}{1.5 \times 10^{11}} \times 1.39 \times 10^9 \\
& =9.2 \times 10^{-4} \mathrm{~m}
\end{aligned}$

Topic: Physics, Ray Optics, NEET, NEET 2008 (Screening)
A boy learns typewriting and harmonium at the same time. He finds the harmonium easier to learn. This is

Options:
A Conditioned reflex
B Short term homeostasis
C Long term homeostasis
D Residual learning

The correct answer is:
Conditioned reflex

Solution:
Reflexes developed by training and learning are called conditioned or acquired reflexes for example after proper learning, we can perform dancing, cycling, swimming singing playing etc. simply by relaxes.

Topic: Biology, Neural Control and Coordination, NEET
A boy of mass 50 kg is climbing a vertical pole at a constant speed. If coefficient of friction between his palms and the pole is 0.75, then the normal reaction between him and the pole is (take g=10 m/s2 )

Options:
A 700 N
B 625.67Â N
C 550Â N
D 666.67 N

The correct answer is:
666.67 N

Solution:
Let the normal reaction is N.
Since, the boy is climbing with constant speed.
â´Fnet=0 (on the boy)
Free body diagram
mg=fr (friction force)
mg=Î¼N âµ fr=Î¼N
âN=mgÎ¼=50Ã100.75
=666.67 N

Topic: Physics, Laws of Motion, NEET
A boy playing on the roof of a 10 m high building throws a ball with a speed of 10 m s-1 at an angle of 30Â° with the horizontal. The distance of the ball in the same horizontal plane of the point of throw is

Options:
A 5.20 m

B 4.33 m

C 2.60 m

D 8.66 m

The correct answer is:
8.66 m

Solution:
It is equal to the range : R = u 2 sin 2 θ g
= 10 2 10 sin 60
=1032=8.66 m

Topic: Physics, Motion In Two Dimensions, NEET
A boy standing at the top of a tower of $20 \mathrm{~m}$ height drops a stone. Assuming $g=10 \mathrm{~ms}^{-2}$, the velocity with which it hits the ground is

Options:
A $20 \mathrm{~m} / \mathrm{s}$
B $40 \mathrm{~m} / \mathrm{s}$
C $5 \mathrm{~m} / \mathrm{s}$
D $10 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$20 \mathrm{~m} / \mathrm{s}$

Solution:
Given, $g=10 \mathrm{~ms}^{-2}$ and $h=20 \mathrm{~m}$
We have
$$
\begin{aligned}
V & =\sqrt{2 g h} \\
& =\sqrt{2 \times 10 \times 20}=\sqrt{400}=20 \mathrm{~ms}^{-1}
\end{aligned}
$$

Topic: Physics, Motion In One Dimension, NEET, NEET 2011 (Screening)
A brass wire 2 m long at 27 Â°C is held taut with negligible tension between two rigid supports. If the wire is cooled to a temperature of - 33 Â°C, then the tension developed in the wire, its diameter being 2 mm, will be (coefficient of linear expansion of brass = 2.0 Ã 10-5 Â°C-1 and Young's modulus of brass = 0.91 Ã 1011 Pa)

Options:
A 3400 N

B 34 kN

C 0.34 kN

D 6800 N

The correct answer is:
0.34 kN

Solution:
Thermal stress that is produced in an elastic wire is Y Î± Î¸ per unit area, wher Y is Young's modulus, Î± coefficient of linear expansion and Î¸ change of temperature.
Thus, tension developed i the wire is
T = Thermal stress x area of cross section
= Y Î± Ã Î T Ï r 2 N
= 0.91 Ã 1 0 1 1 Ã 2 Ã 1 0 - 5 Ã 2 7 - - 3 3 Ï Ã 1 Ã 1 0 - 6 N
= 0.91 Ã 2 Ã 6 Ã 3.14 Ã 1 0 N
= 34.4 Ã 1 0 N = 0.34 kN

Topic: Physics, Thermal Properties of Matter, NEET
A breed developed in Punjab by crossing Bikaneri ewes and Marino rams is

Options:
A Hisardale
B Leghorn
C Marin
D Mule

The correct answer is:
Hisardale

Solution:
Hisardale is a new breed of sheep developed in Punjab by crossing Bikaneri ewes and Marino rams.

