datavorous/entrance-exam-dataset huggingface dataset:
A body of mass 8 kg is suspended through two light springs X and Y connected in series as shown in the figure. The reading in X and Y respectively are

Options:
A 8 kg, zero
B Zero, 8 kg
C 6 kg, 2 kg
D 8 kg, 8 kg

The correct answer is:
8 kg, 8 kg

Solution:
As the springs are light in weight, therefore, the tension in both springs will be the same. So, both springs will show the same reading 8 kg.

Topic: Physics, Laws of Motion, NEET
A body of mass 10 kg is dropped to the ground from a height of 10 m. The work done by the gravitational force is g=9.8 m s-2

Options:
A -490 J
B +490 J
C -980 J
D +980 J

The correct answer is:
+980 J

Solution:
As the body moves in the direction of force therefore work done by gravitational force will be positive
W=mgh
W=10Ã9.8Ã10
W=980 J

Topic: Physics, Work Power Energy, NEET
A body of mass 10 kg is moving on a horizontal surface by applying a force of 10 N in the forward direction. The body moves with a constant velocity of 0.2 m s-1. Work done by the force of friction in the first 10 seconds is

Options:
A -20 J
B 10 J
C 20 J
D -5 J

The correct answer is:
-20 J

Solution:
Since, body moves with constant velocity. So, net force on the body is zero.

Here, N = mg
F = f
∴ W=f→·s→= fs cos 180°
= –fs= –10 × 2 = –20 J

Topic: Physics, Work Power Energy, NEET
A body of mass 50 kg is suspended using a spring balance inside a lift at rest. If the lift starts falling freely, the reading of the spring balance is

Options:
A <50 kg
B =50 kg
C >50 kg
D =0

The correct answer is:
=0

Solution:
Given,
Mass of the body, m=50 kg
When lift is moving down,
Scale reading =mg-ag
If lift starts falling freely, then
a=g
Now, scale reading =mg-gg=0

Topic: Physics, Laws of Motion, NEET
A body of mass 60 g experiences a gravitational force of 3.0 N, when placed at a particular point. The magnitude of the gravitational field intensity at that point is

Options:
A 50 N kg-1
B 20 N kg-1
C 180 N kg-1
D 0.05 N kg-1

The correct answer is:
50 N kg-1

Solution:
∵ Gravitational force, FG=Eg×m, where Eg= Gravitational field at the given point.

Eg=FGm=360×10-3=50 N kg-1

Topic: Physics, Gravitation, NEET, NEET 2022 (Phase 1)
A body of mass m1 collides elastically with another body of mass m2 at rest. If the velocity of m1 after collision becomes 23 times its initial velocity, the ratio of their masses is

Options:
A 1Â :5
B 5 :1
C 5Â :2
D 2Â :5

The correct answer is:
5 :1

Solution:
In elastic collision, if u1 and u2 are initial velocities and v1 and v2 are final velocities of bodies with masses m1 and m2 respectively, then
v1=m1-m2m1+m2u1+2m2m1+m2u2.
If the second ball is at rest, i.e., u2=0, then
v1=m1-m2m1+m2u1
According to the question,
23u1=m1-m2m1+m2u1 âµ v1=23u1
â 2m1+2m2=3m1-3m2
â -m1=-5m2
â m1m2=51

Topic: Physics, Center of Mass Momentum and Collision, NEET
A body of mass m1 moving at a constant speed undergoes an elastic head-on collision with a body of mass m2 initially at rest. The ratio of the kinetic energy of mass m1 after the collision to that before the collision is

Options:
A m1-m2m1+m22
B m1+m2m1-m22
C 2m1m1+m22
D 2m2m1+m22

The correct answer is:
m1-m2m1+m22

Solution:
m1v0=m2v2- m1v1 .........(1)
e=v1+v2v0=1 .........(2)
Solving (1) and (2)
v1v02=m1-m2m1+m22

Topic: Physics, Center of Mass Momentum and Collision, NEET
A body of mass m1 moving with an unknown velocity of v1 i^, undergoes a collinear collision with a body of mass m2 moving with a velocity v2 i^. After the collision, m1 and m2 move with velocities of v3 i^ and v4 i^, respectively. If m2=0.5 m1 and v3=0.5 v1, then v1 is:

