datavorous/entrance-exam-dataset huggingface dataset:
A uniform cylinder of length L and mass M, having cross-sectional area A is suspended with its length vertical from a fixed point by a massless spring of constant K such that it is half-submerged in the liquid of density Ï in the equilibrium position. The extension of the spring x0 in equilibrium position is

Options:
A Mg/K

B (Mg/K) 1-LAσ/M
C Mg/K1-LAσ/2M
D MgLA/Kσ

The correct answer is:
Mg/K1-LAσ/2M

Solution:
Let x0 = extension in equilibrium position
F = Restoring force in the spring
F=Kx0
The net downward force acting on the cylinder
= Mg - upward thrust
=Mg-L/2×A×σ×g
For equilibrium Kxo=Mg-LA σ g/2
Kxo=Mg1-LA σ/2M
xo=Mg/K1-LA σ/2 M

Topic: Physics, Mechanical Properties of Fluids, NEET
A uniform cylinder rolls down from rest, on a track whose vertical cross-section is a parabola given by the equation y=kx2. If the surface is rough from A to B due to which the cylinder doesn't slip but it is frictionless from B to C, then the height of ascent of cylinder towards C is

Options:
A y13
B 2y13
C 3y12
D y1

The correct answer is:
2y13

Solution:
Applying work-energy theorem

Work done by gravity = Change in Kinetic Energy
For part AB, ( friction will provide torque to generate rotational kinetic energy but work done by friction will be zero because the velocity of the point of action of friction is zero)

mgy1=34mv2

For part BC, (there is no friction, hence no change in rotational kinetic energy)

mgy2=12mv2

Dividing both equations we get
∴ y2=23y1

Topic: Physics, Rotational Motion, NEET
A uniform disc of mass M and radius R is hinged at its centre C . A force F is applied on the disc as shown. At this instant, angular acceleration of the disc is

Options:
A 3FMR
B FMR
C 23FMR
D F2MR

The correct answer is:
FMR

Solution:
Given,
Mass of uniform disc =M
Radius of uniform disc =R
Applied force on disc =F
According to the question, we can draw the diagram as following
We know that, Ï=IÎ±
â Fsinâ¡30Â°R=MR22Î±
â FÃ12Ãâ¡R=MR22Î±
â Î±=FMR

Topic: Physics, Rotational Motion, NEET
A uniform disc of mass M and radius R is hinged at its centre C. A force F is applied on the disc as shown. At this instant, the angular acceleration of the disc is

Options:
A 3FMR
B FMR
C 23FMR
D F2MR

The correct answer is:
FMR

Solution:
Given,
Mass of uniform disc =M
Radius of uniform disc =R
Applied force on disc =F
According to the question, we can draw the diagram as following
We know that, Ï=IÎ±
â Fsinâ¡30Â°R=MR22Î±
â FÃ12Ãâ¡R=MR22Î±
â Î±=FMR

Topic: Physics, Rotational Motion, NEET
A uniform disc of radius R, is resting on a table on its rim. The coefficient of friction between the disc and table is mu (Fig 7.12). Now the disc is pulled with a force F as shown in the figure. What is the maximum value of F for which the disc rolls without slipping?

Options:
A $F\ge 2\mu Mg$
B $F\le 3\mu Mg$
C $F\ge 3\mu Mg$
D $F\le 2\mu Mg$

The correct answer is:
$F\ge 3\mu Mg$

Solution:
Correct Option is : (C)
$F\ge 3\mu Mg$

Topic: Chemistry, Rotational Motion, NEET
A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected in the region such that its velocity is pointed along the direction of fields, then the electron

Options:
A speed will decrease
B speed will increase
C will turn towards left of direction of motion
D will turn towards right of direction a motion

The correct answer is:
speed will decrease

Solution:
Field $B$ not applied only force. Field $E$ will apply a force opposite to velocity of the electron hence, speed will decreases.

