Question

A tank of height \(5\,m\) is completely filled with water and a cube of side \(1\,cm\) and density \(1.5\,g\,cm^{-3}\) is placed at the bottom of the tank. The work to be done to lift the cube at the bottom to a height of \(15\,m\) above the surface of water is:

• A. \(25\,mJ\)
• B. \(300\,mJ\)
• C. \(250\,mJ\)
• D. \(225\,mJ\)

Hint:

Inside a liquid, use effective weight: \[ W_{eff}=mg-\rho Vg. \] Above the liquid, buoyancy disappears and only \(mg\) acts.

Answer 1

The Correct Option is D

Concept: The external work required equals the increase in gravitational potential energy minus the work done by buoyancy while the cube remains inside water.

Step 1: Calculate volume and mass of the cube.
Side \[ a=1\,cm=10^{-2}m. \] Volume \[ V=a^3=10^{-6}m^3. \] Density \[ \rho=1.5\,g\,cm^{-3}=1500\,kg\,m^{-3}. \] Mass \[ m=\rho V =1500\times10^{-6} =1.5\times10^{-3}kg. \]

Step 2: Compute actual weight.
\[ W_g=mg =(1.5\times10^{-3})(10) =1.5\times10^{-2}N. \]

Step 3: Calculate buoyant force.
\[ F_b=\rho_w V g. \] Taking \[ \rho_w=1000\,kg\,m^{-3}, \] \[ F_b = 1000\times10^{-6}\times10 = 10^{-2}N. \]

Step 4: Work done inside water.
Cube rises from bottom to surface through \[ 5m. \] Effective force \[ F=W_g-F_b = 1.5\times10^{-2}-10^{-2} = 5\times10^{-3}N. \] Work done \[ W_1=F\times5 = 25\times10^{-3}J. \] \[ W_1=25\,mJ. \]

Step 5: Work done above water.
Distance above water surface \[ 15m. \] Now only weight acts. \[ W_2=mg(15) = 1.5\times10^{-2}\times15. \] \[ W_2=225\,mJ. \]

Step 6: Total work.
\[ W=W_1+W_2 = 25+225 = 250\,mJ. \] Thus \[ \boxed{250\,mJ} \]