Options:
A v4-v24
B v4-v22
C v4+v2
D v4-v2

The correct answer is:
v4-v2

Solution:
From conservation of linear momentum,
m1v1+m2v2=m1v3+m2v4
m1v1+0.5 m1 v2=m10.5 v1+0.5 m1 v4
=0.5 v1+0.5 v2=0.5 v4
v4=v1+v2
v1=v4-v2

Topic: Physics, Center of Mass Momentum and Collision, NEET
A body of mass m= 10 â2 kg is moving in a medium and experiences a frictional force F=âk v 2 . Its initial speed is v 0 =10 ms â1 . After 10âs, its kinetic energy is 1 8 mv 0 2 , then value of k will be:-

Options:
A 10â1kgâmâ1sâ1
B 10â3âkgâmâ1
C 10 −3   kg s −1
D 10â4 kgâmâ1

The correct answer is:
10â4 kgâmâ1

Solution:
After 10âsec, 12mv2=18mv02
âv= v 0 2 =5âm/sec
a=â k v 2 m
â dv dt =â k v 2 m
â â« 10 5 dv v 2 =â k m â« 0 10 dt
-15+110=-k10-2Ã10
k=10â4 kgâmâ1

Topic: Physics, Work Power Energy, NEET
A body of mass m=10 kg is attached to a wire of length 0.3 m. The maximum angular velocity with which it can be rotated in a horizontal circle is (Breaking stress of wire =4.8×107 Nm-2 and area of cross-section of a wire =10-6 m2)

Options:
A 4 rads-1
B 8Â rads-1
C 1Â rads-1
D 2Â rads-1

The correct answer is:
4 rads-1

Solution:
Breaking strength = tension in the wire =mrω2
4.8 ×107×10-6=10×0.3 ×ω2
ω2=480.3 ×10=16
ω=4 rads-1

Topic: Physics, Mechanical Properties of Solids, NEET
A body of mass M at rest explodes into three pieces, two of which of mass M4 each are thrown off in mutually perpendicular directions with speeds of 3 m s -1 and 4 m s-1 respectively. Then the third piece will be thrown off with a speed of

Options:
A 1.5 m s-1
B 2.0 m s-1
C 2.5 m s-1
D 3.0 m s-1

The correct answer is:
2.5 m s-1

Solution:
As there is no external force, hence
pâ=pâ1+pâ2+pâ3= constant
â pâ3=pâ1+pâ2=m432+42=5m4
[since v1 and v2 are mutually perpendicular]
â´ p3=m2v3=5m4
â v3=52=2.5 m s-1

Topic: Physics, Center of Mass Momentum and Collision, NEET
A body of mass m collides against a wall with the velocity v and rebounds with the same speed. Its magnitude of change of momentum is

Options:
A 2m⁡v
B m⁡v
C 12m⁡v
D 0

The correct answer is:
2m⁡v

Solution:
Change in momentum
Δp=pf-pi
Δp=mv--mv
Δp=2mv

Topic: Physics, Center of Mass Momentum and Collision, NEET
A body of mass m collides elastically with another body at rest and then continues to move in the original direction with one half of its original speed. mass of the target body is

Options:
A m
B 23m
C m3
D m2

The correct answer is:
m3

Solution:
Speed of first body after elastic collision is
v1=u1+2m2m1+m2u2-u1

since m1=m and V1=v2
v2=v+2m2m+m20-v

solving above equation we get,
m2=m3

Topic: Physics, Center of Mass Momentum and Collision, NEET
A body of mass $M$ hits normally a rigid wall with velocity $y$ and bounces back with the same velocity. The impulse experienced by the body is

Options:
A $1.5 \mathrm{Mv}$
B $2 \mathrm{Mv}$
C zero
D $M v$

The correct answer is:
$2 \mathrm{Mv}$

Solution:
Impulse
$\begin{aligned}
|J| & =|\Delta p| \\
& =M v-(-Mv) \\
& =2 Mv
\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2011 (Screening)
A body of mass m is attached to one end of a massless spring which is suspended vertically from a fixed point. The mass is held in hand so that the spring is neither stretched nor compressed. Suddenly the support of the hand is removed. The lowest position attained by the mass during oscillation is 4cm below the point where it was held in hand. What is the amplitude of oscillation?