Topic: Biology, Strategies for Enhancement in Food Production, NEET
A brick of mass 2 kg just begins to slide down on an inclined plane at an angle of 45Â° with the horizontal. The force of friction will be (g=9.8âm s-2)

Options:
A 19.62 N
B 9.82 N
C 19.6 N
D 9.8 N

The correct answer is:
19.62 N

Solution:
Friction force =mgsin⁡θ
=2×9.8×sin45°=19.62 N

Topic: Physics, Laws of Motion, NEET
A broad spectrum antibiotic is the one which

Options:
A kills the antibodies
B acts on a specific antigen
C acts on different antigens
D acts on both the antigens and antibodies

The correct answer is:
acts on different antigens

Solution:
Broad spectrum antibiotics act on different antigens.

Topic: Chemistry, Chemistry in Everyday Life, NEET
A bubble is at the bottom of the lake of depth h. As the bubble comes to surface, its radius increases three times. If atmospheric pressure is equal to l metre of water column, then h is equal to

Options:
A 26 l
B l
C 25Â l
D 30Â l

The correct answer is:
26 l

Solution:
From Boyle's law
pV= constant
â´ p1V1=p2V2
Here, p1=Ïgh+l, V1=43Ïr3
p2=Ïgl, V2=43Ï(3r)3
â´ h+l43Ïr3=lÃ43Ï3r3
or h+l=27l
â´ h=26l

Topic: Physics, Kinetic Theory of Gases, NEET
A bubble is at the bottom of the lake of depth h. As the bubble comes to the surface, its radius increases three times. If atmospheric pressure is equal to l metre of the water column, then h is equal to

Options:
A 26 l
B l
C 25Â l
D 30Â l

The correct answer is:
26 l

Solution:
From Boyle's law
pV= constant
â´ p1V1=p2V2
Here, p1=Ïgh+l, V1=43Ïr3
p2=Ïgl, V2=43Ï(3r)3
â´ h+l43Ïr3=lÃ43Ï3r3
or h+l=27l
â´ h=26l

Topic: Physics, Kinetic Theory of Gases, NEET
A bubble of air is underwater at temperature 15°C and the pressure 1.5 bar. If the bubble rises to the surface where the temperature is 25°C and the pressure is 1.0 bar, what will happen to the volume of the bubble?

Options:
A volume will become greater by a factor of 1.6
B volume will become greater by a factor of 1.1
C volume will become smaller by a factor of 0.70
D volume will become greater by a factor of 2.5

The correct answer is:
volume will become greater by a factor of 1.6

Solution:
We know that from ideal equation,
VâTP
Given T1=15+273=288,P1=1.5 bar
T2=25+273=298,P2=1 bar
V1â2881.5i.e.V1â192 and V2â2981
V 2 V 1 = 298 192 =1.55≅1.6

Topic: Chemistry, States of Matter, NEET
A bubble of air released by a diver at the bottom of a pool of water becomes larger as it approaches the surface of the water. Assume the temperature is constant and select the true statement. The pressure inside the bubble is

Options:
A greater near the bottom of the water
B greater near the top of the water
C same at all depths
D cannot be determined from the given data

The correct answer is:
greater near the bottom of the water

Solution:
Pressure inside the bubble is greater near the bottom of water because at bottom more pressure is worked on the bubble by the water.

Topic: Chemistry, States of Matter, NEET
A bucket full of hot water cools from 75 Â°C to 70 Â°C in time T1, from 70 Â°C to 65 Â°C in time T2 and from 65 Â°C to 60 Â°C in time T3, then

Options:
A T1=T2=T3
B T1>T2>T3
C T1<T2<T3
D T1>T2<T3

The correct answer is:
T1<T2<T3

Solution:
According to Newton's law of cooling
Rate of cooling ∝ Mean temperature difference
⇒ Fall in temperature Time ∝ θ 1 + θ 2 2 - θ 0
∵ θ 1 + θ 2 2 1 > θ 1 + θ 2 2 2 > θ 1 + θ 2 2 3
⇒ T1<T2<T3

Topic: Physics, Thermal Properties of Matter, NEET
A bucket full of water weighs 5 kg, it is pulled with constant speed from a well 20 m deep. There is a small hole in the bucket through which water leaks at a constant rate of 0.2 kg m-1. Ignore the force due to leakage. The total work done in pulling the bucket up from the well is g=10 m s-2

Options:
A 600 J
B 400 J
C 100 J
D 500 J

The correct answer is:
600 J

Solution:
The weight of bucket when it has been pulled up a distance x is (5-0.2x).Hence, the required work is
W = â« x = 2 0 x = 0 5 - 0 . 2 x Ã 1 0 Ã dx
= 5 0 x x = 0 x = 2 0 - 2 x 2 2 x = 0 x = 2 0
W = 5 0 Ã 2 0 - 2 0 2 = 6 0 0 J

Topic: Physics, Work Power Energy, NEET
A buffer solution can be prepared from a mixture of
1. Sodium acetate and acetic acid in water
2. Excess sodium acetate and hydrochloric acid in water
3. Ammonia and ammonium chloride in water
4. Ammonia and sodium hydroxide in water.

