datavorous/entrance-exam-dataset huggingface dataset:
A small square loop of wire of side l is placed inside a large square loop of wire of side L(L>>I). The loops are coplanar and their centres coincide. The mutual inductance of the system is proportional to

Options:
A lL

B l2L

C Ll

D L2l

The correct answer is:
l2L

Solution:
Magnetic field produced by a current i in a large square loop at its centre,
B â iL
Say B = K iL

Therefore, Magnetic flux linked with smaller loop,
Ï=B. S '
Ï = k iL(/2)
Therefore, the mutual inductance
M =&phiv;i= Kl2L

or M âl2L

Topic: Physics, Electromagnetic Induction, NEET
A small telescope has an objective of focal length 140 cm and an eye piece of focal length 5.0 cm . The magnifying power of telescope for viewing a distant object is:

Options:
A 28
B 17
C 32
D 34

The correct answer is:
28

Solution:
$f_0=140 \mathrm{~cm}$ and $f_e=5 \mathrm{~cm}$
For distant object,
$m=\frac{f_0}{f_e}=\frac{140}{5}=28$

Topic: Physics, Ray Optics, NEET, NEET 2024
A small tiny lead shot is gently dropped on the surface of a viscous liquid

Options:
A the lead shot will fall with an acceleration equal to g at that place
B the velocity of the lead shot will decrease with time
C the velocity of the lead shot will increase continuously
D the velocity of the lead shot will reach a steady value after sometime

The correct answer is:
the velocity of the lead shot will reach a steady value after sometime

Solution:
When a tiny lead shot is gently dropped on the surface of a viscous liquid, the velocity of the lead shot increases with time until it finally attains a steady value known as terminal velocity

Topic: Physics, Mechanical Properties of Fluids, NEET
A smooth steel ball strikes a fixed smooth steel plate at an angle θ with the vertical. If the coefficient of restitution is e, the angle at which the re-bounce will take place is

Options:
A θ
B tan-1â¡tanâ¡Î¸e
C e tan â¡θ
D tan-1â¡etanâ¡θ

The correct answer is:
tan-1⁡tan⁡θe

Solution:
Since, no force is present along the surface of plate. So, momentum conservation principle for ball is applicable along the surface of plate.
mÏ sin â¡Î¸1=mÏ sin â¡Î¸2
Or Ï sin â¡Î¸1=Ï1 sin â¡Î¸2 â¦(i)
e=Ï1 cos â¡Î¸2Ï cos â¡Î¸1=Ï1 cos â¡Î¸2Ï cos â¡Î¸ (Î¸1=Î¸)
â´ Ï1 cos â¡Î¸2=eÏ cos â¡Î¸ â¦(ii)
â´Ï1 sin â¡Î¸2Ï1 cos â¡Î¸2=Ï sin â¡Î¸eÏ cos â¡Î¸=tanâ¡Î¸e
â´tanâ¡Î¸2=tanâ¡Î¸e
â´Î¸2=tan-1â¡tanâ¡Î¸e

Topic: Physics, Center of Mass Momentum and Collision, NEET
A smooth uniform rod of length L and mass M has two identical beads of negligible size, each of mass m , which can slide freely along the rod. Initially the two beads are at the centre of the rod and the system is rotating with angular velocity ω0 about its axis perpendicular to the rod and passing through its mid point (see figure). There are no external forces. When the beads reach the ends of the rod, the angular velocity of the system is

Options:
A Mω0M+3m
B Mω0M+6m
C M+6mω0M
D ω0

The correct answer is:
Mω0M+6m

Solution:
From the principle of conservation of angular momentum, I0ω0=Iω , where I0 and ω0 are the moment of inertia and angular velocity when the beads are at the centre of the rod and I and ω those when the beads are at the ends of the rod.
I0=ML212 and I=ML212+mL24+mL24=L212 M+6m
∴ML212ω0=M+6mωL212
Or ω=Mω0M+6m

Topic: Physics, Rotational Motion, NEET
A soap bubble A of radius 0.03 m and another bubble B of radius 0.04 m are brought together so that the combined bubble has a common interface of radius r, then the value of r is

