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Physics NEET MCQ
Centre Of Mass And Rotational Motion. Mcq
Quiz 15
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Q.1
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.2
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.3
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.4
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.5
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.6
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.7
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.8
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.9
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.10
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.11
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.12
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.13
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.14
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.15
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.16
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.17
) 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.18
What is the minimum velocity with which a body of mass m must enter a vertical loop of radius R so that it can complete the loop ? [ AIPMT 2016]
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a) √(gR)
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b) √(2gR)
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c) √(3gR)
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d) √(5gR)
Explanation
if the body has minimum velocity of √(5gr) at the lowest point of vertical circle, it will complete the circle. Answer:(d)
Q.19
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)
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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.20
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.21
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)
Q.22
An automobile moves on a road with a speed of54 kmh-The radius of its wheels is 0.45 m and the moment of inertia of the wheel about its axis of rotation is 3kgmIf the vehicle is brought to rest in 15s, the magnitude of average torque transmitted by its brakes to wheel is :-
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a) 2.86 kg m2s-2
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b) 6.66 kg m2s-2
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c) 8.58 kg m2s-2
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d) 10.86 kg m2s-2
Explanation
54kmh-1 = 15m/s v= ωr ∴ω= v /r = 15/0.45 τ = Iα Answer:(b)
Q.23
Point masses m1 and m2 are placed at the opposite ends of a rigid rod of length L, and negligible mass. The rod is to be set rotating about an axis perpendicular to it. The position of point P on this rod through which the axis should pass so that the work required to set the rod rotating with angular velocity ω0 is minimum is given by … [ Re AIPMT 2015]
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a)
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b)
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c)
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d)
Explanation
The position of point P on the rod through which the axis should pass so that the work required to set the rod rotating with minimum angular velocity ωo is their centre of mass m1x = m2(L-x) It can be proved as follows τ ∝ I as in given problem minimum angular velocity ωo For torque to be minimum, we can differentiate it with x, as x is variable and we want to find it Answer:(a)
Q.24
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.25
Two stones of masses m and 2 m are whirled in horizontal circles, the heavier one in a radius r/2andthe lighter one in radius r. The tangential speed of lighter stone is n times that of the value of heavier stone when they experience same centripetal forces. The value of n is : … [ Re AMPT 2015]
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a) 1
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b) 2
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c) 3
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d) 4
Explanation
Same centripetal force V and v’ is tangential velocity and given v = nv, n =2 Answer:(b)
Q.26
From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre?
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a)
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b)
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c)
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d)
Explanation
Disc Moment of inertia =I Momentum of remaining part = I1 Moment of inertia of small disc w.r.t centre of big disc = I2 Now I = I1 + I2; I1 = I – I2 Answer:(b)
Q.27
A uniform circular disc of radius 50 cm at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of 2.0 rad s–Its net acceleration in ms–2 at the end of 2.0 s is approximately :
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a) 8.0
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b) 7.0
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c) 6.0
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d) 3.0
Explanation
We will assume that particle is at the circumference of disc ar is the radial acceleration or centripetal acceleration aT is the tangential acceleration ar = 0.5×16= 8 aT= 2×0.5 =1 Answer:(a)
Q.28
A disk and a sphere of same radius but different masses roll off on two inclined planes of the same altitude and length. Which one of the two objects gets to the bottom of the plane first ?
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a) Disk
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b) Sphere
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c) Both reach at the same time
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d) Depends on their masses
Explanation
More the acceleration lesser is the time Thus more is K2/R2 less is the acceleration For disc K2/R2 = 0.5 For sphere K2/R2 = 0.4 As acceleration of sphere is more than disc, sphere will reach bottom first Answer:(b)
Q.29
A particle of mass 10 g moves along a circle of radius 6.4 cm with a constant tangential acceleration. What is the magnitude of this acceleration if the kinetic energy of the particle becomes equal to 8 × 10–4 J by the end of the second revolution after the beginning of the motion?
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d) 0.2 m/s2
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a) 0.1 m/s2
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b) 0.15 m/s2
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c) 0.18 m/s2
Explanation
v2 = u2+ 2ats (s = 4πR as two revolution) 0.4 = 0 +2 at (4π×6.4×10-2) Answer:(a)
Q.30
In the given figure, a = 15 m/s2 represents the total acceleration of a particle moving in the clockwise direction in a circle of radius R = 2.5 m at a given instant of time. The speed of the particle is … [ NEET II – 2016]
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a) 5.7 m/s
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b) 6.2 m/s
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c) 4.5 m/s
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d) 5.0 m/s
Explanation
Acceleration towards centre = acos30 v2=aRcos30 v= 5.7 m/s Answer:(a)
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