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Physics NEET MCQ
Centre Of Mass And Rotational Motion. Mcq
Quiz 7
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Q.1
A mass m rotating freely in horizontal circle of radius 1m on frictionless smooth table supports a stationary mass 2m, attached to the other end of the string passing through smooth hole O in the table, hanging vertically. Find the angular velocity of rotation
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a)√(2g)
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b)√(g)
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c)√(3g)
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d)√(4g)
Explanation
Tension in the string T=2mg Centrifugal force on the mass m=mω2 r Thus to continue rotational motion of mass m Tension=centrifugal force 2mg=mω2 r ω=√(2g/r) but r=1 m Thus ω=√(2g) Answer:(a)
Q.2
Three particles of the same mass lie in the (X, Y) plane, The (X, Y) coordinates of their positions are (1, 1), (2, 2) and (3, 3) respectively. The (X,Y) coordinates of the centre of mass are
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a) (1, 2)
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b) (2, 2)
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c) (1.5, 2)
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d) (2, 1.5)
Explanation
Answer: (b)
Q.3
A Pulley of radius 2 m is rotated about its axis by a force F = (20 t - 5t2 ) N where t is in sec applied tangentially. If the moment of inertia of the Pulley about its axis of rotation is 10 Kgm2 , the number of rotations made by the pulley before its direction of motion is reversed is
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a) more than 3 but less then 6
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b) more than 6 but less then 9
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c) more than 9
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d) Less then 3
Explanation
Here direction of Motion will be reversed when force F = 0 or 20 t – 5t2 = 0 or t = 4sec. If α is angular acceleration then torque τ = Iα = F.r OR 10 × α = (20 t – 5t2 ) × 2 OR α = 4t - t2 2π = 44/7 rad = 1 rotation Therefore 64/3 rad = 3.4 rotation option "a" is correct Answer: (a)
Q.4
A cubical block of side a is moving with velocity V on a horizontal smooth plane as shown in figure. It hits a ridge at point O. The angular speed of the block after it hits O is ...
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a) 3V/4a
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b) 3V/2a
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c) √(3V)/ √(2a)
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d) zero
Explanation
r = √2 ( a/2) or r2 = a2/2 Li = MV × (a/2) Final momentum = ( momentum of inertia about CM. + Mr2)ω For cube Moment of inertia about C.M = Ma2 /6 Net torque about O is zero Initial angular momentum = momentum × perpendicular distance of velocity vector from O ∴ angular momentum (L) about O will be conserved or Li = Lf MV (a/2) = ( Icm + Mr2)ω Answer: (a)
Q.5
Two uniform rod of equal length but different masses are rigidly joined to form L shaped body which is then pivoted as shown. If in equilibrium the body is in the shown configuration ratio M/m will be ..
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a) 2
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b) 3
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c) √2
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d) √3
Explanation
Gravitational force mg and Mg is acting done ward direction Cenetre of mass of each arm is at (l/2). Thus perpendicular distance between fix point O and line of action of force is (l/2)sin60 and (l/2)sin30. Net torque about O should be zero Thus Answer:(d)
Q.6
A light rod carries three equal masses A, B and C as shown in the figure the velocity of B in vertical position of rod if it is released from horizontal position as shown in the figure is ...
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a)
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b)
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c)
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d)
Explanation
Loss in PE = gain in angular kinetic energy Let I be the moment of inertia at fixed point Loss in PE = E Answer: (d)
Q.7
Two identical hollow spheres of mass M and radius R are joined together and the combination is rotated about an axis tangential to one sphere and perpendicular to the line connecting their centres. The moment of inertia of the combination is ________.
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a)10 MR2
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b) (4/3) MR2
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c)(32/3) MR2
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d)(34/3) MR2
Explanation
Answer: (d)
Q.8
A force F = αi + 3j + 6k is acting at a point r=2i-6j-12k. The value of α for which angular momentum about origin is conserved is
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a) 1
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b) -1
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c) 2
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d) zero
Explanation
Angular momentum gets conserved if no external torque act on it Thus r × F=0 Thus α = 1 Answer:(a)
Q.9
Two point masses of 0.3 kg and 0.7 kg are fixed at the ends of a rod of length 1.4 m and of negligible mass. The rod is set rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that the work required for rotation of the rod is minimum, is located at a distance of …..