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

The correct answer is:
2 cm

Solution:
Correct Option is : (B)
2 cm

Topic: Chemistry, Oscillations, NEET
A body of mass, m is attached to the lower end of a spring whose upper end is fixed. The spring has negligible mass. When the mass m is slightly pulled down and released, it oscillates with a time period of 3 s. When the mass m is increased by 1 kg, the time period of oscillations becomes 5 s. The value of m in kg is

Options:
A 34
B 43
C 169
D 916

The correct answer is:
916

Solution:
T1=3=2πmK.

Then, T2=5=2πm+1K.

By dividing both the equations, we get,

35=mm+1

925=mm+1

9m+9=25 m

16m=9

m=916.

Topic: Physics, Oscillations, NEET, NEET 2016 (Phase 2)
A body of mass $m$ is kept on a rough horizontal surface (coefficient of friction $=\mu$ ). Horizontal force is applied on the body, but it does not move. The resultant of normal reaction and the frictional force acting on the object is given $\mathrm{F}$, where $\mathrm{F}$ is

Options:
A $|\mathrm{F}|=m g+\mu m g$
B $|\mathrm{F}|=\mu m g$
C $|\mathrm{F}| \leq m g \sqrt{1+\mu^2}$
D $|\mathrm{F}|=m g$

The correct answer is:
$|\mathrm{F}| \leq m g \sqrt{1+\mu^2}$

Solution:
$$
\text { The situation can be drawn as }
$$
The frictional force, $\mathrm{f}=\mu \mathrm{N}=\mu \mathrm{mg} \quad[\because \mathrm{N}=\mathrm{mg}]$ From Free body diagram (FBD), the resultant force is
$$
\begin{aligned}
|F| & =\sqrt{\mathrm{N}^2+\mathrm{f}^2} \\
& =\sqrt{(\mathrm{mg})^2+(\mu \mathrm{mg})^2}=\mathrm{mg} \sqrt{1+\mu^2}
\end{aligned}
$$
This is the minimum force required to move the object. But as the body is not moving
$$
\therefore \quad|\mathrm{F}| \leq \mathrm{mg} \sqrt{1-\mu^2}
$$

Topic: Physics, Laws of Motion, NEET, NEET 2019 (Odisha)
A body of mass m is lifted by a man to a height of one meter in 30 s. Another man lifts the same mass to the same height in 60 s. The work done by them are in the ratio

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

The correct answer is:
1:1

Solution:
Work done by both the persons is the same and is given as,
W=mgh
It does not depend upon the time taken to perform the work.

Topic: Physics, Work Power Energy, NEET
A body of mass m is moving with speed v makes a one-dimensional collision with a stationary body of same mass on a horizontal table. They are in contact for a very small time interval Ît . The contact force between them varies as shown in the graph. (Neglect friction)

The coefficient of restitution for the collision will be

Options:
A 34
B 12
C 23
D 14

The correct answer is:
12

Solution:
e=IrestitutionIcompression=Area  of  restitutionrightpartArea  of  compressionleftpart
=12Δt3Fmax122Δt3Fmax=12

Topic: Physics, Center of Mass Momentum and Collision, NEET
A body of mass m is placed at the centre of the spherical shell of radius R and mass M. The gravitational potential on the surface of the shell is

Options:
A -GRM+m
B -GRM-m
C -GRmMM+m
D -GRmMM-m

The correct answer is:
-GRM+m

Solution:
Gravitational potential on the surface of the shell due to the body of mass m is Vb=-GmR
Gravitational potential on the surface of the shell due to shell itself is Vs=-GMR
∴ V=Vb+Vs=-GRM+m

Topic: Physics, Gravitation, NEET
A body of mass $m$ is placed on Earth's surface which is a taken from Earth's surface to a height of $h=3 R$, then the change in the gravitational potential energy is:

Options:
A $\frac{m g R}{4}$
B $\frac{2}{3} m g R$
C $\frac{3}{4} m g R$
D $\frac{m g R}{2}$

