Question

A ball of mass \(1\, kg\) and radius \(0.5\, m,\) starting from test rolls down on a \(30^{\circ}\) inclined plane. The torque acting on the ball at the distance of the \(7\, m\) from the starting point is close to
(Take: Acceleration due to gravity as \(10\, m/s^2\))

• A. 0.25 N-m
• B. 0.7 N-m
• C. 0.5 N-m
• D. 0.4 N-m

Answer 1

The Correct Option is B

Given, ball of mass \(M=1\, kg\),
Radius of ball \(R=0.5\,m\),
Angle of inclination \(\theta=30^{\circ}\)
A ball placed on a inclined plane.
Acceleration of ball,
\(a=\frac{g \sin \theta}{1+\frac{I}{M R^{2}}}\)
Where, \(I=\) moment of inertia of ball
\(\Rightarrow a=\frac{10 \times \sin 30^{\circ}}{1+\frac{2}{5} \frac{M R^{2}}{M R^{2}}}=\frac{5 \times 5}{7}\)
Torque on the ball \(\tau=I \alpha=I \frac{a}{R}\)
\(\Rightarrow \tau =\frac{2}{5} M R^{2}\left[\frac{5 \times 5}{7}\right] \frac{1}{R}\)

\(\Rightarrow \tau =\frac{2}{5} \times 1 \times 0.5 \times \frac{5 \times 5}{7}\)

\(\Rightarrow \tau =\frac{5}{7}=0.71\, N - m\)

So, the correct option is (B): \(0.7 \ N-m\)