Rotation The Basic Definition(12th Grade > Physics ) Questions and answers for exam Preparation

12th Grade > Physics

ROTATION THE BASIC DEFINITION MCQs

Total Questions : 30 | Page 2 of 3 pages

Question 11. 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,

I_{1}, I_{2}, I_{3}, I_{4}

are the moments of inertia of the plates about axes 1,2,3,4 respectively.

I1+I2
I3+I4
I1+I3
I1+I2+I3+I4

Discuss Question

Answer: Option D. -> I1+I2+I3+I4
:
D

I \neq I_{1} + I_{2} + I_{3} + I_{4}

(this equation violates

⊥^{r}

axes theorem)

Question 12. Two circular discs A and B are of equal masses and thickness but made of metals with densities

d_{A}

and

d_{B} (d_{A} > d_{B}) .

If their moments of inertia about an axis passing through centres and normal to the circular faces be

I_{A}

and

I_{B}

, then

IA=IB
IA>IB
IA
IA>=

Discuss Question

Answer: Option C. -> IA
:
C
Moment of inertia of circular disc about an axis passing through centre and normal to the circular face

I = \frac{1}{2} M R^{2} = \frac{1}{2} M (\frac{M}{π t ρ})

[A s M = V ρ = π R^{2} t ρ ∴ R^{2} = \frac{M}{π t ρ}]

\Rightarrow I = \frac{M^{2}}{2 π t ρ}

I \propto \frac{1}{ρ}

If mass and thickness are constant.
So, in the problem

\frac{I_{A}}{I_{B}} = \frac{d_{B}}{d_{A}}

∴ I_{A} < I_{B}

[A s d_{A} > d_{B}]

Question 13. A force of

(2 ˆ i - 4 ˆ j + 2 ˆ k)

N acts at a point

(3 ˆ i + 2 ˆ j - 4 ˆ k)

metre from the origin. The magnitude of torque is

Zero
24.4 N-m
0.244 N-m
2.444 N-m

Discuss Question

Answer: Option B. -> 24.4 N-m
:
B

⃗ F = (2 ˆ i - 4 ˆ j + 2 ˆ k) N

and

⃗ r = (3 i + 2 - 4 ˆ k)

meter
Torque

⃗ τ = ⃗ r \times ⃗ F = ∣ ∣∣∣ ∣ \begin{matrix} ˆ i & ˆ j & ˆ k 3 & 2 & - 4 2 & - 4 & 2 \end{matrix} ∣ ∣∣∣ ∣ \Rightarrow ⃗ τ = - 12 ˆ i - 14 ˆ j - 16 ˆ j - 16 ˆ k a n d | ⃗ τ | = \sqrt{(- 12)^{2} + (- 14)^{2} + (- 16)^{2}} =

24.4 N-m

Question 14. 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

1kgm2
0.1kgm2
2kgm2
0.2kgm2

Discuss Question

Answer: Option B. -> 0.1kgm2
:
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 = 5 m r^{2} = 5 \times 2 \times (0.1)^{2} = 0.1 k g - m^{2} .

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 15. 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

12MR2
14MR2
18MR2
√2MR2

Discuss Question

Answer: Option A. -> 12MR2
:
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

I_{d i s c} = \frac{1}{2} M_{d i s c} R^{2} = \frac{1}{2} (4 M_{s e c t o r}) R^{2}

Now,

I_{s e c t o r} = \frac{I_{d i s c}}{4}

\Rightarrow I_{s e c t o r} = \frac{1}{2} M R^{2} .

Question 16. 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

⃗τ.⃗F=0 and ⃗r.⃗τ=0
⃗τ.⃗F=0 and ⃗F.⃗τ≠0
⃗r.⃗τ≠0 and ⃗F.⃗τ≠0
⃗r.⃗τ≠0 and ⃗F.⃗τ=0

Discuss Question

Answer: Option A. -> ⃗τ.⃗F=0 and ⃗r.⃗τ=0
:
A

⃗ τ = ⃗ r \times ⃗ F,

⃗ τ . ⃗ F = 0

Question 17. Figure below shows a small wheel fixed coaxially on a bigger one of double the radius. The system rotates about the common axis. The strings supporting A and B do not slip on the wheels. If 'x' and 'y' be the distances travelled by A and B in the same time interval, then