Options:
A 0.24 m

B 0.48 m

C 0.12 m

D 0.50 m

The correct answer is:
0.12 m

Solution:
Let the radius of curvature of the common internal film surface of the double bubble formed by two bubbles A and B be r.
Excess of pressure as compared to the atmosphere inside A is
p1=4Tr1=4T0.03
Excess of pressure inside B is
p2=4Tr2=4T0.04
In the double bubble, the pressure difference between A and B on either side of the common surface is
4T0.03-4T0.04=4Tr
â 10.03-10.04=1r
â r=0.03Ã0.040.01=0.12 m

Topic: Physics, Mechanical Properties of Fluids, NEET
A soap bubble having radius of 1 mm is blown from a detergent solution having a surface tension of 2.5×10-2 N m-1. The pressure inside the bubble equals at a point Z0 below the free surface of water in a container. Taking g=10 m s-2, density of water =103 kg m-3, the value of Z0 is

Options:
A 100 cm
B 10 cm
C 1 cm
D 0.5 cm

The correct answer is:
1 cm

Solution:
Pressure inside soap bubble is given by, ∆P=4TR
Pressure below water surface at depth Z0 is given by, ∆P=ρgZ0
According to question,
⇒4TR=ρgZ0
⇒Z0=4Tρg=4×2.5×10-2103×10×10-3
⇒Z0=1100 m
⇒Z0=1 cm

Topic: Physics, Mechanical Properties of Fluids, NEET, NEET 2019 (All India)
A soap bubble of radius r is blown up to form a bubble of radius 2r under isothermal conditions. If σ is the surface tension of the soap solution, the energy spent in the process is

Options:
A 6πσr2
B 24πσr2
C 12πσr2
D 9πσr2

The correct answer is:
24πσr2

Solution:
Energy spent = Work done = T∆A
Here, A is total surface area
∴A=8πr2 for soap bubble
Hence, energy spent =σ8π2r2-8πr2
=σ32πr2-8πr2
=24πσr2

Topic: Physics, Mechanical Properties of Fluids, NEET
A sodium salt on treatment with MgCl2 gives white precipitate only on heating. The anion of the sodium salt is

Options:
A CO32-
B SO42-
C NO3-
D HCO3-

The correct answer is:
HCO3-

Solution:
Since magnesium bicarbonate on heating gives insoluble magnesium carbonate, therefore, the anion is HCO3-
MgCl2+2NaHCO3âMg(HCO3)2Soluble+2NaCl
Mg ( HCO 3 ) 2 → Δ MgCO 3 White ↓+ H 2 O+ CO 2 ↑

Topic: Chemistry, s Block Elements, NEET
A solenoid 1.5 m long and 0.4 cm in diameter possesses 10 turns/cm. A current of 5 A flows through it. The magnetic field at the axis inside the solenoid is

Options:
A 2ÏÃ10-3 T
B 2ÏÃ10-5Â T
C 4ÏÃ10-2Â T
D 4ÏÃ10-3Â T

The correct answer is:
2ÏÃ10-3 T

Solution:
Magnetic field at the axis inside the solenoid
B=Î¼0ni
Here, n=10 turns cm-1=1000 turns m-1, i=5 A
â´ B=4ÏÃ10-7Ã1000Ã5
=2ÏÃ10-3 T

Topic: Physics, Magnetic Effects of Current, NEET
A solenoid 30 cm long is made by winding 2000 loops of wire on an iron rod whose cross-section is 1.5 cm2. If the relative permeability of the iron makes it 600. What is the self-inductance of the solenoid?

Options:
A 1.5 H
B 2.5 H
C 3.5 H
D 0.5 H

The correct answer is:
1.5 H

Solution:
Self-inductance of the solenoid
L=μ0μr N2Al
=600×4π×10-7×20002×1.5×10-40.3
=1.5 H

Topic: Physics, Magnetic Properties of Matter, NEET
A solenoid has a core of a material with relative permeability 400. The windings of the solenoid are insulated from the core and carry a current of 2A. If the number of turns is 1000 per metre, calculate H

Options:
A 2 ×103A/m
B 4 ×103A/m
C 2 ×104A/m
D 4 ×104A/m

The correct answer is:
2 ×103A/m

Solution:
Correct Option is : (A)
2 ×103A/m

Topic: Chemistry, Magnetic Properties of Matter, NEET
A solenoid has a core of a material with relative permeability 400. The windings of the solenoid are insulated from the core and carry a current of 2A. If the number of turns is 1000 per metre, calculate the magnetising current Im

