StreakPeaked· Practice

ExamsJEE MainPhysics

A rectangular block of ice with cross-sectional area A and height h floats on water of density rho0. The density of ice is rho1. Find the work that must be done to push the ice block until it is just completely submerged in the water.

  1. (rho0 - rho1)² g A h² / (2 rho0)
  2. (rho0 - rho1) g A h² / (2 rho0)
  3. rho1² g A h² / (2 rho0)
  4. (rho0 - rho1) g A h / (2 rho0)

Correct answer: (rho0 - rho1)² g A h² / (2 rho0)

Solution

Floating: weight = buoyancy, so rho1*A*h*g = rho0*A*d*g, giving submerged depth d = (rho1/rho0)h. The length above water is x0 = h - d = h(rho0 - rho1)/rho0. To submerge fully you must push down by x0. When pushed down an extra distance x, the net extra upward force is rho0*A*x*g (acts like a spring with k = rho0*A*g). Work = integral from 0 to x0 of rho0*A*g*x dx = (1/2) rho0*A*g*x0² = (1/2) rho0*A*g*[h(rho0 - rho1)/rho0]² = (rho0 - rho1)² g A h² / (2 rho0).

Related JEE Main Physics questions

⚔️ Practice JEE Main Physics free + battle 1v1 →