Topic: Physics, Electrostatics, NEET, NEET 2011 (Screening)
A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected in the region such that its velocity is pointed along the direction of fields, then the electron

Options:
A Speed will decrease
B Speed will increase
C Will turn towards right of direction of motion
D Will turn towards left of direction of motion

The correct answer is:
Speed will decrease

Solution:
Given condition is shown in figure
Force due to magnetic field is given by
$F_m=q v B \sin \theta$
Here $\theta=0 \Rightarrow F_m=0$
Force due to electric field is
$\vec{F}_e=q \vec{E}$
Here, $\vec{F}_e=-e \vec{E}$
$\therefore$ Force due to electric field will be opposite to the direction of motion.
Hence its speed will decrease.

Topic: Physics, Magnetic Effects of Current, NEET, NEET 2023 (Manipur)
A uniform electric field and a uniform magnetic field exist in a region in the same direction. An electron is projected with a velocity pointed in the same direction. Then the electron will be

Options:
A Be deflected to the left without increase in speed
B Be deflected to the right without increase in speed
C Not be deflected but its speed will decrease
D Not be deflected but its speed will increase

The correct answer is:
Not be deflected but its speed will decrease

Solution:
As the electron is moving parallel to both the fields, so magnetic force does not affect the electron's motion. But electric force (qE) acts opposite to the motion of the electron. Hence, the electron will not be deflected but its speed will decrease.

Topic: Physics, Mathematics in Physics, NEET
A uniform electric field and a uniform magnetic field exist in a region in the same direction. An electron is projected with a velocity pointed in the same direction. Then the electron will

Options:
A be deflected to the left without increase in speed
B be deflected to the right without increase in speed
C not be deflected but its speed will decrease
D not be deflected but its speed will increase

The correct answer is:
not be deflected but its speed will decrease

Solution:
As the electron is moving parallel to both the fields, so magnetic force does not affect the electron's motion but electric force qE acts opposite to the motion of the electron. Hence, the electron will not be deflected but its speed decreases.

Topic: Physics, Magnetic Effects of Current, NEET
A uniform electric field pointing in positive x-direction exists in a region. Let A be the origin, B be the point on the x-axis at x=+1 cm and C be the point on the y-axis at y=+1 cm. Then the potentials at the points A, B and C satisfy the condition

Options:
A VA<VB
B VA>VB
C VA<VC
D VA>VC

The correct answer is:
VA>VB

Solution:
As electric field is along positive x-axis and E=-dVdx, hence potential at A must be greater than that at B ie, VA>VB

Topic: Physics, Electrostatics, NEET
A uniform force of $(3 \mathbf{i}+\mathbf{j}) \mathrm{N}$ acts on a particle of mass $2 \mathrm{~kg}$. Hence the particle is displaced from position $(2 \mathbf{i}+\mathbf{k}) \mathrm{m}$ to position $(4 \mathbf{i}+3 \mathbf{j}-\mathbf{k}) \mathrm{m}$. The work done by the force on the particle is

Options:
A $9 \mathrm{~J}$
B $6 \mathrm{~J}$
C $13 \mathrm{~J}$
D $15 \mathrm{~J}$

The correct answer is:
$9 \mathrm{~J}$

Solution:
Given, force $F=3 i+j$
$\begin{array}{l}
r_1=(2 i+k) m \text { and } r_2 =(4 i+3 j-k) m \\
\therefore s=r_2-r_1=(4 i+3 j-k)-(2 i+k) \\
=(2 i+3 j-2 k) m \\
\begin{array}{l}
\therefore W=F \cdot s & =(3 i+j) \cdot(2 i+3 j-2 k) \\
& =3 \times 2+3+0 \\
& =6+3=9 \mathrm{~J}
\end{array} \\
\end{array}$

Topic: Physics, Work Power Energy, NEET, NEET 2013 (All India)
A uniform force of 4 N acts on a body of mass 10 kg for a distance of 2.0 m. The kinetic energy acquired by the body is

