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.
Answer: 3
Solution
The cube of side has volume , so when half is submerged the displaced water volume is and the buoyant force is . The unknown mass hangs from the wedge, while the mass is away on the other side. Balancing torques about the wedge, , which gives and . The unknown mass is .
JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Integer answer