JEE Main 2025 Apr 8 Shift 2, Physics Q25: Centre of Mass

JEE Main2025Apr 8, Shift 2Physics

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 $1$ gm/cm $^{3}$ .) The unknown mass is _____ kg.

Your answer

Solution

Set up the geometry: the rod is 27 cm long with the wedge 25 cm from the 200 gm end, so the unknown mass hangs at 2 cm from the wedge on the opposite side.

Volume of the cube: $V = (10 cm)^{3} = 1000$ cm $^{3} = 1 0^{- 3}$ m $^{3}$ .

Buoyant force when half-submerged: $F_{B} = ρ_{w} \cdot \frac{V}{2} \cdot g = 1000 \cdot \frac{1 0 ^{- 3}}{2} \cdot 10 = 5$ N.

Let the unknown mass be $m$ kg. Balancing torques about the wedge (taking distances in metres): the unknown mass exerts a downward net force $(m g - F_{B})$ at 2 cm from the pivot, while the 200 gm weight exerts $0.2 g$ at 25 cm.

$(m g - F_{B}) \cdot 0.02 = 0.2 g \cdot 0.25$ , i.e. $(10 m - 5) \cdot 0.02 = 0.2 \cdot 10 \cdot 0.25 = 0.5$ .

Solving: $10 m - 5 = 25$ , so $10 m = 30$ and $m = 3$ kg.

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JEE Main 2025 Apr 8 Shift 2, Physics Q25: Centre of Mass

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