Options:
A 16 J

B 32×108 erg
C 8 J
D 32 erg

The correct answer is:
8 J

Solution:
Changeâin K.E = work done
Work done = force Ã distance = 4 N Ã 2 m =8 J

Topic: Physics, Work Power Energy, NEET
A uniform magnetic field exists in a region given by Bâ=3i^+4j^+5k^ T. A rod of length 5 m is placed along the y-axis and it is moved along the x-axis with constant speed 1 m s-1. Then induced e.m.f. in the rod will be

Options:
A zero
B 25 V
C 20 V
D 15 V

The correct answer is:
25 V

Solution:
We know that
Magnetic field vector Bâ=3i^+4j^+5k^
length of rod Lâ=5j^
velocity vector vâ=1i^
Induced emf is given as
E=Lâ.vâÃBâ
E=5j^â1i^Ã3i^+ 4j^+5k^
E=5j^.-5j^+4k^
E=-25 V

Topic: Physics, Electromagnetic Induction, NEET
A uniform magnetic field exists in a region given by B→=3i^+4j^+5k^ T. A rod of length 5 m placed along y-axis is moved along x-axis with a constant speed 1 m s-1. Then induced EMF in the rod will be

Options:
A zero
B 25 V
C 20 V
D 15 V

The correct answer is:
25 V

Solution:
We know that
E=BZlYvX
E=551=25 V

Topic: Physics, Electromagnetic Induction, NEET
A uniform magnetic field exists in region which forms an equilateral triangle of side a . The magnetic field is perpendicular to the plane of the triangle. A charge q enters into this magnetic field perpendicularly with speed v along perpendicular bisector of one side and comes out along the perpendicular bisector of other side. The magnetic induction in the triangle is

Options:
A mvqa
B 2mvqa
C mv2qa
D mv4qa

The correct answer is:
2mvqa

Solution:
The charged particle moves in a circle of radius a2

∴qvB=mv2a2 or B=2mvqa

Topic: Physics, Magnetic Effects of Current, NEET
A uniform magnetic field is restricted within a region of radius, r. The magnetic field changes with time at a rate, dB→dt. Loop one of radius R>r encloses the region, r and loop two of radius, R is outside the region of magnetic field as shown in the figure below. Then the emf generated is

Options:
A 0 in loop one and zero in loop two.
B -dB→dtπr2 in loop one and -dB→dtπr2 in loop two.
C -dB→dtπr2 in loop one and infinite in loop two.
D -dB→dtπr2 in loop one and zero in loop two.

The correct answer is:
-dB→dtπr2 in loop one and zero in loop two.

Solution:
Magnetic flux linked with area of loop one is Bπr2.

So, emf in loop one is, -dB→dtπr2 .

Magnetic flux linked with area of loop two is zero.

So, emf in loop two is equal to zero.

Topic: Physics, Electromagnetic Induction, NEET, NEET 2016 (Phase 2)
A uniform magnetic field of induction B is confined in a cylindrical region of the radius R. If the field is increasing at a constant rate of dBdt=Î± T s-1, then the intensity of the electric field induced at a point P distant r from the axis as shown in the figure is equal to

Options:
A 18r
B 18rα
C 12rα
D r

The correct answer is:
12rα

Solution:
Magnetic flux linked with circular section of radius r of the cylinder , ϕ=πr2B.
Due to this induced emf, an electric field E is established which is equal to potential difference per unit length.
E=e2πr=πr2α2πr
E=12rα

Topic: Physics, Electromagnetic Induction, NEET
A uniform metal rod is used as a bar pendulum. If the room temperature rises by 10 Â°C, and the coefficient of linear expansion of the metal of the rod is 2Ã10-6 Â°C-1, the period of the pendulum will have a percentage increase of

