MCQGeeks
0 : 0 : 1
CBSE
JEE
NTSE
NEET
English
UK Quiz
Quiz
Driving Test
Practice
Games
NEET
Physics NEET MCQ
Centre Of Mass And Rotational Motion. Mcq
Quiz 5
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Q.1
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]
0%
a) (PQ + PR + QR) / 3
0%
b) (PQ + PR) / 3
0%
c)(PQ + QR) / 3
0%
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.2
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]
0%
a)0.8
0%
b) 0.4
0%
c)0.2
0%
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.3
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]
0%
a) water levels in both section A and B goes up
0%
b) water level in section A goes up and then in section B comes down
0%
c) Water level in section A comes down and then in B goes up
0%
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.4
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]
0%
a)(3/4)A0
0%
b) A0
0%
c)4A0
0%
d)12A0
Explanation
τ= dL /dt=(4A0 - A0) / 4τ=(3/4)A0Answer: (a)
Q.5
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]
0%
a) g/3
0%
b) 2g/3
0%
c)5g/7
0%
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.6
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]
0%
a) 40 gm cm2
0%
b) 300 gm cm2
0%
c)200 gm cm2
0%
d)100 gm cm2
Explanation
We know that I diameter =centre / 2 ∴ I diameter =200/2=100 gm cm2 Answer:(d)
Q.7
Moment of momentum is called
0%
a) torque
0%
b) weight
0%
ac) moment of inertia
0%
d) angular momentum
Explanation
Answer: (d)
Q.8
A dancer on ice spins faster when she folds her hand. This is due to [ CPMT 1986]
0%
a)increase in energy and increase in angular momentum
0%
b) decrease in friction at the skates
0%
c)constant angular momentum and increase in kinetic energy
0%
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.9
The rotational K.E. of a body is E and its moment of inertia id I. The angular momentum is [ MPPMT 1993]
0%
a) EI
0%
b) 2 √(EI)
0%
c) √(2EI)
0%
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.10
A spherical solid ball of 1kg mass and radius 3cm is rotating about an axis passing through its centre with an angular velocity of 50 radian/sec. The kinetic energy of rotation is [ CPMT 1989]
0%
a) 4500J
0%
b) 90J
0%
c) 9×10-3 J
0%
d) (9/20) J
Explanation
Rotational kinetic energy E=½ I ω2 Here I=(2/5)mr2 m=1 kg r=3cm=3 × 10-2Thus I=(2/5) × 9 × 10-4 E=½ (2/5)× 9 × 10-4 × 502E=45 × 10-2=45/100E=(9/20) JAnswer: (d)
Q.11
The quantity which remains constant in conservative field if [ Raj.PET 1996]
0%
a)Potential energy
0%
b) Kinetic energy
0%
c)Angular momentum
0%
d)Linear momentum
Explanation
Answer: (c)
Q.12
A body is moving with a constant velocity, then which of the following statement is correct :
0%
a) the body necessarily a constant angular momentum
0%
b) the body has necessarily a constant moment of inertia
0%
c)the body has necessarily a constant angular speed and moment of inertia
0%
d)the angular speed and moment of inertia may vary but their product is constant
Explanation
Answer: (d)
Q.13
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]
0%
a) 3.2 m/s
0%
b) 5.4 m/s
0%
c) 7.6 m/s
0%
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.14
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
0%
a) (g/L)
0%
b) 2(g/L)
0%
c) (2/3)(g/L)
0%
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)
Q.15
Angular momentum of the body is conserved [ MPPET 1995]
0%
a) Always
0%
b) never
0%
c) in presence of external torque
0%
d) in absence of external torque
Explanation
Answer: (d)
Q.16
You are given two circular disc which have equal weight and equal thickness. they are made up of different metals having densities d1 and dthere radii are R1 and R2 respectively. For disc that will have more moment of inertia about the central axis than the other, the condition is [ MPPET 1993]
