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ExamsJEE MainPhysics

A cubical metal block with edge 12 cm floats on mercury so that one-fifth of its height is submerged in the mercury. Water is then added until the top surface of the block is exactly level with the water surface. Find the height of the water column that must be added. (Specific gravity of mercury = 13.6.)

  1. 10.4 cm
  2. 9.6 cm
  3. 8.4 cm
  4. 12 cm

Correct answer: 10.4 cm

Solution

Edge L = 12 cm. Initially 1/5 submerged in mercury: density of block = (1/5)*13.6 = 2.72 g/cc. After water is poured, let x = depth of block in mercury and (12 - x) = depth in water (water just reaches the top of the block). Weight balance per unit area: rho_block*12 = rho_water*(12 - x) + rho_Hg*x. So 2.72*12 = 1*(12 - x) + 13.6*x => 32.64 = 12 - x + 13.6x = 12 + 12.6x => 12.6x = 20.64 => x = 1.638 cm in mercury. Water column height = 12 - x = 12 - 1.638 = 10.36 ~ 10.4 cm.

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