Options:
A 557 A
B 794 A
C 864 A
D 887 A

The correct answer is:
794 A

Solution:
Correct Option is : (B)
794 A

Topic: Chemistry, Magnetic Properties of Matter, NEET
A solenoid has a core of a material with relative permeability 400. The windings of the solenoid are insulated from the core and carry a current of 2A. If the number of turns is 1000 per metre, calculate B

Options:
A 0.5 T
B 2.0 T
C 1.0 T
D 1.5 T

The correct answer is:
1.0 T

Solution:
Correct Option is : (C)
1.0 T

Topic: Chemistry, Magnetic Properties of Matter, NEET
A solenoid has a core of a material with relative permeability 400. The windings of the solenoid are insulated from the core and carry a current of 2A. If the number of turns is 1000 per metre, calculate M

Options:
A 8 ×105A/m
B 16 ×105A/m
C 8 ×104A/m
D 16 ×104A/m

The correct answer is:
8 ×105A/m

Solution:
Correct Option is : (A)
8 ×105A/m

Topic: Chemistry, Magnetic Properties of Matter, NEET
A solenoid has an inductance of 50 mH and a resistance of 0.025 Î© . If it is connected to a battery, how long will it take for the current to reach one half of its final equilibrium value?

Options:
A 1.38 s
B 1.2Â s
C 6.32Â s
D 0.23Â s

The correct answer is:
1.38 s

Solution:
Time constant is,
Ï=LR
Time for current to reach the half of final value is,
t12=Ïlnâ¡2
t12=LRlnâ¡2
t12=50Ã10-30.0250.693
t12=1.38 s

Topic: Physics, Alternating Current, NEET
A solenoid has length 0.4 cm, radius 1 cm and 400 turns of wire. If a current of 5 A is passed through this solenoid, what is the magnetic field inside the solenoid?

Options:
A 6.28×10-4 T
B 6.28×10-3 T
C 6.28×10-7 T
D 6.28×10-6 T

The correct answer is:
6.28×10-3 T

Solution:
The magnetic field inside the solenoid is given by
B=μ0ni0
(n = Number of turns per unit length, I = Current in coil)
Where n=NL.r
n=5000.4×10-2×1×10-2
n=1000
μ0=4π×10-7 Tm/A
B=4×3.14×1000×5×10-7
=62800×10-7
=6.28×10-3 T

Topic: Physics, Magnetic Effects of Current, NEET
A solenoid of 2.5 m length and 2.0 cm diameter possesses 10 turns per cm. A current of 0.5 A is flowing through it. The magnetic induction at axis inside the solenoid is

Options:
A 2ÏÃ10-4 T
B 2ÏÃ10-5Â T
C 2ÏÃ10-6Â T
D 2ÏÃ10-7Â T

The correct answer is:
2ÏÃ10-4 T

Solution:
The magnetic induction along the axis of solenoid
B=μonI
Where, n= number of turns per unit length
and I= current
∵ μ04π=10-7 T A-1
⇒ μ0=4π×10-7 T A-1
Given, n=10 cm-1=1000 m-1
I=0.5 A
∴ B=4π×10-7×1000×0.5
=2π×10-4 T

Topic: Physics, Magnetic Effects of Current, NEET
A solenoid of radius 4 cm and 1000 turns, carries a current of 8 A. If it is equivalent to a magnet of same size and magnetisation Mâ (Magnetic momentVolume) of
20000 Am-1, then the length of the solenoid is

Options:
A 10Ï cm
B 20ÏÂ cm
C 30 cm
D 40 cm

The correct answer is:
40 cm

Solution:
M=Ïr2INÏr2L=NIL â L=NIM
=1000Ã820000Ã100 cm

=40 cm

Topic: Physics, Magnetic Properties of Matter, NEET
A solid AB has rocksalt structure. If the edge length is 400 pm and the radius of the cation is 80 pm , then the radius of the anion is

Options:
A 100 pm
B 120 pm
C 250 pm
D 325 pm

The correct answer is:
120 pm

Solution:
a=2(r++r-) ,
400=2(80 + r-)
r-=120 pm

Topic: Chemistry, Solid State, NEET
A solid ball of density ρ1 and radius r falls vertically through a liquid of density ρ2. Assume that the viscous force acting on the ball is F=krv, where k is a constant and v its velocity. What is the terminal velocity of the ball?

