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Physics NEET MCQ
Quiz 15
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Q.1
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.2
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.3
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.4
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.5
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.6
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.7
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.8
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.9
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.10
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.11
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.12
) 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.13
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.14
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.15
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.16
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.17
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.18
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.19
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.20
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.21
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.22
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.23
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.24
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)
Q.25
Two rotating bodies A and B of masses m and 2mwith moments of inertia IA and IB (IB > IA) have equal kinetic energy of rotation. If LA and LB be their angular momenta respectively, then … [NEET II -2016] a) b) c) d)
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a) LB > LA
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b) LA > LB
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c) LA = LB/2
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d) LA = 2LB
Explanation
Since IB > IA ∴ LB > LA Answer:(a)
Q.26
A solid sphere of mass m and radius R is rotating about its diameter. A solid cylinder of the same mass and same radius is also rotating about its geometrical axis with an angular speed twice that of the sphere. The ratio of their kinetic energies of rotation (Esphere / Ecylinder) will be …[NEET II – 2016]
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a) 1 : 4
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b) 3 : 1
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c) 2 : 3
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d) 1 : 5
Explanation
Answer:(d)
Q.27
A light rod of length l has two masses m1 and m2 attached to its two ends. The moment of inertia ofthe system about an axis perpendicular to the rodand passing through the centre of mass is .. [ NEET II -2016]
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a)
0%
b)
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c)
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d)
Explanation
m1x = m2(l-x) m1x = m2l - m2x Similarly Answer:(c)
Q.28
One end of string of length l is connected to a particle of mass ‘m’ and the other end is connected to a small peg on a smooth horizontal table. If the particle moves in circle with speed ‘v’, the net force on the particle (directed towards center) will be(T represents the tension in the string)
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a) T
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b)
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c)
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d) zero
Explanation
Centripetal force is provided by tension Answer:(a)
Q.29
Two discs of same moment of inertia rotating about their regular axis passing through centre and perpendicular to the plane of disc with angular velocities ω1 and ωThey are brought into contact face to face coinciding the axis of rotation. The expression for loss of energy during this process is … [ NEET 2017]
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a)
0%
b)
0%
c)
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d)
Explanation
According to law of conservation of angular momentum Iω1 + Iω2 = 2Iω ω = (ω1 + ω2)/2 Initial kinetic energy Loss in energy Answer:(b)
Q.30
A rope is wound around a hollow cylinder of mass3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N? [ NEET 2017]
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a) 25 m/s²
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b) 0.25 rad/s²
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c) 25 rad/s²
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d) 5 m/s²
Explanation
τ=Iα F×R=MR2α 30 × 0.4 = 3 (0.4)2α α = 25 rad/s2 Answer:(c)
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