Q101 / 160 · JIPMER 2011 · Physics
A solid sphere and a hollow sphere of the same material and of a same size can be distinguished without weighing
- Aby determining their moments of inertia about their coaxial axes
- Correct answerBby rolling them simultaneously on an inclined plane
- Cby rotating them about a common axis of rotation
- Dby applying equal torque on them
Solution
The acceleration of a body rolling down the plane $=a$.
$\therefore \quad a=\frac{g \sin \theta}{1+\frac{K^2}{R^2}}$
where $K$ is radius of gyration and $R$ the radius of sphere.
For solid sphere,
$\frac{K^2}{R^2}=\frac{2}{5}$
$\therefore \quad a=\frac{5}{7} g \sin \theta$
$=0.7(g \sin \theta)$
For hollow sphere,
$\frac{K^2}{R^2}=\frac{2}{3}$
$=\frac{3}{5} g \sin \theta$
$=0.6(g \sin \theta)$
Since, acceleration of solid sphere is more than of hollow sphere, it rolls faster, and reaches the bottom of the inclined plane earlier.
Hence, solid sphere and hollow sphere can be distinguished by rolling them simultaneously on an inclined plane.
$\therefore \quad a=\frac{g \sin \theta}{1+\frac{K^2}{R^2}}$
where $K$ is radius of gyration and $R$ the radius of sphere.
For solid sphere,
$\frac{K^2}{R^2}=\frac{2}{5}$
$\therefore \quad a=\frac{5}{7} g \sin \theta$
$=0.7(g \sin \theta)$
For hollow sphere,
$\frac{K^2}{R^2}=\frac{2}{3}$
$=\frac{3}{5} g \sin \theta$
$=0.6(g \sin \theta)$
Since, acceleration of solid sphere is more than of hollow sphere, it rolls faster, and reaches the bottom of the inclined plane earlier.
Hence, solid sphere and hollow sphere can be distinguished by rolling them simultaneously on an inclined plane.Hence, the correct answer is option: 'B'.