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Question: Two small balls of mass m each are suspended side by side by two equal threads of length L as shown ...

Two small balls of mass m each are suspended side by side by two equal threads of length L as shown in the figure. If the distance between the upper ends of the threads be a, the angle θ\theta that the threads will make with the vertical due to attraction between the balls is :

A

tan⁡−1(a−x)gmG\tan^{-1} \frac{(a-x)g}{mG}

B

tan⁡−1mG(a−x)2g\tan^{-1} \frac{mG}{(a-x)^2g}

C

tan⁡−1(a−x)2gmG\tan^{-1} \frac{(a-x)^2g}{mG}

D

tan⁡−1(a2−x2)gmG\tan^{-1} \frac{(a^2-x^2)g}{mG}

Answer

B

Explanation

Solution

The problem describes two small balls of mass m suspended by threads of length L. Due to mutual gravitational attraction, the threads make an angle θ with the vertical. The distance between the upper ends of the threads is a, and the distance between the centers of the balls in the equilibrium position is (a-x).

Let's consider the forces acting on one of the balls:

  1. Weight (gravitational force): W = mg, acting vertically downwards.
  2. Tension in the thread: T, acting along the thread.
  3. Gravitational force of attraction from the other ball: F_G, acting horizontally towards the other ball.

The ball is in equilibrium, so the net force on it is zero. We resolve the tension T into its horizontal and vertical components.

  • Vertical component: T cos θ (upwards)
  • Horizontal component: T sin θ (towards the other ball, away from the vertical line of suspension)

Applying the conditions for equilibrium:

1. Vertical Equilibrium:

The upward vertical component of tension balances the downward weight: T cos θ = mg (Equation 1)

2. Horizontal Equilibrium:

The horizontal component of tension balances the gravitational attractive force between the balls: T sin θ = F_G (Equation 2)

Now, let's calculate the gravitational force F_G. According to Newton's Law of Universal Gravitation, the force between two masses m1 and m2 separated by a distance r is F = G * (m1 * m2) / r^2.

In this case, m1 = m, m2 = m, and the distance r between the balls is (a-x). So, F_G = G * (m * m) / (a-x)^2 = Gm^2 / (a-x)^2

Substitute F_G into Equation 2: T sin θ = Gm^2 / (a-x)^2 (Equation 3)

Now, divide Equation 3 by Equation 1 to eliminate T: (T sin θ) / (T cos θ) = (Gm^2 / (a-x)^2) / (mg) tan θ = (Gm^2 / (a-x)^2) * (1 / mg) tan θ = Gm / ((a-x)^2 * g)

Finally, solve for θ: θ = tan⁻¹ [Gm / ((a-x)^2 * g)]

Our derived expression matches option (B).