Question

A ring and a solid sphere are released from rest from the same height on enough rough inclined surface. The ratio of their speed when they reach the bottom is \( \sqrt{\frac{7}{x}} \) m/s, then \( x \) is ______.

• A. 3
• B. 4
• C. 7
• D. 5

Hint:

When comparing the velocities of a rolling solid sphere and a ring, consider the ratio of their moments of inertia. A solid sphere has a lower moment of inertia compared to a ring, leading to a higher speed when it reaches the bottom of the incline.

Answer 1

The Correct Option is B

When a solid sphere and a ring are released from rest on an inclined plane, both undergo rolling motion. The total kinetic energy of an object rolling without slipping is the sum of its translational kinetic energy and rotational kinetic energy. The total kinetic energy of a rolling object is given by: \[ K_{\text{total}} = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 \] Where: - \( m \) is the mass of the object, - \( v \) is the linear velocity, - \( I \) is the moment of inertia, - \( \omega \) is the angular velocity. For rolling motion without slipping, the relation between linear velocity and angular velocity is \( v = r\omega \), where \( r \) is the radius of the object. The moment of inertia for a solid sphere is \( I_{\text{sphere}} = \frac{2}{5} m r^2 \), and for a ring, it is \( I_{\text{ring}} = m r^2 \). Using the principle of conservation of mechanical energy, the potential energy lost by the objects as they roll down the incline is converted into kinetic energy. The potential energy is \( mgh \), and the total energy at the bottom is the sum of translational and rotational kinetic energy. For the solid sphere: \[ mgh = \frac{7}{10} m v_{\text{sphere}}^2 \] For the ring: \[ mgh = m v_{\text{ring}}^2 \] Thus, the ratio of the final velocities is: \[ \frac{v_{\text{sphere}}}{v_{\text{ring}}} = \sqrt{\frac{7}{10}} = \sqrt{\frac{7}{x}} \] Solving for \( x \), we get \( x = 4 \). Therefore, the correct answer is \( x = 4 \).