Options:
A 4πgr2ρ1-ρ23k
B 2πrρ1-ρ23gk
C 2πgρ1+ρ23gr2k
D none of these

The correct answer is:
4πgr2ρ1-ρ23k

Solution:
Net force on the ball =0
(when terminal velocity is attained).
Hence,
Weight=upthrust+viscous force
∴43 πr3 ρ1g=43 πr3ρ2g+krvT
∴ vT=4πgr23k ρ1- ρ2

Topic: Physics, Mechanical Properties of Fluids, NEET
A solid ball of radius 0.2 m and mass 1 kg lying at rest on a smooth horizontal surface is given an instantaneous impulse of 50 N s at point P as shown. The number of rotations made by the ball about its diameter before hitting the ground is

Options:
A 62532π
B 250032π
C 312532π
D 125032π

The correct answer is:
312532π

Solution:
Impulse = change in momentum
Along vertical
VV=50cos⁡30°1ms=253 m s-1
Along horizontal
VH=50cos⁡60°1ms=25 m s-1
Time of flight T=2VVg=53 s
Also ω=50×0.225mr2 rad s-1
=625 rad s-1
2πn=ω T
⇒n=ωT2π=312532π

Topic: Physics, Rotational Motion, NEET
A solid ball of radius R has a charge density ρ given by ρ=ρo1-rR for 0≤r≤R. The electric ﬁeld outside the ball is

Options:
A ρ0R3ε0r2
B 4ρ0R33ε0r2
C 3ρ0R34ε0r2
D Ï0R312Îµ0r2

The correct answer is:
ρ0R312ε0r2

Solution:
ρ=ρ01-rR
dq=ρdv
qin=∫dq=ρdv

⇒qin=∫0Rρ01-rR4πr2dr

=4πρ0∫0R1-rRr2dr

=4πρ0∫0Rr2dr-r3Rdr

=4πρ0r330R-r44R0R

=4πρ0 R33-R44R

=4πρ0R312

⇒qin=πρ0R33
E×4πr2=πρ0R33ε0
⇒E=ρ0R312ε0r2

Topic: Physics, Electrostatics, NEET
A solid body rotates an angle θ about a stationary axis according to the law θ=6t-2t3. What is the mean value of angular velocity over the time interval between t=0 and the time when the body comes to rest?

Options:
A 1 rad s-1
B 2 rad s-1
C 3 rad s-1
D 4 rad s-1

The correct answer is:
4 rad s-1

Solution:
Given, Î¸=6t-2t3 ...(i)
â´dÎ¸dt=Ï=6-6t2
As, Ï=0 (given)
6-6t2=0
t2=Â±1
As, t=-1 s is not possible, hence t=1 s.
Substituting t=1s in Equation (i), we get
Î¸1=6-2=4 rad
and Î¸0=6Ã0-2Ã02t=0=0
â´ Mean value of Ï=Î¸1-Î¸0t1-t0=4-01-0=4 rad s-1
Hence, option 4 is correct.

Topic: Physics, Laws of Motion, NEET
A solid compound $\mathrm{X}$ on heating gives $\mathrm{CO}_2$ gas and a residue. The residue mixed with water forms Y. On passing an excess of $\mathrm{CO}_2$ through $\mathrm{Y}$ in water, a clear solution $\mathrm{Z}$ is obtained. On boiling $\mathrm{Z}$, compound $\mathrm{X}$ is reformed. The compound $\mathrm{X}$ is:

Options:
A $\mathrm{Ca}\left(\mathrm{HCO}_3\right)_2$
B $\mathrm{CaCO}_3$
C $\mathrm{Na}_2 \mathrm{CO}_3$
D $\mathrm{K}_2 \mathrm{CO}_3$

The correct answer is:
$\mathrm{CaCO}_3$

Solution:
$\begin{aligned}
& \mathrm{CaCO}_3 \stackrel{\Delta}{\longrightarrow} \mathrm{CaO}+\mathrm{CO}_2 \uparrow \\
& \mathrm{X} \quad \text { residue } \\
& \mathrm{CaO}+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{Ca}(\mathrm{OH})_2 \\
& \mathrm{Y} \\
& \mathrm{Ca}(\mathrm{OH})_2+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{Ca}\left(\mathrm{HCO}_3\right)_2 \\
& \mathrm{Z} \\
& \mathrm{Ca}\left(\mathrm{HCO}_3\right)_2 \stackrel{\Delta}{\longrightarrow} \mathrm{CaCO}_3+\mathrm{CO}_2 \uparrow +\mathrm{H}_2 \mathrm{O}
\end{aligned}$