0%
a) d1 > d2
0%
b) R1 > R2
0%
c)d1> d2 and R1 > R2
0%
d)d1 < d2 and R1 < R2
Explanation
given that mass of both the disc is same thus I1 / I2=R1 / R2 ∴ If I1 > I2 , R1 >R2 Answer: (b)
Q.17
If the moment of inertia of disc about the tangent in its plane is I, its moment of inertia about the tangent perpendicular to the plane will be [ Raj. PET 1996]
0%
a)6I /5
0%
b)3I/4
0%
c)3I/2
0%
d)5I/4
Explanation
moment of inertia of the disc about tangent in its plane I=(5/4)MR2 MR2=4I/5 Moment of inertia of the disc in perpendicular plane to the disc=(3/2)MR2 Substituting value of MR2 we get Moment of inertia of the disc in perpendicular plane to the disc=6I/5 Answer:(a)
Q.18
The rotational and translation kinetic energy of rolling body are same, the body is [ RajPET 1996]
0%
a) Disc
0%
b) Sphere
0%
c) Cylinder
0%
d) Ring
Explanation
given ½ mv2=½ I ω2 angular velocity ω=v/R substituting in above equation we get ½ mv2=½ I ( v/R)2 I=MR2 above formula for moment of inertia is for ring Answer: (d)
Q.19
A body having moment of inertia about its axis of rotation equal to 3 kg m2 i rotating with angular velocity equal to 3rad/s. Kinetic energy of this rotating body is the same as that of body of mass 27kg moving with a speed of [ SCRA 1994]
0%
a)1.0 ms2
0%
b) 0.5 ms2
0%
c)1.5 ms2
0%
d)2.0 ms2
Explanation
given :Rotational kinetic energy=Translation kinetic energy of 27kg mass ∴ ½ I ω2=½ Mv2 ½ 3 × 32=½ 27 v2 ∴ v=1 Answer: (a)
Q.20
A cord is wound round the circumference of wheel of radius r. The axis of the wheel is horizontal and moment of inertia about it is I. A weight mg is attached to the end of the cord and falls from rest. After falling through a distance h, the angular velocity of the wheel will be [ MPPMT 1994]
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
Loss in potential energy of mass=Rotational kinetic energy of wheel + Translation energy of mass mgh=½I ω2 + ½ ½ mv2 we know that v=ω R 2mgh=Iω2 + mR2ω2 ∴ ω2 (I + mR2)=2mgh Answer: (b)
Q.21
A rigid body rotates about a fixed axis with variable angular velocity equal to ω=α - βt, at time t, α and β are constants. find the angle turned through by the body before coming to rest
0%
a)
0%
b)
0%
c)
0%
d)
Explanation
When body comes to rest ω=0 0=α - βt or time taken to stop is t=α/βNow dθ / dt=ω=α - βt dθ=(α - βt)dt ∴ Angle turned in this time is Answer:(a)
Q.22
Before jumping in water from above a swimmer bends his body to [ MPPMT 1994]
0%
a) increase moment of inertia
0%
b) decrease moment of inertia
0%
c) decrease the angular momentum
0%
d) reduce the angular velocity
Explanation
Answer: (b)
Q.23
A particle of mass m=5 is moving with a uniform speed v=3√2 in the XY plane along the straight line Y=X +The magnitude of the angular momentum about origin is
0%
a)zero
0%
b) 30 units
0%
c)75 units
0%
d)40√2 units
Explanation
Equation Y=X + 4 represent line with intercept on the y axis at 4 units According to formula L=mvr [ Here r is the perpendicular distance between the origin and line of action of moment of inertia as shown in figure by red line ] Since slope = 1 therefor tanθ=1 or θ= 45. From geometry of figure r=4sin45=4/√2 L=5 × (3/√2) × (4/√2) L=30 unitsAnswer: (b)
Q.24
A tap can be operated easily using two fingers because : [ UGET 1995]
0%
a) the force available for the operation will be more
0%
b) this helps application of angular force
0%
c)the rotational effect is caused by the couple formed
0%
d)the force formed by one finger overcomes friction and the other finger the force for the operation
Explanation
Answer: (c)
Q.25
Two solid spheres each of mass M and radius R/2 are attached to weightless rod of length 2R, the moment of inertia about the axis passing through the centre of one of the sphere and perpendicular to the length of the rod be [ raj PMT 1996]