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

The correct answer is:
1, 2, 3

Solution:
In all the 1, 2, 3 buffer solution can be prepared. In choice 2, when CH3COONa reacts with HCl it forms CH3COOH+NaCl .

Topic: Chemistry, Ionic Equilibrium, NEET
A buffer solution contains equal concentration of X- and HX . The kb for X- is 10-10 . The pH of the buffer solution is

Options:
A 4
B 7
C 10
D 14

The correct answer is:
4

Solution:
pOH=pKb+log[salt][base]
pOH=10+log[HX][X-]
[HX]=[X-]
Log1=0
pOH=10
pH+pOH=pKw
pH=4 .

Topic: Chemistry, Ionic Equilibrium, NEET
A buffer solution is prepared by mixing 20 ml of 0.1 M CH3COOH and 40 ml of 0.5 M CH3COONa and then diluted by adding 100 ml of distilled water. The pH of resulting buffer solution is (Given pKa CH3COOH=4.76 )

Options:
A 5.76
B 4.67
C 3.48
D 5.9

The correct answer is:
5.76

Solution:
pH=pKa+log⁡saltacid
=4.76+log40×0.520×0.1
=4.76+1=5.76
pH of a buffer does not change on dilution

Topic: Chemistry, Ionic Equilibrium, NEET
A buffer solution is prepared in which the concentration of $\mathrm{NH}_3$ is $0.30 \mathrm{M}$ and the concentration of $\mathrm{NH}_4^{+}$is $0.20 \mathrm{M}$. If the equilibrium constant, $K_b$ for $\mathrm{NH}_3$ equals $1.8 \times 10^{-5}$, what is the $\mathrm{pH}$ of this solution?
$(\log 2.7=0.43)$

Options:
A 9.43
B 11.72
C 8.73
D 9.08

The correct answer is:
9.43

Solution:
$$
\begin{aligned}
\mathrm{pOH} & =\mathrm{p} K_{\mathrm{b}}+\log \frac{[\text { Salt }]}{[\text { Base }]} \\
& =-\log K_{\mathrm{b}}+\log \frac{[\text { Salt }]}{[\text { Base }]} \\
& =-\log 1.8 \times 10^{-5}+\log \frac{0.20}{0.30} \\
& =5-0.25+(-0.176) \\
& =4.75-0.176=4.57 \\
\therefore \mathrm{pH} & =14-4.57=9.43
\end{aligned}
$$

Topic: Chemistry, Ionic Equilibrium, NEET, NEET 2011 (Screening)
A bulb is located on a wall. Its image is to be obtained on a parallel with the help of a convex lens. If the distance between parallel walls is ' $d$ ' then the required focal length of the lens placed in between the walls is:

Options:
A only $\frac{d}{4}$
B only $\frac{d}{2}$
C more than $\frac{d}{4}$ but less than $\frac{d}{2}$
D less than or equal to $\frac{d}{4}$

The correct answer is:
less than or equal to $\frac{d}{4}$

Solution:
Since $v+u=d \Rightarrow v=d-u$
From lens formula
$$
\begin{aligned}
\frac{1}{f} & =\frac{1}{v}-\frac{1}{u} \\
\Rightarrow \quad \frac{1}{f} & =\frac{1}{d-u}-\frac{1}{-u} \\
& =\frac{d}{(d-u) u} \\
\Rightarrow \quad f & =\frac{d u-u^2}{d} \ldots(\mathrm{i})
\end{aligned}
$$
Differentiating eq. (i) we have
$$
\frac{d f}{d u}=\frac{1}{d}(d-2 u)
$$
for maximum value of $f_1=\frac{d f}{d u}=0$
$$
\begin{aligned}
& \Rightarrow \frac{1}{d}(d-2 u)=0 \Rightarrow d=24 \\
& \Rightarrow u=\frac{d}{2}
\end{aligned}
$$
putting these value in eq. (i)
$$
f=\frac{d}{4}
$$

Topic: Physics, Ray Optics, NEET, NEET 2002
A bullet fired into a fixed target loses half its velocity after penetrating 3 cm. How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion?

