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NEET PHYSICSGeneralHard

Question

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×(4020) cm=0.5 kg×(10040) cm+m×(16040) 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=120mm=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).

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from General. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSuniformlengthbalancedplacedsuspended

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