Options:
A -2×10-3
B -1×10-3
C 2×10-3
D 1×10-3

The correct answer is:
1×10-3

Solution:
The fractional change in period
ΔTT=12αΔθ
ΔTT=12×2×10-6×10=10-5
% change in time period = ΔTT×100=10-5×100=10-3

Topic: Physics, Thermal Properties of Matter, NEET
A uniform metal wire of length / has $10 \Omega$ resistance. Now this wire is stretched to a length $2 /$ and then bent to form a perfect circle. The equivalent resistance across any arbitrary diameter of that circle is

Options:
A $10 \Omega$
B $5 \Omega$
C $40 \Omega$
D $20 \Omega$

The correct answer is:
$10 \Omega$

Solution:
$R_0=10 \Omega$
After stretching its length upto 2l
$R_1=n^2 R_0$
$=4 R_0$
$=40 \Omega$
$R_{A B}=\frac{20 \times 20}{20+20}=10 \Omega$

Topic: Physics, Current Electricity, NEET, NEET 2024 (Re-NEET)
A uniform meter scale of mass 1 kg is placed on a table such that a part of the scale is beyond the edge. If a body of mass 0.25 kg is hung at the end of the scale then the minimum length of scale that should lie on the table so that it does not tilt is

Options:
A 90 cm

B 80 cm
C 70 cm
D 60 cm

The correct answer is:
60 cm

Solution:
If the distance from one end is x then
50-x1=x0.25=x4
200-4x=xâx=40 cm
â´ Length on the table =100-40=60 cm

Topic: Physics, Rotational Motion, NEET
A uniform rod $\mathrm{AB}$ of length $I$ and mass $m$ is free to rotate about point $A$. The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about $\mathrm{A}$ is $m l^2 / 3$, the initial angular acceleration of the rod will be

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

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

Solution:
Given that moment of inertia about \(A\) is
\(I=m l^3 / 3\)
Now, torque about \(A\) is given as
\(\begin{aligned}
& \tau=F \times r \\
& \tau=m g \frac{1}{2} \\
& I \alpha=m g \frac{1}{2} \\
& \frac{m l^2}{3} \alpha=m g \frac{1}{2} \\
& \alpha=\frac{3 g}{2 l}
\end{aligned}\)

Topic: Physics, Rotational Motion, NEET, NEET 2007
A uniform rod AB of mass m and length l is at rest on a smooth horizontal surface. An impulse J is applied to the end B perpendicular to the rod in horizontal direction. Speed of particle P at a distance l6 from the centre towards A of the rod after time t=Ïml12J is

Options:
A 2 J m
B J 2 m
C J m
D 2 J m

The correct answer is:
2 J m

Solution:
Let v and Ï be the linear and angular speeds of the rod
after applying an impulse J at B. Then from :
impulse = change in momentum
we have
mv = J or v = J m ...(1)
I Ï = J Â· l 2
or m l 2 1 2 Ï = J Â· l 2 or Ï = 6 J m l ...(2)
After the given time t = Ï m l 1 2 J , the rod will rotate an angle
Î¸ = Ï t = 6 J m l Ï m l 1 2 J = Ï 2
l 6 . Ï = l 6 6 J m l = J m = v
â´ v P â = 2 v = 2 J m

Topic: Physics, Rotational Motion, NEET
A uniform rod is held at an angle of 45o to the horizontal and released from rest as shown. The minimum coefficient of friction between plane and rod required so that the rod does not slip on the plane upon release, will be

Options:
A 35
B 25
C 37
D 47

The correct answer is:
35

Solution:
Mgl2cos⁡θ=Ml23α
Mg – Ny = Mαl2cos⁡45o
μNy = Mαl2sin⁡45o

Topic: Physics, Rotational Motion, NEET
A uniform rod is rotating about a horizontal axis as shown. The rod is hinged at one of the ends. The rod is released from a vertical position by slightly pushing it. As the rod moves from A to B

Options:
A both the direction and magnitude of angular momentum about the axis change
B the direction of L→ changes but magnitude does not
C the direction of L→ does not change but magnitude does
D neither the direction nor the magnitude changes

The correct answer is:
the direction of L→ does not change but magnitude does

Solution:
As the angular velocity increases without any change in its direction, we can say that the magnitude of angular momentum increases without any change in direction.