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a) 0.4 m from mass of 0.3 kg
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b) 0.98 m from mass of 0.3 kg
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c)0.7 m from mass of 0.7 kg
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d)0.98 m from mass of 0.7 kg
Explanation
let x be the distance from 0.3 kg mass I=0.3x2 + 0.7 (1.4 – x)2For minimum work moment of inertia of the system should be minimum is dI/dt=0dI/dt=0.3 × (2x) – (0.7) × 2 × (1.4 – x) 0.3 ×(2x) – (0.7) × 2 × (1.4 – x)=0 0.6x - 1.96 + 1.4x=0 2x=1.96x=0.98 m from mass 0.3kg Answer: (b)
Q.10
A smooth sphere A is moving on a friction-less horizontal plane with angular speed ωand centre of mass velocity v. It collides elastically and head on with an identical sphere B at rest. Neglect friction everywhere. After the collision, their angular speeds are ωA and ωB respectively, Then
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a) ωA < ωB
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b) ωA=ωB
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c) ωA=ω
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d) ω=ωB
Explanation
As it is head-on elastic collision between two identical balls there fore they will exchange their linear velocity is A comes to rest and B starts moving with linear velocity V.As there is no friction any where, torque on both the spheres about their centre of mass is zero and their angular velocities remains unchanged.Therefore ωA=ω and ωB=0 Answer:(c)
Q.11
Consider a body as shown in figure, consisting of two identical balls, each of mass M connected by a light rigid rod. If an impulse J=MV is imparted to the body at one of its ends, what would be its angular velocity. What is V ?
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a) V / L
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b) 2V / L
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c) V / 3L
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d) V / 4L
Explanation
Given system of two particles will rotate about its centre of mass. Initial angular momentum=momentum × Perpendicular distance between direction of momentum and centre of mass=mV(L/2) Final angular momentum=2Iω=2M(L/2)2 ω By law of conservation of angular momentum Answer: (a)
Q.12
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.
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a)
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b)
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c)
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d)
Explanation
From law of conservation of energy mgh=½Mv2 + ½ I ω2 Use following formula to find ω V=ω × R and For cylinder I=½ MR2Answer: (c)
Q.13
Two disc of same thickness but of different radii are made of two different materials such thattheir masses are same. The densities of the materials are in the ratio 1:The moment of inertiaof these disc about the respective axes passing through their centres and perpendicular to theirplanes will be in the ratio.
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a) 1 : 3
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b) 3 : 1
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c)1 : 9
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d)9 : 1
Explanation
Moment of inertia of disc I=½ MR2 Given that masses are same, and thickness is same , if ρ is mass per unit area Now mass=π R2 ρ Thus R2 ∝ (1/ρ) And I ∝ R2 thus I ∝ 1/ρ Thus correct option is "b" Answer: (b)
Q.14
Identify the correct statement for the rotational motion of a rigid body
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a) Individual particles of the body do not undergo accelerated motion
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b) The centre of mass of the body remains unchanged.
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c)The centre of mass of the body moves uniformly in a circular path
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d) Individual particle and centre of mass of the body undergo an accelerated motion
Explanation
Answer:(b)
Q.15
A cylinder of mass 5 kg and radius 30 cm, and free to rotate about its axis, receives an angularimpulse of 3 kg m2s-1 initially followed by a similar impulse after every 4 sec. what is the angularspeed of the cylinder 30 sec after initial impulse ? The cylinder is at rest initially.
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a) 106.7 rad s-1
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b) 206.7 rad s-1
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c) 7.6 rad s-1
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d) 07.6 rad s-1
Explanation
Initial angular momentum=0angular momentum after initial impulse=3 kg m2s-1angular momentum after initial 4 sec=3 + 3=6kg m2s-1 angular momentum after initial 8 sec=6 + 3=9kg m2s-1 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - angular momentum after initial 28 sec=24 kg m2s-1angular momentum after initial 30 sec=24 kg m2s-1 Iω=24 here I=½ MR2 I=½ × 5 × (0.3) 2=0.225 kgm2 ∴ ω=24/I ω=24/0.225=106.7 rad s-1Answer: (a)
Q.16
A uniform rod of length L is suspended from one end such that it is free to rotate about anax is passing through that end and perpendicular to the length, what maximum angular speed must be imparted to the lower end so that the rod completes one full revolution?