The correct answer is:
$\frac{3}{4} m g R$

Solution:
$\begin{aligned}
& \therefore \text { change in GPE }=\text { final } \\
& \text { energy }- \text { initial energy } \\
& =-\frac{G M m}{4 R}+\frac{\mathrm{GMm}}{R} \\
& =-\frac{G M m}{4 R}\left[1-\frac{1}{4}\right]=\frac{3}{4} \frac{G M m}{4 R} \\
& =\frac{3}{4} \frac{G M}{R^2} m R=\frac{3}{4} g m R
\end{aligned}$

Topic: Physics, Gravitation, NEET, NEET 2002
A body of mass m is placed on the earth's surface. It is taken from the earth's surface to a height h=3R when R is the radius of the earth. The change in gravitational potential energy of the body is

Options:
A 32mgR
B 34mgR
C 12mgR
D 14mgR

The correct answer is:
34mgR

Solution:
PE at the surface of earth
PE=-GMmR
PE at a height 3Re
PE=-GMmR+3 R=-GMm4R
ÎPE=PEf-PEi=-GMm4R --GMmR =34 GMmR
âÎPE=34 mgR

Topic: Physics, Gravitation, NEET
A body of mass m is suspended from vertical spring and is set into simple harmonic oscillations of time period T . Next the spring is fixed at one end on a smooth horizontal table and same body is attached at the other end. The body is pulled slightly and then released to produce horizontal oscillations of the spring. The time period of horizontal oscillatioins is

Options:
A T
B 2T
C T2
D Infinite as the spring does not oscillate

The correct answer is:
T

Solution:
Time period of spring is given by :
T=2Ïmk
Time period of a spring and mass does not depend, on it being horizontal or vertical.
So in both cases, it will be same i.e. T.

Topic: Physics, Oscillations, NEET
A body of mass m is taken from the surface of the earth to the height h, equal to the radius of the Earth. The increase in potential energy will be

Options:
A mgR
B 12mgR
C 2mgR
D 14mgR

The correct answer is:
12mgR

Solution:
Increase in potential energy is,
ΔU=mgh1+hR
∵ h=R
So ΔU=mgR2

Topic: Physics, Gravitation, NEET
A body of mass m kg collides elastically with another body at rest and then continues to move in the original direction with one half of its original speed. The mass of the target body is

Options:
A m kg
B 23m kg
C m3 kg
D m2 kg

The correct answer is:
m3 kg

Solution:
Velocity of the colliding body after elastic collision is,
v1=u1+2m2m1+m2u2-u1
u12=u1+2m2m+m20-u1
m2m+m2=14
m2=m3 kg

Topic: Physics, Center of Mass Momentum and Collision, NEET
A body of mass M moving with speed v collides elastically with another block of same mass M which is at rest. After the collision, the first block moves at an angle θ to its initial direction and has a speed v3. The speed of the second block after the collision will be

Options:
A 34Â v
B 34Â v
C 223 v
D 35v

The correct answer is:
223 v

Solution:
According to the question,
According to the law of conservation of kinetic energy, we have
12Mv2+0=12Mv32+12Mv12
v2=v29+v12
v12=v2-v29=8v29
∴v1=223v

Topic: Physics, Center of Mass Momentum and Collision, NEET
A body of mass m slides down an incline and reaches the bottom with a velocity v. If the same mass was in the form of a ring which rolls down this incline, the velocity of the ring at the bottom would have been

Options:
A v
B 2v
C v2
D 25 v

The correct answer is:
v2

Solution:
When body slides, v=2gh ....(1)
When ring rolls, vâ²=2gh1+k2/r2=2gh2=v2

Topic: Physics, Rotational Motion, NEET
A body of mass $m$ taken from the earth's surface to the height equal to twice the radius $(R)$ of the earth. The change is potential energy of body will be

Options:
A $m g 2 R$
B $\frac{2}{3} m g R$
C $3 \mathrm{mgR}$
D $\frac{1}{3} m g R$

The correct answer is:
$\frac{2}{3} m g R$

Solution:
Change in potential energy
$\begin{aligned}
\Delta U & =-\frac{G M m}{R+2 R}-\left(-\frac{G M m}{R}\right) \\
& =-\frac{G M m}{3 R}+\frac{G M m}{R} \\
& =\frac{2 G M m}{3 R}=\frac{2}{3} m g R \quad\left[\because g=\frac{G M}{R^2}\right]
\end{aligned}$

