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Physics NEET MCQ
Quiz 12
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Q.1
A man is sitting with folded hands on a revolving table. Suddenly, he stretches his arms. Angular speed of the table would [ CPMT 1996]
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a) increase
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b) decrease
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c)remain the same
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d)nothing can be said
Explanation
On stretching arms, distance K increases I=MK2 increases. As Iω=constant, therefore, angular velocity ω decreases. Answer:(b)
Q.2
When a body rolls down an inclined plane, its potential energy is converted into [ CBSE 1996]
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a) translational K.E only
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b) rotational K.E only
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c) translational and rotational K.E
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d) none of above
Explanation
On rolling, the body has both translational as well as rotational motion. Therefore, potential energy of the body is converted into translational and rotational energy Answer: (c)
Q.3
A raw egg and a hard boiled egg are made to spin on a table with the same angular speed about the same axis. The ratio of the time taken by the two to stop is [ CPMT 1996]
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a) =1
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b) < 1
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c) > 1
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d) none of the above
Explanation
A raw egg has some fluid in it so on spinning the fluid is thrown outwards. Where as boiled egg is solid from inside. Therefore Ir > Ib As I× ω = constant ∴ ωr < ωb Thus ωr / ωb < 1 ∴ tr / tb < 1 Answer: (b)
Q.4
A couple produces a ... [ pre-medical dental, 1997]
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a) pure linear motion
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b) pure rotational motion
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c) both linear and rotational motion
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d) no motion
Explanation
A couple has turning effect and produce pure rotational motion Answer: (b)
Q.5
If sphere is rolling , the ratio of the rotational energy to the total kinetic energy is given by [ MEE BHU 1997]
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a) 7:10
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b) 2:5
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c) 10:7
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d) 2:7
Explanation
Total kinetic energy = KE. linear + K.E. Rotational Total K.E. E= ½ mv2 + ½ I ω2 For sphere I = (2/5) mr2 and v = ω r On taking the ratio of rotational kinetic energy to total kinetic energy we get Answer:(d)
Q.6
The moment of inertia of a solid sphere and a spherical shell of equal masses about their diameters are equal. the ration of their radii are .. [ CBSE 1997]
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a) 5:3
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b) 3:5
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c) √5 : √3
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d) √3 : √5
Explanation
Moment of Inertia of solid sphere I1= (2/5)m r21 Moment of Inertia of Hollow sphere I2= (2/3) m r22 Given I1 = I2 ∴ (2/5) m r21 = (2/3) m r22 ∴ r1 / r2 = √5 / √ 3 Answer: (c)
Q.7
A body rolls down an inclined plane. If its kinetic energy of rotational motion is 40% of its kinetic energy of translational motion, then the body is a
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a)ring
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b) cylinder
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c)spherical shell
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d)solid sphere
Explanation
In case of solid sphere K.E=½ Iω2 but I=(2/5) m r2Substituting value of I in equation for energy we get K.E=(1/5) mv2 which is 40% of kinetic energy ½m v2Answer: (d)
Q.8
The least coefficient of friction for an inclined plane at an angle α with horizontal, in order that a solid cylinder will roll down it without slipping is [ CET 1997 punjab]
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a) 2/3( tan α)
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b) 2/7 (tan α)
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c)1/3 (tanα)
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d)4/3( tan α)
Explanation
Answer: (c)
Q.9
A mass m is moving with a constant velocity along a line parallel to the x-axis away from the origin. Its angular momentum with respect to the origin. Its angular momentum with respect to the origin is [ IIT 1997]
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a) zero
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b) constant
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c)goes on decreasing
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d)goes on increasing
Explanation
Angular momentum of a body about a given axis is the product of linear momentum and perpendicular distance of line of action of linear momentum vector from the axis of rotation.in given problem object is moving parallel to x-axis away from origin thus perpendicular distance will be along y-axis and will not change thus r is constant Given that the object moves with constant velocity thus momentum is also constant Angular momentum=momentum × perpendicular distance between line of action of momentum Since momentum and perpendicular distance is constant Angular momentum is constant Answer:(b)
Q.10
Three identical metal balls each of radius r are placed touching each other on a horizontal surface such that an equilateral triangle is formed when such centres of three balls are joined. The centre of mass of system is located at [CBSEPMT 1995]
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a) horizontal surface
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b) centre of one of the balls
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c) line joining centre of any two balls
