Question

A body of m kg slides from rest along the curve of vertical circle from point A to B in friction less path. The velocity of the body at B is :

(Given, \( R = 14 \, \text{m}, \, g = 10 \, \text{m/s}^2 \, \text{and} \, \sqrt{2} = 1.4 \))

• A. 19.8 m/s
• B. 21.9 m/s
• C. 16.7 m/s
• D. 10.6 m/s

Answer 1

The Correct Option is B

Using the Work-Energy Theorem (W.E.T.) from point A to point B:

1. Apply W.E.T. from A to B:
\[ W_{mg} = K_B - K_A. \] Since the body starts from rest at A, \( K_A = 0 \). Thus,
\[ mg \times \left( \frac{R}{\sqrt{2}} + R \right) = \frac{1}{2} mv_B^2. \]

2. Substitute Values and Solve for \( v_B \):
Simplifying, we get:
\[ mgR \frac{(\sqrt{2} + 1)}{\sqrt{2}} = \frac{1}{2} mv_B^2. \] Solving for \( v_B \), we find:
\[ v_B = \sqrt{\frac{2gR(\sqrt{2} + 1)}{\sqrt{2}}}. \] Substitute \( g = 10 \, \text{m/s}^2 \), \( R = 14 \, \text{m} \), and \( \sqrt{2} = 1.4 \):
\[ v_B = \sqrt{\frac{2 \times 10 \times 14 \times 2.4}{1.4}} = 21.9 \, \text{m/s}. \]
Answer: 21.9 m/s