Topic: Physics, Gravitation, NEET, NEET 2013 (All India)
A body of radius R and mass m is rolling horizontally without slipping with speed v. It then rolls up a hill to a maximum height h=3v24g. The body might be a

Options:
A solid sphere
B hollow sphere
C disc
D ring

The correct answer is:
disc

Solution:
Let I bt the moment of inertia of the body. Then total KE=12 mv2+12 Iω2
or KE=12 mv2+12 Iv2R2ω=vR
According to energy conservation loss in KE=gain in PE.
or 12m+IR2v2=mgh=mg3v24g
Solving this, we get I=12mR2
i.e., the solid body is a disc

Topic: Physics, Rotational Motion, NEET
A body oscillates with SHM according to the equation (in SI units), x = 5 cos [2π t + π/4]. At t = 1.5 s, calculate the acceleration of the body.

Options:
A 120$m{{s}^{-2}}$
B 140 $m{{s}^{-2}}$
C 130$m{{s}^{-2}}$
D 110$m{{s}^{-2}}$

The correct answer is:
140 $m{{s}^{-2}}$

Solution:
Correct Option is : (B)
140 $m{{s}^{-2}}$

Topic: Chemistry, Oscillations, NEET
A body oscillates with SHM according to the equation (in SI units), x = 5 cos [2π t + π/4]. At t = 1.5 s, calculate the displacement

Options:
A 4.125m
B 3.535m
C -3.535m
D -4.125m

The correct answer is:
-3.535m

Solution:
Correct Option is : (C)
-3.535m

Topic: Chemistry, Oscillations, NEET
A body oscillates with SHM according to the equation (in SI units), x = 5 cos [2π t + π/4]. At t = 1.5 s, calculate the speed

Options:
A 44m/s
B 33m/s
C 11m/s
D 22m/s

The correct answer is:
22m/s

Solution:
Correct Option is : (D)
22m/s

Topic: Chemistry, Oscillations, NEET
A body projected vertically upwards crosses a point at a height h twice in its journey at instants t1 and t2 second. The maximum height reached by the body is

Options:
A g4t1+t22
B gt1+t242
C 2gt1+t242
D g4t1t2

The correct answer is:
2gt1+t242

Solution:
Time taken by the body to reach the point A is t1 (During upward journey).
The body crosses this point again (during the downward journey) after t2, ie, the body takes the time t2-t1 to come again at the point A.
So, the time taken by the body to reach at point B (a maximum height).
t=t1+t2-t12
∵Time of ascending=Time of descending
t=t1+t22
So, maximum height H=12 gt2
=12gt1+t222
=2gt1+t242

Topic: Physics, Motion In One Dimension, NEET
A body rolls down an inclined plane. If its kinetic energy of rotation is 40% of its kinetic energy of translation, then the body is

Options:
A Solid cylinder
B Solid sphere
C Disc
D Ring

The correct answer is:
Solid sphere

Solution:
12Iω2=40% of12mv2
12 Iω2=4010012mr2ω2
I=25mr2
So, the body is solid sphere.

Topic: Physics, Rotational Motion, NEET
A body slides down on a frictionless track which ends in a circular loop of diameter D. The minimum height h in terms of D so that the body may just complete the circular loop, is

Options:
A h=52D
B h=32D
C h=54D
D h=2D

The correct answer is:
h=54D

Solution:
By conservation of energy,
1 2 m v 2 =mgh
v= 2gh ...(i)
For looping the loop, the lower velocity must be greater than 5gr
v min = 5gr = 5gD 2 ...(ii)
From equation (i) and (ii),
2gh= 5gD 2
h= 5D 4

Topic: Physics, Laws of Motion, NEET
A body starts from rest and moves with constant acceleration for t seconds. It travels a distance x1 in the first half of time and x2 in the next half of time, then

Options:
A x2=x1
B x2=2x1
C x2=3x1
D x2=4x1

The correct answer is:
x2=3x1

Solution:
We know that a=DisplacementTime2
a=x2-x1t2 if x2>x1
by Newton's second law of motion
s=ut+12at2
Where u = 0 (Body is at rest)
So, x1=0+12×x2-x1t2×t2
x1=x2-x12
⇒2x1=x2-x1
3x1=x2