Topic: Physics, Rotational Motion, NEET
A uniform rod of length 8a and mass 6m lies on a smooth horizontal surface. Two-point masses m and 2m moving in the same plane with speed 2v and v respectively strike the rod perpendicular at distances a and 2a from the midpoint of the rod in the opposite directions and stick to the rod. The angular velocity of the system immediately after the collision is

Options:
A 6v32a
B 6v33a
C 6v40a
D 6v41a

The correct answer is:
6v41a

Solution:
Applying conservation of angular momentum about point O,
ma2v+2m2av=IÏ
or Ï=6mavI ...(i)
Now,
I=6m8a212+ma2+2m2a2
=32ma2+ma2+8ma2=41ma2
Hence, from Eq.(i)
Ï=6mav41ma2=6v41a

Topic: Physics, Rotational Motion, NEET
A uniform rod of length 50 cm is released in the vertical plane from the position shown in the figure. The rod is hinged smoothly at O. The angular speed of rod when it becomes horizontal is ( take g=10 m s-2 )

Options:
A 30 rad s-1
B 30 rad s-1
C 40 rad s-1
D 32 rad s-1

The correct answer is:
30 rad s-1

Solution:
From the conservation of mechanical energy
mgl2sinâ¡30=12IÏ2
mgl2sinâ¡30=12ml23Ï2
Ï=3g2l=3Ã102Ã50Ã10-2=30 rad s-1

Topic: Physics, Work Power Energy, NEET
A uniform rod of length 200 cm and mass 500 g is balanced on a wedge placed at 40 cm mark. A mass of 2 kg is suspended from the rod at 20 cm and another unknown mass m is suspended from the rod at 160 cm mark as shown in the figure. Find the value of m such that the rod is in equilibrium. (g=10 m s-2

Options:
A 12 kg
B 13 kg
C 112 kg
D 16 kg

The correct answer is:
112 kg

Solution:
Balancing torque about the wedge, we get

2g×20=0.5g×60+mg×120

⇒2=1.5+6m

⇒m=0.56=112 kg

Topic: Physics, Rotational Motion, NEET, NEET 2021
A uniform rod of length L and mass M has been placed on a rough horizontal surface as shown in the figure. The rod is pulled by applying a horizontal force. The coefficient of friction Î¼ between the surface and the rod is given by Î¼=Î¼0x, 0â¤xâ¤L0,x>L, where Î¼0 is a positive constant. The heat generated due to friction as the rod moves by a distance L is

Options:
A Î¼0MgL22
B μ0MgL2
C μ0MgL23
D μ0MgL26

The correct answer is:
μ0MgL22

Solution:
Heat generated = - (Work done by friction)
Wf=-∫0Lμ0xML-xLgdx
Wf=-μ0Mg∫0Lx-x2Ldx
Wf=-μ0MgL22
Heat generated =μ0MgL22

Topic: Physics, Work Power Energy, NEET
A uniform rod of length $l$ and mass $m$ is free to rotate in a vertical plane about $\mathrm{A}$. The rod initially in horizontal position is released. The initial angular acceleration of the rod is (moment of inertia of the rod about $\mathrm{A}$ is $\frac{m l^2}{3}$ ).

