Question

A ball of mass 1 kg and radius 0.5 m, starting from rest rolls down on a $30^\circ$ inclined plane. The torque acting on the ball at the distance of 7 m from the starting point is close to (Take $g = 10 \text{ m/s}^2$) (A) 0.25 N-m

• A. 0.7 N-m
• B. 0.5 N-m
• C. 0.4 N-m
• D. 1.4 N-m
• E.

Hint:

Torque remains constant on uniform incline.

Answer 1

The Correct Option is B

Concept: \[ \tau = fR \]

Step 1: Acceleration \[ a = \frac{g\sin\theta}{1 + \frac{2}{5}} = \frac{5g\sin\theta}{7} \] \[ a = \frac{5 \cdot 10 \cdot 0.5}{7} = \frac{25}{7} \]

Step 2: Friction \[ f = m(g\sin\theta - a) = 5 - 3.57 = 1.43 N \]

Step 3: Torque \[ \tau = fR = 1.43 \times 0.5 \approx 0.7 \, \text{N-m} \] \[ \boxed{\tau \approx 0.7 \, \text{N-m}} \]