A cube having a side of 10 cm with unknown mass and 200 gm mass were hung at two ends of a 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.
Answer
3
Solution
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Distance of the cube from the wedge point = 27 - 25 = 2 cm (the weight of the rod is not given, so it is neglected).
Volume of cube = 103 = 1000 cm3; volume inside water = 500 cm3
Buoyant force FB = 500 cm3 1 g/cm3 g = 0.5g N (with masses in kg)
Taking moments about the wedge point:
(Mg - 0.5g) 2 = 0.2g 25
M - 0.5 = 2.5
M = 3 kg
Answer: 3