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Physics Question 46 – JEE-MAIN 2025

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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/cm3.) The unknown mass is _______ kg.

For the rod to be in balance, the total clockwise torque about the wedge must equal the total counter-clockwise torque. The unknown mass experiences an upward buoyant force when submerged in water.

Step 1: Determine distances and submerged volume✦ Active

The total length of the rod is L=27 cm. The distance from the wedge to the 200 gm mass is d2=25 cm. Therefore, the distance from the wedge to the unknown mass is d1=L−d2=27 cm−25 cm=2 cm.

The unknown mass is a cube with side s=10 cm. Its total volume is V=s3=(10 cm)3=1000 cm3. When balanced, half of its volume is submerged, so the submerged volume is Vsub=12V=12(1000 cm3)=500 cm3.

💡 Teacher's Secret Hint

Ensure correct identification of pivot point and lever arms for torque calculation.

Step 2: Apply Archimedes' Principle and Torque Balance○ Expand

The density of water is ρw=1 gm/cm3. The buoyant force acting on the unknown mass is FB=ρwVsubg. The effective weight of the unknown mass is Weff=Mg−FB=(M−ρwVsub)g.

For the rod to be in equilibrium, the torques about the wedge must balance:

Weff⋅d1=m2g⋅d2 (M−ρwVsub)g⋅d1=m2g⋅d2 (M−ρwVsub)d1=m2d2
💡 Teacher's Secret Hint

Remember to account for the buoyant force reducing the effective weight of the submerged object.

Step 3: Substitute values and solve for unknown mass○ Expand

Substitute the known values into the torque balance equation:

(M−1 gm/cm3⋅500 cm3)⋅2 cm=200 gm⋅25 cm (M−500)⋅2=5000 2M−1000=5000 2M=6000 M=3000 gm

Convert the mass to kilograms:

M=3000 gm=3 kg
💡 Teacher's Secret Hint

Pay attention to units and ensure the final answer is in the requested unit (kg).

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