Topic: Chemistry, s Block Elements, NEET, NEET 2004
A solid cube and a solid sphere are made out of the same material and they have equal surface areas. Both are at the same temperature of 120 Â°C, then

Options:
A both of them will cool down at the same rate
B the cube will cool down slower than the sphere
C the sphere will cool down slower than the cube
D the information is insufficient

The correct answer is:
the sphere will cool down slower than the cube

Solution:
As Q=ms∆T

dQdt=msdTdt

⇒dTdt=1msdQdt
Here, dTdt is the rate of cooling

dQdt∝A [Rate of heat transfer is directly proportional to the surface area]

∴dTdt∝1m⇒ Heavier will cool slower.
For the same surface area, the volume of the sphere will be more than that of the cube, hence it will have more mass

Topic: Physics, Thermal Properties of Matter, NEET
A solid cube of the edge a is molten and moulded in eight identical small solid cubes and are placed on one other on a straight line with the edge of the bottom cube on the same horizontal plane on which big cube was placed, then the vertical shift in the centre of mass is

Options:
A 3a2
B 2a
C 5a2
D 3a

The correct answer is:
3a2

Solution:
Initial height of centre of mass =a2
keeping volume same, before and after
8 L 3 = a 3
⇒L= a 2 (side of samller cubes)
new height of center of mass =4L=2a
Δh=2a− a 2 = 3 2 a

Topic: Physics, Center of Mass Momentum and Collision, NEET
A solid cube of the edge a is molten and moulded in eight identical small solid cubes and are placed on one other on a straight line with the edge of the bottom cube on the same horizontal plane on which big cube was placed, then the vertical shift in the centre of mass is

Options:
A 3a2
B 2a
C 5a2
D 3a

The correct answer is:
3a2

Solution:
Initial height of centre of mass =a2
keeping volume same, before and after
8 L 3 = a 3
âL= a 2 (side of samller cubes)
new height of center of mass =4L=2a
Îh=2aâ a 2 = 3 2 a

Topic: Physics, Center of Mass Momentum and Collision, NEET
A solid cylinder is rolling without slipping down an inclined plane. Then its angular momentum is:

Options:
A Conserved about COM of the cylinder
B Conserved about point of contact
C Conserved about all the points
D Not conserved about any point

The correct answer is:
Not conserved about any point

Solution:
The forces are not concurrent, hence a finite torque will act about any of the point and hence angular momentum is not conserved about any point.

Topic: Physics, Rotational Motion, NEET
A solid cylinder of mass 2 kg and radius 4 cm is rotating about its axis at the rate of 3 rpm. The torque required to stop after 2π revolutions is

Options:
A 2×10-6 N m
B 2×10-3 N m
C 12×10-4 N m
D 2×106 N m

The correct answer is:
2×10-6 N m

Solution:
From rotational kinematic equations,
ωf2-ωi2=2αθ
02-3 2π602=2 α 2π2π
α=-1800 rad s-2
∴ Torque required τ=Iα
=24×10-222×1800
=16×10-4800
=2×10-6 N m

Topic: Physics, Rotational Motion, NEET, NEET 2019 (All India)
A solid cylinder of mass 2 kg and radius 0.2 m is rotating about its own axis without friction with an angular velocity of 3 rad s-1. Angular momentum of the cylinder is

Options:
A 0.2 J s

B 1.12 J s

C 0.12 J s
D 12 J s

The correct answer is:
0.12 J s

Solution:
L=IÏ=12mr2Ï=12Ã20.22Ã3=0.12 J s

Topic: Physics, Rotational Motion, NEET
A solid cylinder of mass $2 \mathrm{~kg}$ and radius $50 \mathrm{~cm}$ rolls up an inclined plane of angle inclination $30^{\circ}$. The centre of mass of cylinder has speed of $4 \mathrm{~m} / \mathrm{s}$. The distance travelled by the cylinder on the inclined surface will be : (Take $g=10 \mathrm{~m} / \mathrm{s}^2$ )

Options:
A $2.2 \mathrm{~m}$
B $1.6 \mathrm{~m}$
C $1.2 \mathrm{~m}$
D $2.4 \mathrm{~m}$

