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Physics · GRAVITATION
हिन्दीIn a Cavendish experiment, a light bar of length \( L = 1.0\text{ m} \) carries two lead spheres of mass \( m = 1.0\text{ kg} \) each at its ends. Two large spheres of mass \( M = 100\text{ kg} \) each are placed at a separation distance \( d = 0.1\text{ m} \) from each small sphere. If the restoring couple per unit angle of the suspension wire is \( \tau = 6.67 \times 10^{-7}\text{ N}\cdot\text{m/rad} \), what is the equilibrium angle of twist \( \theta \)?
Correct answer · Option 2
\( 1.0 \times 10^{-1}\text{ rad} \)
Explanation
Gravitational torque is \( \frac{G M m}{d^2} L = \frac{(6.67 \times 10^{-11})(100)(1)}{(0.1)^2} (1) = 6.67 \times 10^{-8}\text{ N}\cdot\text{m} \). Equating to \( \tau \theta \) gives \( \theta = \frac{6.67 \times 10^{-8}}{6.67 \times 10^{-7}} = 0.1\text{ rad} = 1.0 \times 10^{-1}\text{ rad} \).