Topic: Physics, Laws of Motion, NEET
A body starts from rest with uniform acceleration. If its velocity after n seconds is υ, then its displacement in the last two seconds is

Options:
A 2υn+1n
B υn+1n
C υn-1n
D 2υn-1n

The correct answer is:
2υn-1n

Solution:
Displacement in last 2 sec .
S=12an2-12an-22
S=a22n-22
S=2an-1
But v=an
∴ S=2n-1vn

Topic: Physics, Motion In One Dimension, NEET
A body takes T minutes to cool from 62 Â°C to 61 Â°C when the surrounding temperature is 30 Â°C. The time taken by the body to cool from 46 Â°C to 45.5 Â°C is

Options:
A Greater than T minutes
B Equal to T minutes
C Less than T minutes
D None of these

The correct answer is:
Equal to T minutes

Solution:
According to Newton's law of cooling
θ1-θ2t ∝θ1+θ22-θ
For the first condition
62-61T ∝62+612-30 (i)
And for the second condition
46-45.5t∝46+45.52-30 (ii)
By solving equation (i) and (ii), we get t = T minutes.

Topic: Physics, Thermal Properties of Matter, NEET
A body, under the action of a force $\overrightarrow{\mathrm{F}}=6 \hat{\mathrm{i}}-8 \hat{\mathrm{j}}+10 \hat{\mathrm{k}}$, acquires an acceleration of 1 $\mathrm{ms}^{-2}$. The mass of this body must be

Options:
A $2 \sqrt{10} \mathrm{~kg}$
B $10 \mathrm{~kg}$
C $20 \mathrm{~kg}$
D $10 \sqrt{2} \mathrm{~kg}$

The correct answer is:
$10 \sqrt{2} \mathrm{~kg}$

Solution:
Key Idea According to Newton's second law of motion, force $=$ mass $\times$ acceleration.
Here,
$$
\text { Here, } \begin{aligned}
\quad \overrightarrow{\mathrm{F}} & =6 \hat{\mathrm{i}}-8 \hat{\mathrm{j}}+10 \hat{\mathrm{k}} \\
|\mathrm{F}| & =\sqrt{36+64+100} \\
& =10 \sqrt{2} \mathrm{~N} \\
\mathrm{a} & =1 \mathrm{~ms}^{-2} \\
\therefore \quad \mathrm{m} & =\frac{10 \sqrt{2}}{1}=10 \sqrt{2} \mathrm{~kg}
\end{aligned}
$$

Topic: Physics, Laws of Motion, NEET, NEET 2009 (Screening)
A body weighs 72 N on the surface of the earth. What is the gravitational force on it, at a height equal to half the radius of the earth?

Options:
A 32 N
B 30 N
C 24 N
D 48 N

The correct answer is:
32 N

Solution:
Ws=mgs=72 N

Wh=mgh=mgs1+hR2=72 N1+R/2R2=729/4

Wh=32 N

Topic: Physics, Gravitation, NEET, NEET 2020 (Phase 1)
A body weighs 200 N on the surface of the earth. How much will it weigh half way down to the centre of the earth?

Options:
A 150 N
B 200 N
C 250 N
D 100 N

The correct answer is:
100 N

Solution:
Weight =mg=mGMR2=W
At half way down to the centre, W'=mGMR3R23R22
=mGM2R2
=100 N

Topic: Physics, Gravitation, NEET, NEET 2019 (All India)
A body weights 8 g when placed in one pan and 18 g when placed on the other pan of a false balance. If the beam is horizontal when both the pans are empty, then the true weight of the body is

Options:
A 15 g
B 13 g
C 10 g
D 12 g

The correct answer is:
12 g

Solution:
From figure
8p=wq
⇒ pq=w8 .... (i)
Now, from figure,
wp=18q
⇒ pq=18w ..... (ii)
From equations (i) and (ii), we get
pq=w8=18w
w=18×8=12 g

Topic: Physics, Laws of Motion, NEET
A Bohr hydrogen atom undergoes a transition n=5ân=4 and emits a photon of frequency Ï. Frequency of circular motion of electron in n = 4 orbit is v4. The ratio Î½Î½4 is