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a)√(g/l)
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b) √(2g/l)
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c)√(6g/l)
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d)2√(g/l)
Explanation
we should impart kinetic energy sufficient to move centre of mass from lower side to upper side thus change in potential energy=mgLKinetic energy imparted=Increase in PE ½ I ω2=mgL Answer: (c)
Q.17
Consider a two-particle system with the particles having masses M1 , and M2 . If the firstparticle is pushed towards the centre of mass through a distance d, by what distance shouldthe second particle be moved so as to keep the centre of mass at the same position?
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a)
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b)
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c)
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d)
Explanation
From figure M1 × x1=M2 × x2 When M1 is moved to wards centre , then let us consider M2 is moved by d' towards centre such that centre of mass remain constant M1 ×( x1 - d)=M2 × (x2 - d') Thus M1d=M2d' d'=(M1d )/M2Answer: (d)
Q.18
The moment of inertia of a thin rod of mass M and length L about an axis passing through thepoint at a distance L/4 from one of its ends and perpendicular to the rod is _____
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a)(7ML2)/48
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b)(ML2)/ 12
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c)(ML2)/ 9
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d)(ML2)/ 3
Explanation
Here L/4 is the distance of axis of rotation from end Use parallel axis theorem to find solution Answer:(a)
Q.19
The height of a solid cylinder is four times that of its radius. It is kept vertically at time t=0 on a belt which is moving in the horizontal direction with a velocity v=2.45t2 where v in m/sand t is in second. If the cylinder does not slip, it will topple over a time t=____
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a) 1 second
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b) 2 sec.
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c) 3 sec.
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d) 4 sec.
Explanation
The cylinder will topple when the torque mgr equals the torque ma(h/2) Thus mgr=ma(h/2) mgr=ma( 4r/2) g=2a now a=dv/dt=(2) × (2.45)t g=2 ×(2) × (2.45)t t=1 sec Answer: (a)
Q.20
A uniform rod of length 2L is placed with one end in contact with horizontal and is then inclined at an angle α to the horizontal and allowed to fall without slipping at contact point. When it becomes horizontal, its angular velocity will be…..
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a)
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b)
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c)
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d)
Explanation
By Conservation of Energy P.E. of rod=Rotational K.E.Length of the rod is 2L Centre of mass is at the mid point(L) of the rod thus it will fall by the distance of Lsinα Moment of inertia of the rod axis of of rotation is at the end I=M (2L)2 / 3 Answer: (a)
Q.21
In a bicycle the radius of rear wheel is twice the radius of front wheel. If Rf and Rr are the radius, Vf and Vr are speed of top most points of wheel respectively then...
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a) Vr=2Vf
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b) Vf=2Vr
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c)Vf=Vr
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d)Vf > Vr
Explanation
Angular speed for both wheels are different but linear speed for both same so Vf=Vr Answer: (c)
Q.22
The M.I. of a body about the given axis is 1.2 kgm2 initially the body is at rest. In order to produce a rotational kinetic energy of 1500 J. an angular acceleration of 25rad sec-2 must be applied about that axis for duration of ….
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a) 4 sec
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b) 2 sec
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c)8 sec
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d)10 sec
Explanation
Rotational K.E.=½ Iω2 1500=½ ×1.2 × ω2ω=50 rad/sec and ω=ωo + αt 50=0 + 25 × t t=2 sec Answer:(b)
Q.23
Let I be the moment of inertia of a uniform square plate about an axis AB that passes through its centre and is parallel to two of its sides. CD is a line in the plane of the plate that passes through the centre of the plate and makes an angle of θ with AB. The moment of inertia of the plate about the axis CD is then equal to ...