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d) point of intersection of medians
Explanation
The centre of mass of three identical balls will be at the point of intersection of medians of the triangle. Answer: (d)
Q.11
The moment of inertia of disc of mass M and radius R about an axis, which is tangential to circumference of disc and parallel to its diameter [ CBSE PMT 1999]
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a)(3/2) MR2
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b) (2/5) MR2
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c)(5/4) MR2
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d)(4/4) MR2
Explanation
Moment of inertia of disc about its diameter=(1/4)MR2Using theorem of parallel axes I=(1/4)MR2 + MR2 I=(5/4) MR2Answer: (c)
Q.12
The radius of gyration of a uniform rod of length L about on axis passing through its centre of mass [AMU 1999]
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a) L/ √12
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b) L2 / 12
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c)L/√3
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d)L/√2
Explanation
moment of inertia for uniform rod of length L=(1/12)M L2 Monet of inertia in terms of radius of gyration I=MK2 On comparing above equations we get Radius of gyration K=L / √12 Answer: (a)
Q.13
A wheel is rotating at 900 r.p.m about its axis. When power is cut off it comes to rest in 1 minute. The angular retardation is rad/s2 [ MP CEE 1999]
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a) π/2
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b) π/4
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c)π/6
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d)π/8
Explanation
Initial angular velocity 900 r.p.m One rotation=2π Thus 900 rotation per minute=900 × 2πInitial angular velocity per second ω=900 × 2π / 60=30π Final angular velocity=0Angular retardation=( Final angular velocity - Initial angular velocity) / 60 Angular retardation=π/2 Answer:(a)
Q.14
A solid cylinder rolls down a smooth inclined plane 4.8m high without slipping what is itslinear speed at the bottom of the plane, if it starts rolling from the top of the plane?(take g=10 m/s2)
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a) 4 m/s
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b) 2 m/s
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c) 10 m/s
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d) 8 m/s
Explanation
Linear velocity for solid cylinder K=R/√2 Answer: (d)
Q.15
A cylindrical solid of mass M has radius R and length L. Its moment of inertia about a generator is [ MP PMT 2009]
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a)
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b)
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c)½ MR2
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d)(3/2) MR2
Explanation
Generator is axis touching surface of cylinder and parallel to axis of cylinder. Using theorem of parallel axes I=I0 + MR2 I=½ MR2 + MR2I=(3/2) MR2Answer: (d)
Q.16
An athlete throws a discus from rest to a final angular velocity of 15rad/s in 0.27s before releasing it. During acceleration, discuss moves in a circular arc of radius 0.81m. Acceleration of discuss before it is released is .... ms2 [ kerala PET 2004]
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a)45
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b) 182
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c)187
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d)192
Explanation
Here Initial angular velocity ω1=0 and final angular velocity ω2=15rad/sec time t=0.27 s and radius r=0.81m Angular acceleration=Change in angular velocity / time α=(ω2 - ω1 ) /t α =( 15 - 0) / 0.27=55.55 formula for tangential acceleration=r × angular accelerationa=0.81 × 55.55=45 m/s2Answer: (a)
Q.17
Three identical spheres, each of mass 1 kg are kept as shown in figure, touching each other, with their centre on a straight line. If their centre are marked P,Q,R respectively, the distance of centre of mass of the system from P is [ kerala C.E.T 2005]
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a) (PQ + PR + QR) / 3
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b) (PQ + PR) / 3
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c)(PQ + QR) / 3
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d)(PQ + QR + PR) / 3
Explanation
As is clear from symmetry of figure, centre of mass of the system is at Q. Its distance from P is Answer: (b)
Q.18
A solid cylinder of mass 20kg has length 1m and radius 0.2m. Then its moment of inertia in kgm2 about its geometrical axis is [ kerala PMT 2005]
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a)0.8
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b) 0.4
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c)0.2
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d)20.4
Explanation
Here m=20kg, l=1m , r=0.2 m momentum of inertia about its geometrical axis is I=½ mr2 I=½ × 20 × (0.2)2 =0.4kg m2 Answer:(b)
Q.19
A given shaped glass tube having uniform cross section is filed with water and is mounted on a rotating shaft as shown in figure. If the tube is rotated with constant angular velocity ω, then [ AIIMS 2005]
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a) water levels in both section A and B goes up
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b) water level in section A goes up and then in section B comes down
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c) Water level in section A comes down and then in B goes up
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d) water level remains the same in both the sections
Explanation
When the tube is rotated with constant angular velocity, centrifugal force acting radially outwards towards A=mLω2 and towards B m(2L)ω2. Therefore, water level in both section A and section B goes up Answer: (a)
Q.20
A constant torque is acting on a uniform circular wheel changes its angular momentum from A0 to 4A0 in 4 seconds. The magnitude of this torque is [ BHU 1998]
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a)(3/4)A0
