JEE Physics 2025 Question with Solution

Given: A uniform rod has mass $250 \, \text{g}$ and length $100 \, \text{cm}$. It 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, and an unknown mass $m$ is to be suspended at the $90 \, \text{cm}$ mark.

Find: The value of $m$ for rotational equilibrium.

For torque balance about the sharp edge,

The $400 \, \text{g}$ mass is $30 \, \text{cm}$ from the pivot, the rod’s weight acts at its center at $50 \, \text{cm}$, so it is $10 \, \text{cm}$ from the pivot, and the unknown mass is $50 \, \text{cm}$ from the pivot.

Using the balance equation shown in the solution,

Therefore, the mass to be suspended is $190 \, \text{g}$. The correct option is D.

Given: The pivot is at the $40 \, \text{cm}$ mark. The rod is uniform, so its center of gravity is at the $50 \, \text{cm}$ mark. A mass of $400 \, \text{g}$ is at the $10 \, \text{cm}$ mark and an unknown mass $M_3$ is at the $90 \, \text{cm}$ mark.

Find: The mass required at the $90 \, \text{cm}$ mark to maintain balance.

The extracted explanation identifies the relevant distances from the pivot as:

• Rod’s center of gravity: $50 - 40 = 10 \, \text{cm}$
• $400 \, \text{g}$ mass: $40 - 10 = 30 \, \text{cm}$
• Unknown mass: $90 - 40 = 50 \, \text{cm}$

the solution finally concludes that the required mass is $190 \, \text{g}$ and marks D as correct. There is an internal inconsistency in the longer working because one intermediate equation leads to $290 \, \text{g}$, but the page’s stated final conclusion is $190 \, \text{g}$.

Therefore, following the solution’s final stated answer, the correct option is D and the required mass is $190 \, \text{g}$.