Options:
A $m g \frac{l}{2}$
B $\frac{3 g}{2 l}$
C $\frac{2 l}{3 g}$
D $\frac{3 g}{2 l^2}$

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

Solution:
Here torque
$$
\tau=m g\left(\frac{1}{2}\right)
$$
M. I. of $\operatorname{rod}$ about $A$ is:
$$
I=\frac{m l^2}{3}
$$
$\therefore$ Angular acceleration of the rod is
$$
\alpha=\frac{\tau}{I}
$$
$$
\alpha=\frac{m g\left(\frac{1}{2}\right)}{\frac{m l^2}{3}}=\frac{3 g}{2 l} .
$$

Topic: Physics, Rotational Motion, NEET, NEET 2006
A uniform rod of mass m and length L is at rest on a smooth horizontal surface. A ball of mass m, moving with velocity v0, hits the rod perpendicularly at its one end and sticks to it. The angular velocity of rod after collision is

Options:
A 6 v05 L
B 12Â v05Â L
C 2Â v05Â L
D 3Â v07Â L

The correct answer is:
6 v05 L

Solution:
mv0ÃL4=748 m L2+mÃL216ÃÏ
âÏ=6 v05 L

Topic: Physics, Rotational Motion, NEET
A uniform rod of the length of 1 m and mass of 2 kg is attached to side support at O as shown in the figure. The rod is at equilibrium due to upward force T acting at P. Assume the acceleration due to gravity as 10 m s-2. The value of T is

Options:
A 0
B 2 N
C 5 N
D 10 N

The correct answer is:
10 N

Solution:
Torque about o
τ=mg l 2 −Tl=0
T=mg2=10 N

Topic: Physics, Rotational Motion, NEET
A uniform rope having mass m hangs vertically from a rigid support. A transverse wave pulse is produced at the lower end. The speed (v) of wave pulse varies with height h from the lower end as shown in the figure.

Options:
A
B
C
D

The correct answer is:

Solution:
Let m be the total mass of the rope of lengthl. Tension in the rope at a height h from lower end= weight of rope of length h is T=mgl(h)
As v2=gh
Which is a parabola. Therefore, h versus v graph is a parabola option (a) is correct.

Topic: Physics, Waves and Sound, NEET
A uniform rope of length L and mass m1, hangs vertically from a rigid support. A block of mass m2 is attached to the free end of the rope. A transverse pulse of wavelength λ1 is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is λ2. The ratio λ2λ1 is:

Options:
A m2m1
B m1+m2m2
C m1m2
D m1+m2m1

The correct answer is:
m1+m2m2

Solution:
The situation in the rope is given below,

The velocity of the wave at the bottom of the rope is given by,

v=Tμ; where T is tension and μ is linear mass density of the string.

v1=m2gLm1

Hence, the wavelength for frequency f will be,

λ1=v1f=1fm2gLm2

Similarly, at top the velocity will be,

v1=(m1+m2)gLm1

And, the wavelength will be,

λ2=1f(m1+m2)gLm1

Hence, the ratio of both wavelengths will be,

λ 2 λ 1 = m 1 + m 2 m 2 .

Topic: Physics, Waves and Sound, NEET, NEET 2016 (Phase 1)
A uniform rope of length l lies on a table. If the coefficient of friction is μ, then the maximum length l1 of the part of this rope which can overhang from the edge of the table without sliding down is

Options:
A 1μ
B lμ+1
C μl1+μ
D μl1-μ

The correct answer is:
μl1+μ

Solution:
Weight of hanging part is balanced by friction
ml1lg=μN
ml1lg=μml-l1lg
l1=μl-l1
l1=μlμ+1

Topic: Physics, Laws of Motion, NEET
A uniform solid cylindrical roller of mass m and R is rolled on the ground, without slipping, by applying a constant horizontal force F . Find angular acceleration of cylinder.