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b) A0
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c)4A0
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d)12A0
Explanation
τ= dL /dt=(4A0 - A0) / 4τ=(3/4)A0Answer: (a)
Q.21
An inclined plane makes an angle of 30° with the horizontal. A solid sphere rolling down this inclined plane from rest without slipping has a linear acceleration equal to [ MPPMT 1987]
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a) g/3
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b) 2g/3
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c)5g/7
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d)5g/14
Explanation
Formula for acceleration along the inclined plane is For solid sphere I=(2/5)mr2 and θ=30°on substituting the values we get Answer: (d)
Q.22
The moment of inertia of a circular ring about an axis passing through its centre and normal to its plane passing through its centre and normal to its plane is 200gm/cmThen its moment of inertia about a diameter is [ MPPMT 1987]
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a) 40 gm cm2
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b) 300 gm cm2
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c)200 gm cm2
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d)100 gm cm2
Explanation
We know that I diameter =centre / 2 ∴ I diameter =200/2=100 gm cm2 Answer:(d)
Q.23
Moment of momentum is called
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a) torque
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b) weight
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ac) moment of inertia
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d) angular momentum
Explanation
Answer: (d)
Q.24
A dancer on ice spins faster when she folds her hand. This is due to [ CPMT 1986]
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a)increase in energy and increase in angular momentum
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b) decrease in friction at the skates
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c)constant angular momentum and increase in kinetic energy
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d)increase in energy and decrease in angular momentum
Explanation
No external torque acts on the dancer. But reduces the moment of inertia thus increased angular acceleration τ=I α Thus because of increased angular acceleration, increased energyAnswer: (c)
Q.25
The rotational K.E. of a body is E and its moment of inertia id I. The angular momentum is [ MPPMT 1993]
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a) EI
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b) 2 √(EI)
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c) √(2EI)
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d)E/I
Explanation
Rotational kinetic energy E=½ Iω2 Thus ω=√(2E/I) Angular Momentum L=I× ω L=I × √(2EI)=√(2EI)Answer: (c)
Q.26
A homogeneous disc of mass 2kg and radius 15cm is rotating about its axis ( which is fixed) with an angular velocity of 4 radian/sec with an angular velocity of 4 radian/sec. The linear momentum of the disc is ..[ CPMT 1989]
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a) 1.2 kg m/s
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b) 1.0 kg m/s
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c)0.6 kg m/s
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d)none of the above
Explanation
Disc does not have translatory motion so linear momentum is zero Answer:(d)
Q.27
The quantity which remains constant in conservative field if [ Raj.PET 1996]
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a)Potential energy
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b) Kinetic energy
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c)Angular momentum
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d)Linear momentum
Explanation
Answer: (c)
Q.28
A body is moving with a constant velocity, then which of the following statement is correct :
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a) the body necessarily a constant angular momentum
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b) the body has necessarily a constant moment of inertia
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c)the body has necessarily a constant angular speed and moment of inertia
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d)the angular speed and moment of inertia may vary but their product is constant
Explanation
Answer: (d)
Q.29
A meter stick is held vertically with one end on the floor and is then allowed to fall. The speed of the other end when it hits the floor assuming that the end at floor does not slip is [ g=9.8 m/s2] [ JIPMER 1998]
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a) 3.2 m/s
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b) 5.4 m/s
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c) 7.6 m/s
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d) 9.2 m/s
Explanation
When stick falls, the potential energy changes to kinetic energy of rotation centre of mass is at l/2 and A is axis of rotation Moment of inertia of rod at point of rotation A = ml2/3 ( we can calculate it using theorem of parallel axis)∴ mg(l/2)=½ I ω2 ∴ ω=√(3g/l)linear velocity v=ω × l V=√(3gl) v=√(3 × 9.8 × 1)=5.4 m/sec Answer:(b)
Q.30
A rod pQ of mass M and length L is hinged at one end O. The rod is kept in horizontal position by a massless string tied to point Q as shown in figure. When the string is cut the initial angular acceleration of the rod is
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a) (g/L)
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b) 2(g/L)
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c) (2/3)(g/L)
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d) (3/2)(g/L)
Explanation
centre of mass of rod is at l/2 from p and moment of inertia about P is ML2 /3 Taking angular moment about P Torque by definition=Force × perpendicular distance between line of action of force and centre τ=Mg(L/2) --eq(1) Also τ=Iα --eq(2) from equation 1 and 2 we get Mg(L/2)=Iα on substituting value of I in above equation we get Mg(L/2)=(ML2/3)α α=(3/2)(g/L) Answer: (d)
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