The correct answer is:
$2.4 \mathrm{~m}$

Solution:
When a body rolls i.e. have rotational motion, the total kinetic energy of the system will be
$$
\mathrm{KE}=\frac{1}{2} \mathrm{mv}^2\left(1+\frac{\mathrm{k}^2}{\mathrm{R}^2}\right)
$$
where, $\mathrm{m}=$ mass of body, $\mathrm{v}=$ velocity and $\mathrm{k}=$ radius of gyration
Given, $\mathrm{m}=2 \mathrm{~kg}, \theta=30^{\circ}, \mathrm{v}=4 \mathrm{~ms}^{-1}$
Let $h$ be the height of the inclined plane, then from law of conservation of energy,
$$
\begin{aligned}
\mathrm{KE} & =\mathrm{PE} \\
\frac{1}{2} \mathrm{mv}^2\left(1+\frac{\mathrm{k}^2}{\mathrm{R}^2}\right) & =\mathrm{mgh}
\end{aligned}
$$
Substituting the given values in the above equation, we get
$$
\begin{aligned}
& \frac{1}{2} \times 2 \times 16\left(1+\frac{1}{2}\right)=2 \times 10 \times \mathrm{h}\left[\because \text { For cylinder } \frac{\mathrm{k}^2}{\mathrm{R}^2}=\frac{1}{2}\right] \\
& \Rightarrow \quad 8 \times \frac{3}{2}=10 \mathrm{~h} \Rightarrow \mathrm{h}=1.2 \mathrm{~m}
\end{aligned}
$$
From the above diagram
$$
\begin{gathered}
\sin \theta=\frac{\mathrm{h}}{\mathrm{x}} \\
\Rightarrow \mathrm{x}=\frac{\mathrm{h}}{\sin \theta}=\frac{1.2}{\sin 30^{\circ}}=1.2 \times 2=2.4 \mathrm{~m}\left[\because \sin 30^{\circ}=\frac{1}{2}\right]
\end{gathered}
$$

Topic: Physics, Rotational Motion, NEET, NEET 2019 (Odisha)
A solid cylinder of mass $3 \mathrm{~kg}$ is rolling on a horizontal surface with velocity $4 \mathrm{~ms}^{-1}$. It collides with a horizontal spring of force constant $200 \mathrm{Nm}^{-1}$. The maximum compression produced in the spring will be

Options:
A $0.5 \mathrm{~m}$
B $0.6 \mathrm{~m}$
C $0.7 \mathrm{~m}$
D $0.2 \mathrm{~m}$

The correct answer is:
$0.6 \mathrm{~m}$

Solution:
Loss in $\mathrm{KE}=$ Gain in spring energy
$\frac{1}{2} m v^2\left[1+\frac{K^2}{R^2}\right]=\frac{1}{2} k x_{\max }^2$
where $k$ is the force constant.
Given, $v=4 \mathrm{~m} / \mathrm{s}, m=3 \mathrm{~kg}, k=200 \mathrm{~N} / \mathrm{m}$
For solid cylinder, $\frac{K^2}{R^2}=\frac{1}{2}$
$\therefore \frac{1}{2} \times 3 \times(4)^2\left[1+\frac{1}{2}\right]=\frac{1}{2} \times 200 \times x_{\max }^2$
The maximum compression in the spring
$x_{\text {max }}=0.6 \mathrm{~m}$

Topic: Physics, Laws of Motion, NEET, NEET 2012 (Screening)
A solid cylinder of mass 20 kg rotates about its axis with the angular speed 100 rad s-1. The radius of the cylinder is 0.25 m. What is the magnitude of angular momentum of the cylinder about its axis?

Options:
A 62.5 J s
B 70.4 J s
C 79.6 J s
D 60.5 J s

The correct answer is:
62.5 J s

Solution:
If I be the moment of inertia of the cylinder about its axis, then
I=12 MR2=12×20×0.252=0.625 kg m2
∴ KE associated with the rotating cylinder is given by,
KE=12Iω2=12×0.625×1002=3125 J
Also using the relation,
L=2IKE
We have L=2×0.625×3125
L=62.5 J s

Topic: Physics, Rotational Motion, NEET
A solid cylinder of mass 50 kg and radius 0.5 m is free to rotate about horizontal axis. A massless string is wound round the cylinder with one end attached to it and other hanging freely. Tension in the string required to produce an angular acceleration of 2 revolutions s-2