Options:
A FmR
B 2FmR
C 2F3mR
D F3mR

The correct answer is:
2F3mR

Solution:
F-fr=ma â¦i
frR=IÎ±=mR22Î± â¦(ii)
For pure rolling
a=Î±R â¦(iii)
From (i), (ii) and (iii)
F-mRâÎ±2=mÎ±R
F=32mRââÎ±
Î±=2F3mR

Topic: Physics, Rotational Motion, NEET
A uniform solid right circular cone of the base radius r is joined to a uniform solid hemisphere of radius r and of the same density, so as to have a common face. The centre of mass of the composite solid lies on the common face. The height of the cone is

Options:
A 2 r
B 3 r
C 3 r
D 6 r

The correct answer is:
3 r

Solution:
Volume of cone=13πr2h
Mass of cone, m1=ρ×13πr2h
Mass of hemisphere, m2=ρ×12×43πr3
=ρ×23πr3
Now, Y=m1y1+m2y2m1+m2
⇒0=ρ×13πr2h×h4+23πr3(−3r8)ρ×13πr2h+ρ×23πr2
⇒ρ×13πr2h24−2r×3r8=0
h24−3r24=0  or h=3 r

Topic: Physics, Center of Mass Momentum and Collision, NEET
A uniform solid sphere A of mass m is rolling without sliding on a smooth horizontal surface. It collides elastically and head-on with another stationary hollow sphere B of the same mass and radius. Assuming friction to be absent everywhere, the ratio of the kinetic energy of B to that of A just after the collision is

Options:
A 5:2
B 1:1
C 2:3
D 3:2

The correct answer is:
5:2

Solution:
Since the two bodies have the same mass and collide head-on elastically, the linear velocities get exchanged.
Hence, just after the collision B will move with velocity v0 and A becomes stationary but continues to rotate at the same initial angular velocity v0R.

So after the collision, the kinetic energies of A and B are

KB=12mv02

KA=12Iω2=1225mR2·v0R2

⇒KBKA=52

Topic: Physics, Rotational Motion, NEET
A uniform solid sphere of mass M and radius R is placed on a smooth horizontal surface. It is given a horizontal impulse J at a height h above the centre of mass and sphere starts rolling. Then the value of h and speed of centre of mass are

Options:
A h=25R and v=JM
B h=25RÂ andÂ v=25JM
C h=75RÂ andÂ v=75JM
D h=75RÂ andÂ v=JM

The correct answer is:
h=25R and v=JM

Solution:
Let the force producing impulse J is F.
⇒F×h=25mR2×a
Also, F=ma (where a=Rα)
∴ mah=25mRa
⇒h=25R
Also impulse = change in momentum
or, J=Mv
⇒v=JM

Topic: Physics, Rotational Motion, NEET
A uniform solid sphere of mass m, radius R moving with velocity v0 is rolling without slipping on a frictionless surface. It collides elastically with a frictionless vertical wall. Ratio of magnitude of angular momentum of the sphere before and after the collision about its bottommost point is

Options:
A 35
B 53
C 37
D 73

The correct answer is:
73

Solution:
Lin=mv0R+25mR2w
Lin=75mv0R
Lfinal=mv0R-25mR2ω
=35mvR
Ratio=73

Topic: Physics, Rotational Motion, NEET
A uniform sphere of mass 500 g rolls without slipping on a plane horizontal surface with its centre moving at a speed of 5.00 cm s-1. Its kinetic energy is:

Options:
A 8.75×10-4 J
B 8.75×10-3 J
C 6.25×10-4 J
D 1.13×10-3 J

The correct answer is:
8.75×10-4 J

Solution:
K.E of the sphere = Translational K.E+ Rotational K.E
=12mv21+K2R2K= Radius of gyration
= 12×12×210021+25
= 354×10-4 J
= 8.75×10-4 J

Topic: Physics, Rotational Motion, NEET
A uniform sphere of weight w and radius 3 m is being held by a string of length 2 m and attached to a frictionless wall, as shown in the figure. The tension in the string will be

Options:
A 5w4
B 15w4
C 15w6
D none of these

The correct answer is:
5w4

Solution:
From geometry:
cos⁡θ=35
sin⁡θ=45
As the sphere is at equilibrium
Tsin⁡θ=w
T=5w4

Topic: Physics, Laws of Motion, NEET