Rotation The Laws(11th And 12th > Physics ) Questions and answers for exam Preparation

11th And 12th > Physics

ROTATION THE LAWS MCQs

Total Questions : 30 | Page 1 of 3 pages

Question 1.

If net external force on a body adds up to zero, which of the following statements are not necessarily true?

Linear acceleration of the centre of mass is zero
Angular acceleration of the body has to be zero
Net External torque about any point need not be zero
Angular momentum about any axis may not be conserved

Discuss Question

Answer: Option B. -> Angular acceleration of the body has to be zero
:
B

Angular acceleration of the body may not necessarily be zero about any axis about which net external torque is non-zero.

Question 2.

A uniform cylinder of radius R is spinned about its axis to the angular velocity $ω_{0}$ and then placed into a corner, see the figure. The coefficient of friction between the corner walls and the cylinder is $μ_{k}$ . How many turns will the cylinder accomplish before it stops?

$\frac{{(1 + μ_{k})}_{k}^{2} ω_{0}^{2} R}{8 π μ_{k} (1 + μ_{k}) g}$
$\frac{(1 + μ_{k}^{2}) ω_{0}^{2} R}{8 π μ_{k} (1 + μ_{k}) g}$
$\frac{(1 + μ_{k}^{2}) ω_{0}^{2} R}{4 π μ_{k} (1 + μ_{k}) g}$
$\frac{(1 + μ_{k}^{2}) ω_{0}^{2} R}{8 π μ_{k} (1 + μ_{k}^{2}) 6 g}$

Discuss Question

Answer: Option B. ->

\frac{(1 + μ_{k}^{2}) ω_{0}^{2} R}{8 π μ_{k} (1 + μ_{k}) g}

:
B

As the centre of mass of the cylinder does not accelerate, hence $\sum F = 0$

$\sum F_{x} = 0, N_{2} - μ_{k} N_{1} = 0$ ...(1)
$\sum F_{x} = 0, N_{1} - μ_{k} N_{2} - m g = 0$ ...(2)
Solving these equations: $N_{1} = \frac{m g}{1 + μ_{k}^{2}}, N_{2} = \frac{μ_{k} m g}{1 + μ_{k}^{2}}$

The torque on the cylinder about the axis of rotation

The moment of inertia about axis of rotation $1_{c m} = \frac{1}{2} m R^{2}$

The torque equation $T = 1 α$

Using equation $ω^{2} = ω_{0}^{2} + 2 α θ$ , Calculate the angular displacement $θ$ ,

Revolution accomplished,

Question 3.

Consider a body shown in the figure below, consisting of two identical balls, each of mass 'M' connected by a light rigid rod of length 'L'. If an impulse, J = Mv is imparted to the body at one of its ends, what would be its angular velocity?

$\frac{v}{L}$
$\frac{2 v}{L}$
$\frac{v}{3 L}$
$\frac{v}{4 L}$

Discuss Question

Answer: Option A. ->

\frac{v}{L}

:
A
Initial angular momentum of the system about point O
= Linear momentum

\times

Perpendicular distance of linear momentum from the axis of rotation

= M v (\frac{L}{2}) . . . (i)

Final angular momentum of the system about point

O = I_{1} ω + I_{2} ω = (I_{1} + I_{2}) ω

= [M {(\frac{L}{2})}^{2} + M {(\frac{L}{2})}^{2}] ω

.....(ii)
Applying the law of conservation of angular momentum

\Rightarrow M v (\frac{L}{2}) = 2 M {(\frac{L}{2})}^{2} ω

\Rightarrow ω = \frac{v}{L}

Consider A Body Shown In The Figure Below, Consisting Of Two...

Question 4.

The position of a particle is given by $⃗ r = (ˆ i + 2 ˆ j - ˆ k)$ and momentum $⃗ P = (3 ˆ i + 4 ˆ j - 2 ˆ k)$ . The direction of angular momentum is perpendicular to

X-axis
Y-axis
Z-axis
Line at equal angles to all the three axes

Discuss Question

Answer: Option A. -> X-axis
:
A

⃗ L = ⃗ r \times ⃗ p = ∣ ∣∣∣ ∣ \begin{matrix} ˆ i & ˆ j & ˆ k 1 & 2 & - 1 3 & 4 & - 2 \end{matrix} ∣ ∣∣∣ ∣ = 0 ˆ i - ˆ j - 2 ˆ k = - ˆ j - 2 ˆ k

and the X-axis is given by

i + 0 ˆ j + 0 ˆ k

Question 5.

A solid cylinder of mass 2 kg and radius 0.2 m is rotating about its own axis without friction with angular velocity 3 rad/s. A particle of mass 0.5 kg moving with a velocity 5 m/s strikes the cylinder and sticks to it as shown in figure below. The angular momentum of the cylinder before collision will be

0.12 J-s
12 J-s
1.2 J-s
1.1.2 J-s

Discuss Question

Answer: Option A. -> 0.12 J-s
:
A
Angular momentum of the cylinder before collision

L = I ω = \frac{1}{2} M R^{2} ω = \frac{1}{2} (2) (0.2)^{2} \times 3 = 0.12 J - s .

Question 6.

If the angular momentum of a rotating body is increased by $200 %,$ then its kinetic energy of rotation will be increased by

$400 %$
$800 %$
$200 %$
$100 %$

Discuss Question

Answer: Option B. ->

800 %

:
B
As

E = \frac{L^{2}}{2 I} \Rightarrow \frac{E_{2}}{E_{1}} = {(\frac{L_{2}}{L_{1}})}^{2} = {(\frac{3 L_{1}}{L_{1}})}^{2}

[A s L_{2} = L_{1} + 200 % . L_{1} = 3 L_{1}]

\Rightarrow E_{2} = 9 E_{1} = E_{1} + 800 % o f E_{1}

Question 7.

An automobile engine develops 100 kW when rotating at a speed of 1800 rev/min. What torque does it deliver

350 N-m
440 N-m
531 N-m
628 N-m

Discuss Question

Answer: Option C. -> 531 N-m
:
C

P = τ ω \Rightarrow τ = \frac{100 \times 10^{3}}{2 π \frac{1800}{60}} = 531 N - m

Question 8.

A ring of radius 0.5 m and mass 10 kg is rotating about its diameter with an angular velocity of 20 rad/s. Its kinetic energy is

10 J
100 J
500 J
250 J

Discuss Question

Answer: Option D. -> 250 J
:
D
Rotational kinetic energy

\frac{1}{2} I ω^{2} = \frac{1}{2} (\frac{1}{2} M R^{2}) ω^{2} = \frac{1}{2} (\frac{1}{2} \times 10 \times (0.5)^{2}) (20)^{2} = 250 J

Question 9.

A hollow sphere of radius 'R' rests on a horizontal surface of finite coefficient of friction. A point object of mass 'm' moved horizontally and hits the sphere at a height of 'R/2' above its center. The collision is instantaneous and completely inelastic. Which of the following is/are correct ?

Total linear momentum of the system is not conserved
Total angular momentum about center of mass of the system remains conserved
The sphere gets finite angular velocity immediately after collision
The sphere moves with finite speed immediately after collision

Discuss Question

Answer: Option D. -> The sphere moves with finite speed immediately after collision
:
A, C, and D
Impulse due to normal reaction is finite. So friction force gives finite impulse. Therefore frictional torque causes a finite angular impulse about center of mass of system so angular momentum about center of mass of system will change. All other options are correct.

Question 10.

A particle of mass 'm' is projected with velocity 'v' making an angle of $45^{\circ}$ with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when the particle is at its maximum height 'h' is