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A uniform rod of mass 250 g and length 100 cm is balanced on a sharp edge at the 40 cm mark. A mass of 400 g is suspended at the 10 cm mark. To keep the rod balanced, the mass that must be suspended at the 90 cm mark is

Asked in JEE Main 28th Jan 2nd Shift 2025 · Static equilibrium

• (1) 290 g
• (2) 200 g
• (3) 300 g
• (4) 190 g

Step-by-step solution

Why the other options are wrong

• (1) comes from putting the rod's own weight on the same side as the 400 g. The rod's centre is at the 50 cm mark, which is on the far side of the edge from the 400 g, so its moment helps rather than hinders.
• (2) comes from no single slip reproduces this. Sanity check: moments about the 40 cm mark must match, 400×30=12000 on one side against 250×10+m×50 on the other; m=200 gives 12500, so that side would drop.
• (3) comes from the same check rules this out even faster: m=300 gives 250×10+300×50=17500 against 12000.

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