Question

A solid sphere is rolling down without slipping on an inclined plane of length 21 m with an acceleration of $5\,ms^{-2}$. The time taken by a circular disc to roll down without slipping to reach the bottom from the top of the same inclined plane is}

• A. 5 s
• B. 9 s
• C. 3 s
• D. 6 s

Hint:

For rolling bodies: \[ a=\frac{g\sin\theta}{1+\frac{I}{mR^2}} \] Smaller moment of inertia gives larger acceleration.

Answer 1

The Correct Option is C

Concept: For rolling bodies, \[ a=\frac{g\sin\theta} {1+\frac{I}{mR^2}} \] Acceleration depends on the moment of inertia.

Step 1: Write acceleration of solid sphere.
For sphere, \[ I=\frac25 mR^2 \] Therefore, \[ a_s=\frac{g\sin\theta}{1+\frac25} \] \[ a_s=\frac57 g\sin\theta \] Given \[ a_s=5 \] Hence, \[ g\sin\theta=7 \]

Step 2: Find acceleration of disc.
For disc, \[ I=\frac12mR^2 \] Thus, \[ a_d= \frac{g\sin\theta}{1+\frac12} \] \[ a_d=\frac23(7) \] \[ a_d=\frac{14}{3} \]

Step 3: Calculate time of descent.
\[ s=\frac12 at^2 \] \[ 21=\frac12\times\frac{14}{3}\times t^2 \] \[ 21=\frac73 t^2 \] \[ t^2=9 \] \[ t=3\,s \] Hence, \[ \boxed{3\,s} \]