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Question: Two blocks of masses m each are hung from a balance as shown in the figure. The scale pan A is at he...

Two blocks of masses m each are hung from a balance as shown in the figure. The scale pan A is at height H₁ whereas scale pan B is at height H2. Net torque of weights acting on the system about point 'C', will be (length of the rod is l and H₁ & H2 are << R) (H₁ > H2)

A

mg(1−2H1R)lmg(\frac{1-2H_1}{R})l

B

mgR(H1−H2)l\frac{mg}{R}(H_1-H_2)l

C

2mgR(H1+H2)l\frac{2mg}{R}(H_1+H_2)l

D

2mgH2H1H1+H2l2mg\frac{H_2H_1}{H_1+H_2}l

Answer

(B) mgR(H1−H2)l\frac{mg}{R}(H_1-H_2)l

Explanation

Solution

The problem asks for the net torque of weights acting on the system about point 'C'. The system consists of a balance with two masses 'm' placed in scale pans A and B. The pivot point is 'C', which is at the center of the rod, so the distances from C to A and C to B are both l/2l/2.

The key to this problem is to consider the variation of acceleration due to gravity (g) with height. The acceleration due to gravity at a height 'h' above the Earth's surface is given by:

gh=g0(1−2hR)g_h = g_0 \left(1 - \frac{2h}{R}\right)

where g0g_0 is the acceleration due to gravity at the Earth's surface (which we denote as 'g' in the solution), and R is the radius of the Earth. This approximation is valid because H1H_1 and H2H_2 are given to be much smaller than R (H1,H2≪RH_1, H_2 \ll R).

  1. Weight of mass in pan A:

The scale pan A is at height H1H_1. So, the acceleration due to gravity at H1H_1 is gH1=g(1−2H1R)g_{H_1} = g \left(1 - \frac{2H_1}{R}\right).

The weight of mass 'm' in pan A is WA=mgH1=mg(1−2H1R)W_A = m g_{H_1} = mg \left(1 - \frac{2H_1}{R}\right).

This force acts vertically downwards.

  1. Weight of mass in pan B:

The scale pan B is at height H2H_2. So, the acceleration due to gravity at H2H_2 is gH2=g(1−2H2R)g_{H_2} = g \left(1 - \frac{2H_2}{R}\right).

The weight of mass 'm' in pan B is WB=mgH2=mg(1−2H2R)W_B = m g_{H_2} = mg \left(1 - \frac{2H_2}{R}\right).

This force also acts vertically downwards.

  1. Torque due to WAW_A about C:

The distance from C to A is l/2l/2. The force WAW_A acts downwards, creating a clockwise torque about C.

τA=WA×l2=mg(1−2H1R)l2\tau_A = W_A \times \frac{l}{2} = mg \left(1 - \frac{2H_1}{R}\right) \frac{l}{2} (clockwise)

  1. Torque due to WBW_B about C:

The distance from C to B is l/2l/2. The force WBW_B acts downwards, creating an anti-clockwise torque about C.

τB=WB×l2=mg(1−2H2R)l2\tau_B = W_B \times \frac{l}{2} = mg \left(1 - \frac{2H_2}{R}\right) \frac{l}{2} (anti-clockwise)

  1. Net Torque:

We are given H1>H2H_1 > H_2. This implies that 2H1/R>2H2/R2H_1/R > 2H_2/R.

Therefore, 1−2H1R<1−2H2R1 - \frac{2H_1}{R} < 1 - \frac{2H_2}{R}.

This means gH1<gH2g_{H_1} < g_{H_2}, and consequently WA<WBW_A < W_B.

Since the lever arms are equal (l/2l/2), the torque τA\tau_A will be less than τB\tau_B.

The net torque will be in the direction of the larger torque, which is anti-clockwise.

Net torque τnet=τB−τA\tau_{net} = \tau_B - \tau_A.

τnet=mg(1−2H2R)l2−mg(1−2H1R)l2\tau_{net} = mg \left(1 - \frac{2H_2}{R}\right) \frac{l}{2} - mg \left(1 - \frac{2H_1}{R}\right) \frac{l}{2}

Factor out mgl2\frac{mgl}{2}:

τnet=mgl2[(1−2H2R)−(1−2H1R)]\tau_{net} = \frac{mgl}{2} \left[ \left(1 - \frac{2H_2}{R}\right) - \left(1 - \frac{2H_1}{R}\right) \right]

τnet=mgl2[1−2H2R−1+2H1R]\tau_{net} = \frac{mgl}{2} \left[ 1 - \frac{2H_2}{R} - 1 + \frac{2H_1}{R} \right]

τnet=mgl2[2H1R−2H2R]\tau_{net} = \frac{mgl}{2} \left[ \frac{2H_1}{R} - \frac{2H_2}{R} \right]

τnet=mgl22R(H1−H2)\tau_{net} = \frac{mgl}{2} \frac{2}{R} (H_1 - H_2)

τnet=mgR(H1−H2)l\tau_{net} = \frac{mg}{R} (H_1 - H_2) l

The net torque is mgR(H1−H2)l\frac{mg}{R}(H_1-H_2)l.