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Physics · GRAVITATION

हिन्दी

In Cavendish's experiment to measure \( G \), a bar of length \( L \) connects two small lead spheres each of mass \( m \). Two large spheres each of mass \( M \) are placed at a separation \( d \) from the small spheres. If the restoring couple per unit angle of twist of the suspension wire is \( \tau_0 \) and the angle of deflection is \( \theta \), what is the equilibrium relation?

  1. \( 2G\frac{Mm}{d^2} L = \tau_0 \theta \)
  2. \( G\frac{Mm}{d^2} L = \tau_0 \theta \)
  3. \( G\frac{Mm}{d} L^2 = \tau_0 \theta \)
  4. \( G\frac{Mm}{d^2} = \tau_0 \theta L \)

Correct answer · Option 2

\( G\frac{Mm}{d^2} L = \tau_0 \theta \)

Explanation

The gravitational torque is F × L = (G M m / d^2) L, which balances the restoring torque tau_0 * theta at equilibrium.
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