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Physics · WORK, ENERGY AND POWER

हिन्दी

The bob of a simple pendulum of length \( L = 1.5\text{ m} \) is released from rest from a horizontal position. During its downward swing to the lowermost point, \( 5\% \) of its initial mechanical energy is dissipated against air resistance. Taking \( g = 10\text{ m s}^{-2} \), the speed of the bob at the lowest point is:

  1. \( 5.48\text{ m s}^{-1} \)
  2. \( 5.34\text{ m s}^{-1} \)
  3. \( 5.12\text{ m s}^{-1} \)
  4. \( 4.85\text{ m s}^{-1} \)

Correct answer · Option 2

\( 5.34\text{ m s}^{-1} \)

Explanation

Initial potential energy \( E_i = mgL \). Final kinetic energy \( K_f = 0.95 mgL \implies \frac{1}{2}mv^2 = 0.95 mgL \implies v = \sqrt{2 \times 0.95 \times g L} = \sqrt{1.9 \times 10 \times 1.5} = \sqrt{28.5} \approx 5.34\text{ m s}^{-1} \).
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