Q 11-06-075JEE MainJEE Main 2025 (28 Jan, Shift 2)Easy
A uniform rod of mass $250\ \text{g}$ having length $100\ \text{cm}$ is balanced on a sharp edge at the $40\ \text{cm}$ mark. A mass of $400\ \text{g}$ is suspended at the $10\ \text{cm}$ mark. To maintain the balance of the rod, the mass to be suspended at the $90\ \text{cm}$ mark is
Answer: (A) $190\ \text{g}$
Take moments about the $40\ \text{cm}$ mark:
- $400\ \text{g}$ at $10\ \text{cm}$: arm $30\ \text{cm}$, one side.
- The rod's own weight ($250\ \text{g}$) acts at $50\ \text{cm}$: arm $10\ \text{cm}$, other side.
- Unknown $m$ at $90\ \text{cm}$: arm $50\ \text{cm}$, same side as the rod's weight.
$$400\times30 = 250\times10 + m\times50 \Rightarrow 12000 = 2500 + 50m \Rightarrow m = 190\ \text{g}$$
Solution by Sreeraj P, M.Sc Physics