Options:
A 1825
B 1625
C 925
D 825

The correct answer is:
1825

Solution:
En=-mz2e48ε02n2h2
So, hv=+mz2e48ε02h2116-125
∴ v=mz2e48ε02h3916×25 ...(1)
And frequency v4=1T=v2πr
=Ze22ε0nh12π πmze2ε0h2n2=z2e4m4ε02n3h3 ...(2)
∴vv4=1825=0.72

Topic: Physics, Atomic Physics, NEET
A bomb is dropped from an aeroplane flying horizontally with a velocity 469 m s-1 at an altitude of 980 m. The bomb will hit the ground after a time (use g=9.8âm s-2)

Options:
A 2 s

B 2 s
C 52 s
D 102 s

The correct answer is:
102 s

Solution:
t=2hg=29809.8=102m s-1

Topic: Physics, Motion In Two Dimensions, NEET
A bomb of mass 9 kg explodes into two parts. One part of mass 3 kg moves with a velocity of 16 m s-1, then the kinetic energy of the other part is

Options:
A 192 J
B 200Â J
C 300Â J
D 400Â J

The correct answer is:
192 J

Solution:
From momentum conservation principle
0=316+6v2 â v2= â8 m/s
K 2 = 1 2 m 2 v 2 2 = 1 2 ( 6 ) ( â8 ) 2 =192âJ

Topic: Physics, Center of Mass Momentum and Collision, NEET
A bomb of mass 9 kg explodes into two pieces of masses 3 kg and 6 kg. The velocity of mass 3 kg is 16 m s-1. The kinetic energy of mass 6 kg in joule is

Options:
A 96
B 384
C 192
D 768

The correct answer is:
192

Solution:
The momentum of 6 kg piece p2= momentum of 3 kg piece
p1=m1v1=3×16=48 kg m s-1
The kinetic energy of 6 kg piece K2=p222m2=48×482×6=192 J

Topic: Physics, Center of Mass Momentum and Collision, NEET
A bomb of mass 30 kg at rest, explodes into two pieces of masses 18 kg and 12 kg. The velocity of 18 kg mass is 6 m s-1. The kinetic energy of the other mass is

Options:
A 324 J

B 486 J

C 256 J

D 523 J

The correct answer is:
486 J

Solution:
Form conservation of linear momentum
m1v1+m2v2=0
v2=- m1m2v1=-18126=-9 m s-1
KE=12m2v22=12Ã12Ã92=486 J

Topic: Physics, Center of Mass Momentum and Collision, NEET
A bomb of mass 30 kg at rest explodes into two pieces of masses 18 kg and 12 kg. The velocity of 18 kg mass is 6 m s-1. The kinetic energy of the other mass is

Options:
A 324 J
B 486 J
C 256 J
D 524 J

The correct answer is:
486 J

Solution:
From conservation of linear momentum
m1Î½1+m2Î½2=0
Î½2=-m1m2Î½1=-18126=-9 ms -1
K.E.=12m2Î½22=12Ã12Ã92=486 J

Topic: Physics, Laws of Motion, NEET
A bomb of mass $30 \mathrm{~kg}$ at rest explodes into two pieces of masses $18 \mathrm{~kg}$ and 12 $\mathrm{kg}$. The velocity of $18 \mathrm{~kg}$ mass is $6 \mathrm{~ms}^{-1}$. The kinetic energy of the other mass is:

Options:
A $324 \mathrm{~J}$
B $486 \mathrm{~J}$
C $256 \mathrm{~J}$
D $524 \mathrm{~J}$

The correct answer is:
$486 \mathrm{~J}$

Solution:
According to law of conservation of conservation of angular moments

$\begin{aligned}
30 \times 0 & =18 \times 6+12 \times V \\
\Rightarrow V & =-9 \mathrm{~m} / \mathrm{s}
\end{aligned}$
Negative sign indicates that both fragments move in opposite direction.
$\begin{aligned}
\text {K.E. of } 12 \mathrm{~kg} & =\frac{1}{2} \mathrm{mv}^2 \\
& =\frac{1}{2} \times 12 \times 81 \\
& =486 \mathrm{~J}
\end{aligned}$

Topic: Physics, Center of Mass Momentum and Collision, NEET, NEET 2005