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Physics · WORK, ENERGY AND POWER

हिन्दी

A bob of mass \( m \) is attached to a light string of length \( L \) and oscillates in a vertical circle. If the horizontal velocity imparted at the lowest point \( A \) is just sufficient for the string to become slack at the topmost point \( C \), the ratio of the tension in the string at the lowest point \( A \) to that at the horizontal position \( B \) is:

  1. \( 2 : 1 \)
  2. \( 3 : 1 \)
  3. \( 5 : 3 \)
  4. \( 6 : 1 \)

Correct answer · Option 1

\( 2 : 1 \)

Explanation

At the lowest point, \( T_A = mg + \frac{m v_0^2}{L} = mg + 5mg = 6mg \). At horizontal position \( B \), \( v_B = \sqrt{3gL} \), so \( T_B = \frac{m v_B^2}{L} = 3mg \). Therefore, \( T_A / T_B = 6mg / 3mg = 2 : 1 \).
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