JEE Main 2025 — Rotational Motion Question with Solution
JEE Main 2025 (8 Apr Shift 2)
Question
A cube having a side of 10 cm with unknown mass and 200 gm mass were hung at two ends of an uniform rigid rod of 27 cm long. The rod along with masses was placed on a wedge keeping the distance between wedge point and 200 gm weight as 25 cm. Initially the masses were not at balance. A beaker is placed beneath the unknown mass and water is added slowly to it. At given point the masses were in balance and half volume of the unknown mass was inside the water.
(Take the density of unknown mass is more than that of the water, the mass did not absorb water and water density is .) The unknown mass is ________ kg.
(Take the density of unknown mass is more than that of the water, the mass did not absorb water and water density is .) The unknown mass is ________ kg.
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Show full solutionCorrect answer: 3
Correct answer
3
Step-by-step explanation
Given, volume of block
Let density of block
mass of block
Buoyant Force
F.B.D. of blocks
Balancing torque about point O , we get
$\begin{aligned}
& \operatorname{mg}\left(2 \times 10^{-2}\right)-\mathrm{F}_{\mathrm{B}}\left(2 \times 10^{-2}\right)=0.2 \mathrm{~g}\left(25 \times 10^{-2}\right) \\ & \rho \times 10^{-3} \times 10 \times 2-10=50 \\ & \rho=3000 \mathrm{~kg} / \mathrm{m}^3
\end{aligned}=\rho \times 10^{-3}=3000 \times 10^{-3}=3 \mathrm{~kg}$
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This is a previous-year question from JEE Main 2025, covering the Rotational Motion chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.