
Balancing torque about the wedge, we get
2g×20=0.5g×60+mg×120
⇒2=1.5+6m
⇒m=60.5=121kg
NEET UG 2021 — Physics Mechanics
A uniform rod of length 200cm and mass 500g is balanced on a wedge placed at 40cm mark. A mass of 2kg is suspended from the rod at 20cm and another unknown mass m is suspended from the rod at 160cm mark as shown in the figure. Find the value of m such that the rod is in equilibrium. (g=10ms−2)

Held on 30 Apr 2021 · Verified 9 Jul 2026.
21kg
31kg
121kg
61kg
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