0%
a)(21MR2) / 5
0%
b)(2MR2) / 5
0%
c)(5MR2) / 2
0%
d)(5MR2) / 21
Explanation
Formula for moment of inertia about axis of sphere=(2/5)Mr2given r=R/2 Thus Moment of inertia about YY"=(2/5) M (R/2)2 =MR2 / 10 Moment of inertia of the other sphere about the axis YY"=(1/10)MR2 + M(2R)2=(41/10) MR2 / 10 Thus total moment of inertia about YY'=(MR2) /10 + (41MR2) /10 total moment of inertia about YY'=(42MR2) / 10=(21MR2)/ 5 Answer:(a)
Q.26
Two masses M and m are attached to a vertical axis by weightless threads of combined length l.They are set in rotational motion in a horizontal plane about this axis with constant angular velocity ω. If the tension in the threads are same during motion, the distance of M from the axis is [ MPPMT 1995]
0%
a) Ml/ (M + m)
0%
b) ml / (M + m)
0%
c) (M + m)l / M
0%
d) (M + m)l / m
Explanation
As shown in figure M and m are the two masses attached to axis of rotation O by mass less string Since tension is same centrifugal force due to both the masses is same Mxω2=m(l - x)ω2 x=ml / (M + m) Answer: (b)
Q.27
Two rings of the same radius and mass are placed such that their centres are at a common point and their planes are perpendicular to each other. The moment of inertia of the system about an axis passing through the centre and perpendicular to the plane of one of the ring is ( mass of ring=m, radius of ring=r) [ MPPMT 1994]
0%
a)½ mr2
0%
b) mr2
0%
c)(3/2) mr2
0%
d)2mr2
Explanation
The axis is passing through the centre and perpendicular to the plane of one ring is the diameter of other ring. Therefore M.I. of system I=Mr2 + ½ Mr2 I=3Mr2 / 2Answer: (c)
Q.28
A disc of radius 33cm is hanged by a point at circumference by horizontal nail. Period of oscillation is 1.42 seconds, value of g by this experiment will be [ Raj. PMT 1997]
0%
a) 9.25 m/s2
0%
b) 9.68 m/s2
0%
c)9.86 m/s2
0%
d)100 m/s2
Explanation
periodic time is given by formula ( refer oscillation) Inertial factor = I , and restoring factor = mgd as point of suspension is at circumference By theorem of parallel axis By substituting value of I in equation for periodic time we get here d=33cm=0.33m , T=1.49 seconds by substituting the values and on simplification we get g=9.68m/s2Answer: (b)
Q.29
A smooth wire is bent into a vertical circle of radius a. A bead P can slide smoothly on the wire. The circle is rotated about diameter AB as axis with speed ω as shown in figure . the bead P is at rest with respect to the circular ring in the position shown. Then ω2 is equal to
0%
a)2g/a
0%
b)2g/(a√3)
0%
c)g√3 /a
0%
d)2a/ (g√3)
Explanation
Let N b the normal reaction on the bead. resolving it into two components. we get Ncosθ=mg and Nsinθ=m(a/2)ω2 Thus tanθ=aω2 / 2g ω2=(2g/a ) tanθ From figure sinθ=a/2a=1/2 or θ=30° ∴ ω2=(2g/a) tan30 ω2=2g / (a√3) Answer:(b)
Q.30
The moment of inertia of rod ( length l, mass m) about an axis perpendicular to the length of the rod and passing through a point equidistant from its mid-point and one end is [ MPPMT 1999]
0%
a) ml2 / 12
0%
b) 7ml2 / 48
0%
c) 13ml2 / 48
0%
d) 19ml2 / 48
Explanation
Momentum of inertia along the axis passing through the centre=Ml2 / 12 Axis is passing through the point equidistant from mid-point thus distance between axis passing through the centre and given axis is l/4 By applying parallel axis theorem Moment of Inertia=Ml2 / 12 + M(l/4)2 M.I=(7/48)Ml2 Answer: (b)
0 h : 0 m : 1 s
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
Report Question
×
What's an issue?
Question is wrong
Answer is wrong
Other Reason
Want to elaborate a bit more? (optional)
Support mcqgeeks.com by disabling your adblocker.
×
Please disable the adBlock and continue.
Thank you.
Reload page