Options:
A 1.5 cm

B 1.0 cm
C 3.0 cm
D 2.0 cm

The correct answer is:
1.0 cm

Solution:
For first part of penetration, by equation of motion,
( u⁡ 2 ) 2 = ( u⁡ ) 2 - 2 f⁡ ( 3 )
or 3 u⁡ 2 = 2 4 f⁡
For latter part of penetration,
0=(u⁡2)2-2f⁡x
or u⁡2=8f⁡x
Therefore,
3×(8f⁡x)=24f⁡
or x=1 cm

Topic: Physics, Laws of Motion, NEET
A bullet from a gun is fired on a rectangular wooden block with velocity u. When bullet travels 24 cm through the block along its length horizontally, velocity of bullet becomes u3. Then it further penetrates into the block in the same direction before coming to rest exactly at the other end of the block. The total length of the block is :

Options:
A 30 cm
B 27 cm
C 24 cm
D 28 cm

The correct answer is:
27 cm

Solution:
For the first part of the motion of the bullet into the block, it can be written that

u32=u2-2a×24⇒48a=8u29 ...1

For the second part of the motion of the bullet through the block, it can be written that

0=u32-2aS⇒2aS=u29 ...2

Divide equation (2) by equation (1) to obtain the distance travelled by the bullet during the second part of its motion.

2aS48a=u298u29⇒S=3

Hence, the total length of the block is given by

L=3+24 cm= 27 cm

Topic: Physics, Motion In One Dimension, NEET, NEET 2023 (All India)
A bullet is fired from a gun at the speed of 280 m s-1 in the direction 30° above the horizontal. The maximum height attained by the bullet is (g=9.8 m s-2, sin30°=0.5):

Options:
A 3000 m
B 2800 m
C 2000 m
D 1000 m

The correct answer is:
1000 m

Solution:
Maximum height of the projectile is given by,

H=u2sin2θ2g=2802×sin230°2×9.8=1000 m

Topic: Physics, Motion In Two Dimensions, NEET, NEET 2023 (All India)
A bullet of mass 0.1 kg is fired with a speed of 100 m s-1, the mass of gun is 50 kg. The velocity of recoil is

Options:
A 0.2 m s-1
B 0.1 m s-1
C 0.5 m s-1
D 0.05 m s-1

The correct answer is:
0.2 m s-1

Solution:
According to the principle of conservation of linear momentum, mGvG=mBvB
vG=mBvBmG=0.1×10250=0.2 m s-1

Topic: Physics, Laws of Motion, NEET
A bullet of mass $2 \mathrm{~g}$ is having of $2 \mu \mathrm{C}$. Through what potential difference must it be accelerated, starting from rest, to acquire a speed of of $10 \mathrm{~m} / \mathrm{s}$ ?

Options:
A $5 \mathrm{kV}$
B $50 \mathrm{kV}$
C $5 \mathrm{~V}$
D $50 \mathrm{~V}$

The correct answer is:
$50 \mathrm{kV}$

Solution:
Here we apply $\frac{1}{2} m v^2=q V$
$$
\begin{gathered}
\Rightarrow V=\frac{1}{2} \times \frac{2 \times 10^{-3} \times 10 \times 10}{2 \times 10^{-6}} \\
=50 \mathrm{kV}
\end{gathered}
$$

Topic: Physics, Electrostatics, NEET, NEET 2004
A bullet of mass 0.02kg travelling horizontally with velocity 250 m sâ1 strikes a block of wood of mass 0.23kg which rests on a rough horizontal surface. After the impact, the block and bullet move together and come to rest after travelling a distance of 40 m. The coefficient of sliding friction of the rough surface is (g=9.8 m sâ2)

Options:
A 0.75
B 0.61
C 0.51
D 0.30

The correct answer is:
0.51

Solution:
After impact the bullet and block move together and come's to rest after covering a distance of 40 m.
By conservation of momentum,
m1u1+m2u2=m1v1+m2v2
or 0.02×250+0.23×0=0.0 2 v+0.23 v

5+0=v0.25
v=50025=20 m s-1
Now, by conservation of energy
12M⁡v⁡2=μ N.d
or 12×0.25×400=μ×0.25×9.8×40
⇒ μ=2009.8×40=0.51

