# Quiz

#### Topics11th And 12thPhysicsRotation The Basic Definition

Question: A force of (2ˆi4ˆj+2ˆk)N acts at a point (3ˆi+2ˆj4ˆk) metre from the origin. The magnitude of torque is
Options:
 A. Zero B. 24.4 N-m C. 0.244 N-m D. 2.444 N-m
: B

F=(2ˆi4ˆj+2ˆk)N and r=(3i+24ˆk) meter
Torque τ=r×F=∣ ∣ ∣ˆiˆjˆk324242∣ ∣ ∣τ=12ˆi14ˆj16ˆj16ˆkand|τ|=(12)2+(14)2+(16)2=
24.4 N-m

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### More Questions on This Topic :

Question 1. Five particles of mass 2 kg are attached to the rim of a circular disc of radius 0.1 m and negligible mass. Moment of inertia of the system about the axis passing through the centre of the disc and perpendicular to its plane is
1.    1kgm2
2.    0.1kgm2
3.    2kgm2
4.    0.2kgm2
: B

We will not consider the moment of inertia of ring because it doesn't have any mass. So, moment of inertia of five particle system I=5mr2=5×2×(0.1)2=0.1kgm2.
Note: The masses are concentrated at fixed distance from the axis similar to that of ring. That is why you simply need to add all the individual contributions.
Question 2. One quarter sector is cut from a uniform disc of radius 'R'. This sector has mass 'M'.  It is made to rotate about a line perpendicular to its plane and passing through the center of the original disc. Its moment of inertia about the axis of rotation is

1.    12MR2
2.    14MR2
3.    18MR2
4.    √2MR2
: A

If we assume complete disc to be present, then it would have a mass 4 times the mass of the sector. Then, moment of inertia of the complete disc is
Idisc=12MdiscR2=12(4Msector)R2
Now, Isector=Idisc4
Isector=12MR2.
Question 3. If F be a force acting on a particle having the position vector r  and τ be the torque of this force about the origin, then
1.    ⃗τ.⃗F=0 and ⃗r.⃗τ=0
2.    ⃗τ.⃗F=0 and ⃗F.⃗τ≠0
3.    ⃗r.⃗τ≠0 and ⃗F.⃗τ≠0
4.    ⃗r.⃗τ≠0 and ⃗F.⃗τ=0
: A

τ=r×F, so τ.F=0
Question 4. Two circular discs A and B are of equal masses and thickness but made of metals with densities dA and dB(dA>dB). If their moments of inertia about an axis passing through centres and normal to the circular faces be IA and IB, then
1.    IA=IB
2.    IA>IB
3.    IA
4.    IA>=
: C

Moment of inertia of circular disc about an axis passing through centre and normal to the circular face
I=12MR2=12M(Mπtρ) [AsM=Vρ=πR2tρR2=Mπtρ]
I=M22πtρ or I1ρ If mass and thickness are constant.
So, in the problem IAIB=dBdA IA<IB [AsdA>dB]
Question 5. The moment of inertia of a thin square plate ABCD of uniform thickness about an axis passing through the centre O and perpendicular to the plane of the plate can NOT be given by :

where, I1,I2,I3,I4 are the moments of inertia of the plates about axes 1,2,3,4 respectively.
1.    I1+I2
2.    I3+I4
3.    I1+I3
4.    I1+I2+I3+I4
: D

II1+I2+I3+I4
(this equation violates raxes theorem)
Question 6. The the system in the figure is in equilibrium. The value of PR in terms of RQ is equal to

1.    14RQ
2.    38RQ
3.    35RQ
4.    25RQ