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a) I
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b) I sin2θ
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c) I cos2θ
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d) I cos2(θ/2)
Explanation
Let Iz be the moment of inertia about an axis passing perpendicularly through plane of plate hence according to Perpendicular axis theorem.Iz=IAB + IA'B' Iz=ICD + IC'D' As axis are symmetric IAB=IA'B'=Iz / 2 ICD=IC'D'=Iz / 2 So we can say that IAB=IA'B'=ICD=IC'D'=IAnswer: (a)
Q.24
An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to each other. The first part of mass 1 kg moves with a speed of 12 ms-1 and the second part of mass 2 kg moves with 8 ms-1 speed. If the third part flies off with 4 ms–1 speed, then its mass is … [ NEET 2013]
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a) 3 kg
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b) 5 kg
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c) 7 kg
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d) 17 kg
Explanation
Given that two parts moves right angles with each other. Let first part of 1kg move along x-axis with 12 ms-1 velocity vector is (12i )m/s Let second part of 2kg move along y-axis with velocity 8ms-1 velocity vector is (8j)m/s Let vv be the velocity vector of third part of mass m, having velocity 4m/s Now according to law of conservation of momentum Momentum before explosion = momentum after explosion Momentum = mass × velocity 20= m×4 ∴ m = 5 kg Answer:(b)
Q.25
A small object of uniform density rolls up a curvedsurface with an initial velocity v. It reaches up to amaximum height of with respect to the initialposition. The object is …. [ NEET 2013]
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a) Ring
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b) Solid sphere
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c) Hollow sphere
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d) Disc
Explanation
We can also think of a small object with zero initial zero velocity rolled down from height and attend the speed of v at the bottom. Now from the formula for velocity of object at bottom By substituting value of h in above equation K is radius of gyration which for disc Answer:(d)
Q.26
) A solid cylinder of mass 50 kg and radius0.5mis free to rotate about the horizontal axis. A massless string it 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 is … [ AIPMT 2014]
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a) 78.5 N
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b) 157 N
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c) 25 N
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d) 50 N
Explanation
Torque = T×R and Torque = Iα Answer:(c)
Q.27
The ratio of the accelerations for a solid sphere(mass 'm' and radius 'R') rolling down an incline of angle 'θ'with out slipping and slipping down the incline without rolling is
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a) 2 : 5
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b) 7 : 5
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c) 5 : 7
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d) 2 : 3
Explanation
Acceleration of slipping down = gsinθ Acceleration without slipping = Answer:(c)
Q.28
A mass m moves in a circle on a smooth horizontal plane with velocity v0 at a radius RThe mass is attached to a string which passes through a smooth hole in the plane as shown. The tension in the string is increased gradually and finally m moves in a circle of radius R0/The final value of the kinetic energy
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a)
0%
b)
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c)
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d)
Explanation
No external torque acts on the system thus according to law of conservation of angular momentum V=2v New kinetic energy Answer:(d)
Q.29
Three identical spherical shells, each of mass m and radius r are placed as shown in figure. Consider an axis XX’ which is touching to two shells and passing through diameter of third shell. Moment of inertia of the system consisting of these three spherical shells about XX’ axis is :
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a) 4mr2
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b)
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c) 3mr2
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d)
Explanation
Momentum of inertia of ‘a’ Momentum of inertia of ‘b’ Momentum of inertia of ‘c’ Total momentum =Ia + Ib + Ic Answer:(a)
Q.30
Two spherical bodies of mass M and 5 M and radii R and 2 R are released in free space with initial separation between their centres equal to 12 R. If they attract each other due to gravitational force only, then the distance covered by the smaller body before collision is :
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a) 1.5 R
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b) 2.5 R
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c) 4.5 R
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d) 7.5 R
Explanation
Center of mass before collision let it be at (0,0) thus Mr = 5Mr’ r = 5r’ ratio is 5:1 for M:5M Thus centre of mass of m is at 10R from the centre of mass of system After collision same ratio will be maintained as no external force is applied. Now new distance between the centers is 3R Thus r=2.5R and r’=0.5r Thus cenre of mass of m changed from position of 10R to 2.5R Thus distance travelled = 10R-2.5R =7.5R Answer:(d)
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