Topic: Physics, Center of Mass Momentum and Collision, NEET
A bullet of mass 0.02kg travelling horizontally with velocity 250 msâ1 strikes a block of wood of mass 0.23kg which rests on a rough horizontal surface. After the impact, the block and bullet move together and come to rest after travelling a distance of 40 m. The coefficient of sliding friction of the rough surface is (g=9.8 msâ2)

Options:
A 0.75
B 0.61
C 0.51
D 0.30

The correct answer is:
0.51

Solution:
After impact the bullet and block move together and come's to rest after covering a distance of 40âm.
By conservation of momentum,
m1u1+m2u2=m1v1+m2v2
or 0.02Ã250+0.23Ã0=0.0 2v+0.23 v

5+0=v0.25
V = 5 0 0 2 5 = 2 0 ms - 1
Now, by conservation of energy
12Mâ¡vâ¡2=Î¼ N.d
or 12Ã0.25Ã400=Î¼Ã0.25Ã9.8Ã40
â Î¼=2009.8Ã40=0.51

Topic: Physics, Center of Mass Momentum and Collision, NEET
A bullet of mass 10 g moving horizontally with a velocity of 400 m s-1 strikes a wooden block of mass 2 kg which is suspended by a light inextensible string of length 5 m. As a result, the centre of gravity of the block is found to rise a vertical distance of 10 cm. The speed of the bullet after it emerges out horizontally from the block will be

Options:
A 100 m s-1
B 80Â m s-1
C 120 m s-1
D 160Â m s-1

The correct answer is:
120 m s-1

Solution:
CM rises through height h, so its velocity after collision =2gh
Apply conservation of linear momentum.
0.01Ã400=2Ã2gh+0.01Ãv
âv=120 m/s

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2016 (Phase 2)
A bullet of mass 10 g moving horizontally with a velocity of 400 m s-1 strikes a wooden block of mass 2 kg which is suspended by a light inextensible string of length 5 m. As a result, the centre of gravity of the block is found to rise a vertical distance of 10 cm. The speed of the bullet after it emerges out horizontally from the block will be

Options:
A 100 m s-1
B 80Â m s-1
C 120 m s-1
D 160Â m s-1

The correct answer is:
120 m s-1

Solution:
CM rises through height h, so its velocity after collision =2gh
Apply conservation of linear momentum.
0.01Ã400=2Ã2gh+0.01Ãv
âv=120 m/s

Topic: Physics, Center of Mass Momentum and Collision, NEET
A bullet of mass 10 gm is fired from a gun of mass 1 kg. If the recoil velocity of the gun is 5 m/s, what is the velocity of the bullet?

Options:
A 0.05 m/s
B 5 m/s
C 50 m/s
D 500 m/s

The correct answer is:
500 m/s

Solution:
mGνG=mBνB
⇒νB=mGνGmB=1×510×10-3=500 m/s

Topic: Physics, Center of Mass Momentum and Collision, NEET
A bullet of mass 20 g has an initial speed of 1 m s-1, just before it starts penetrating a mud wall of thickness 20 cm. If the wall offers a mean resistance of 2.5Ã10-2 N, the speed of the bullet after emerging from the other side of the wall is close to:

Options:
A 0.7 m s-1
B 0.3Â mÂ s-1
C 0.1Â mÂ s-1
D 0.4Â mÂ s-1

The correct answer is:
0.7 m s-1

Solution:
a=-2.5×10-220×10-3=-54 m s-2
From v2-u2=2as
v2=u2+2as
v=12-2×54×20×10-2
=0.5
=0.7 m s-1

Topic: Physics, Laws of Motion, NEET
A bullet of mass 50 g is fired from a gun of mass 2 kg. If the total kinetic energy produced is 2050 J, the kinetic energy of the bullet and the gun respectively are

Options:
A 200 J, 5 J

B 2000 J, 50 J
C 5 J, 200 J
D 50 J, 2000 J

The correct answer is:
2000 J, 50 J

Solution:
Since there is no external force acting on rifle bullet system, hence
pb=pg and hence KbKg=mgmb=2kg50g=401
Or Kg=Kb40
Now total energy Kb+Kg = K⁡ b⁡ +Kb40= 4 1 K⁡ b⁡ 4 0 =2050
⇒ Kb=2050×4041=2000 J
And Kg=2050-2000=50 J.

Topic: Physics, Center of Mass Momentum and Collision, NEET