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

12th Grade > Physics

ROTATION THE LAWS MCQs

Total Questions : 28 | Page 1 of 3 pages

Question 1. A solid cylinder of mass m = 4 kg and radius R = 10 cm has two ropes wrapped around it, one near each end. The cylinder is held horizontally by fixing the two free ends of the cords to the hooks on the ceiling such that both the cords are exactly vertical. The cylinder is released to fall under gravity. Find the tension in the cords when they unwind and the linear acceleration of the cylinder.

6.5ms2,6.5N
5.5ms2,6.5N
6.5ms2,5.5N
5.5ms2,5.5N

Discuss Question

Answer: Option A. -> 6.5ms2,6.5N
:
A
Let a = linear acceleration and

α

= angular acceleration of the cylinder.
for the linear motion of the cylinder : mg - 2T = ma
for the rotational motion : net torque = 1

α

Also, the linear acceleration of cylinder is same as the tangential acceleration of any point on its surface. A = R

α

Combining the three equations, we get :

m g = m a + \frac{m}{2} a

\Rightarrow a = \frac{2 g}{3} = 6.53 \frac{m}{s^{2}} a n d T = \frac{m g - m a}{2} = 6.53 N

Question 2. 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

Zero
mv34g
mv3√2g
m√2gh3

Discuss Question

Answer: Option D. -> m√2gh3
:
D

L = m v_{x} h

\Rightarrow L = m (v cos 45^{\circ}) \frac{v^{2} {sin}^{2} 45^{\circ}}{2 g}

\Rightarrow L = \frac{m v^{3}}{4 \sqrt{2} g}

Further

L = m v_{x} h

\Rightarrow L = m (v cos 45^{\circ}) h o

But

h = \frac{v^{2} {sin}^{2} 45^{\circ}}{2 g}

\Rightarrow v = 2 \sqrt{g h}

\Rightarrow L = \frac{m v}{\sqrt{2}} 2 \sqrt{g h} h

\Rightarrow L = m \sqrt{2 g h^{3}} .

Question 3. 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 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. For the toppling of the hexagon as shown in the figure, the coefficient of friction must be

> 0.21
< 0.21
= 0.21
< 0.21

Discuss Question

Answer: Option A. -> > 0.21
:
A
To prevent slipping

For The Toppling Of The Hexagon As Shown In The Figure, The ...

F = μ N = μ m g

the body will topple about point O, if

F . h \geq m g \frac{a}{2}

here h = 2a

cos

30^{\circ}

a \sqrt{3}

μ m g a \sqrt{3} \geq m g \frac{a}{2}

μ \geq \frac{1}{2 \sqrt{3}}

μ \geq 0.21

Question 6. A constant force 'F' is applied at the top of a ring as shown in figure. Mass of the ring is 'M' and radius is 'R'. Angular momentum of particle about point of contact at time 't' is

Is constant
Increases linearly with time
Is Zero
Decreases linearly with time

Discuss Question

Answer: Option B. -> Increases linearly with time
:
B
Angular impulse = change in angular momentum
we have L =

τ

*t
or, L = F*(2R)*t
i.e.L increases linearly with time.

Question 7. A uniform wheel of moment of inertia 'I' is pivoted on a horizontal axis through its centre so that its plane is vertical as shown in the figure. A small mass 'm' is stuck on the rim of the wheel as shown. The angular acceleration of the wheel when mass is at point A is

α_{A}

and when mass is at point B is

α_{B}

. Then,

A Uniform Wheel Of Moment Of Inertia 'I' Is Pivoted On A Hor...

αBαA=sinθ
αBαA=sin2θ
αBαA=cosθ
αBαA=cos2θ

Discuss Question

Answer: Option C. -> αBαA=cosθ
:
C
The torque due to the weight of m at A is

τ_{A}

and that when it is at B is

τ_{B}

\Rightarrow \frac{τ_{B}}{τ_{A}} = \frac{I α_{B}}{I α_{A}} = \frac{b cos θ}{b} = cos θ .

Question 8. A uniform slender bar AB of mass m is suspended as shown from a small cart of the same mass m. Neglecting the effect of friction, determine the accelerations of points A and B immediately after a horizontal force F has been applied at B

2F5mleftwards and 16F5mleftwards
2F5mrightwards and 5F16leftwards
2Fmrightwards and 16F5mleftwards
2F5mrightwards and 16F5mleftwards

Discuss Question

Answer: Option D. -> 2F5mrightwards and 16F5mleftwards
:
D
Let the acceleration of bar and cart be a and a' respectively; alsoαbe the angular acceleration of the bar.
Writing newton's equation of motion
For the bar: F - F' =ma ...(i)
For the cart: F' = ma' ...(ii)
Torque equation for rod about center of mass