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A uniform rod of length 200 cm200 \text{ cm} and mass 500 g500 \text{ g} is balanced on a wedge placed at 40 cm40 \text{ cm} mark. A mass of 2 kg2 \text{ kg} is suspended from the rod at 20 cm20 \text{ cm} and another unknown mass 'mm' is suspended from the rod at 160 cm160 \text{ cm} mark as shown in the figure. Find the value of 'mm' such that the rod is in equilibrium. (g=10 m/s2g = 10 \text{ m/s}^2)

A

1/12 kg1/12 \text{ kg}

B

1/2 kg1/2 \text{ kg}

C

5/12 kg5/12 \text{ kg}

D

1/6 kg1/6 \text{ kg}

Step-by-Step Solution

The rod is in rotational equilibrium about the wedge at 40 cm40 \text{ cm}. The center of mass of the rod is at 100 cm100 \text{ cm}. Torque balance: 2 kg×(40−20) cm=0.5 kg×(100−40) cm+m×(160−40) cm2 \text{ kg} \times (40-20) \text{ cm} = 0.5 \text{ kg} \times (100-40) \text{ cm} + m \times (160-40) \text{ cm}. 2×20=0.5×60+m×1202 \times 20 = 0.5 \times 60 + m \times 120. 40=30+120m40 = 30 + 120m. 10=120m⇒m=10/120=1/12 kg10 = 120m \Rightarrow m = 10/120 = 1/12 \text{ kg}. (Note: The provided options might contain a typo based on the calculation, but following the torque balance principle).

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