Options:
A 25 N
B 50 N
C 78.5 N
D 157 N

The correct answer is:
157 N

Solution:
In this question we have to calculate tension

Given M = 50 kg

R = 0.5 m

angular acceleration = 2 revolution/s2

Now, Balancing the Torque

T×R=MR22α
T=MR2α=50×0.522×2π=157 N

So, the tension will be 157 N

Topic: Physics, Rotational Motion, NEET, NEET 2014
A solid cylinder of mass M and radius R rolls down an inclined plane of height h. The angular velocity of the cylinder when it reaches the bottom of the plane will be

Options:
A 2Rgh
B 2Rgh2
C 2Rgh3
D 12Rgh

The correct answer is:
2Rgh3

Solution:
Loss in PE = Gain in KE
mgh=12mv2+12Tω2 v=Rω
mgh=12mRω2+12mR22ω2
gh=34R2ω2
⇒ ω2=hgh3R2 ⇒ω=2Rgh3

Topic: Physics, Rotational Motion, NEET
A solid cylinder of mass M and radius R rolls from rest down a plane inclined at an angle θ to the horizontal. The velocity of the centre of mass of the cylinder after it has rolled down a distance d is -

Options:
A 23gdtan⁡θ
B gd tan⁡θ
C 34gdsin⁡θ
D 43gdsin⁡θ

The correct answer is:
43gdsin⁡θ

Solution:
v=2ad=2gsinâ¡Î¸1+K2R2.d
Here K2R=12
â´ v=4gdsinâ¡Î¸3

Topic: Physics, Rotational Motion, NEET
A solid cylinder of mass M and radius R rolls on a flat surface. Find its moment of inertia about the line of contact.

Options:
A 32MR2
B MR2
C 2 MR2
D 23MR2

The correct answer is:
32MR2

Solution:
Moment of inertia of solid cylinder about O,
ICM=12MR2
Moment of inertia about point of contact P, IP=ICM+MR2 (using parallel axis theorem)
=12MR2+MR2=32MR2

Topic: Physics, Rotational Motion, NEET
A solid cylinder of mass $M$ and radius $R$ rolls without slipping on an inclined plane of length $L$ and height $h$. What is the speed of its centre of mass when the cylinder reaches its bottom?

Options:
A $\sqrt{2 g h}$
B $\sqrt{\frac{3}{4} g h}$
C $\sqrt{\frac{4}{3} g h}$
D $\sqrt{4 g h}$

The correct answer is:
$\sqrt{\frac{4}{3} g h}$

Solution:
Here K.E. of centre of mass when cylinder reached the bottom $=\frac{1}{2} m v^2+\frac{1}{2} \mathrm{I} \omega^2 \ldots(\mathrm{i})$
Here $\mathrm{I}=\frac{1}{2} M K^2$ and $\omega=\frac{V}{R}$
$\begin{aligned}
& \therefore \mathrm{KE}=\frac{1}{2} m v^2+\frac{1}{2} M K^2 \cdot \frac{V^2}{r^2} \\
& =\frac{1}{2} m v^2\left[1+\frac{K^2}{R^2}\right]
\end{aligned}$
But for solid cylinder $K=\frac{R}{\sqrt{2}}$ or $\frac{K^2}{R^2}=\frac{1}{2}$
$\therefore K E=\frac{3}{4} m v^2$
PE of the solid cylinder at height $h$ is
$\mathrm{PE}=m g h$
From eq. (ii) and (iii)
$\begin{aligned}
& m g h=\frac{3}{4} m v^2 \\
& \Rightarrow v=\sqrt{\frac{4}{3} g h}
\end{aligned}$

Topic: Physics, Rotational Motion, NEET, NEET 2003
A solid cylinder rolls down from an inclined plane of height h. What is the velocity of the cylinder when it reaches at the bottom of the plane?

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

The correct answer is:
4gh 3

Solution:
Since, we know that when a body rolls down from an inclined plane of height h, its velocity at the bottom of the plane,
v=2gh1+K2R2
Where, K= radius of gyration of the body and R= radius of the body.
For the solid cylinder,
K=R2
∴v=2gh1+R22R2=2gh1+12=4gh3

Topic: Physics, Rotational Motion, NEET
A solid has a bcc structure. If the distance of closest approach between the two atoms is 1.73 â« . The edge length of the cell is:

Options:
A 200 pm
B 32Â pm
C 142.2Â pm
D 2Â pm

The correct answer is:
200 pm

Solution:
Also closest appraoch in bcc lattice is 12 of body diagonal,
i.e.32a
Or a=1.73Ã23=1.996 â«=199.6 pm

Topic: Chemistry, Solid State, NEET