Question

A tank filled with water to a height of \(1.5\,\text{m}\) contains a cube of side \(10\,\text{cm}\) and density \(1.5\,\text{g cm}^{-3}\) resting at the bottom. The tank is placed in a lift moving downward with an acceleration \(2\,\text{m s}^{-2}\). The work required to pull the cube vertically upward through a distance of \(1\,\text{m}\) is \[ (g=10\,\text{m s}^{-2}) \]

• A. \(5\,\text{J}\)
• B. \(15\,\text{J}\)
• C. \(12\,\text{J}\)
• D. \(4\,\text{J}\)

Hint:

In an accelerating lift, \[ \boxed{ g_{\text{eff}}=g-a } \] Work done while lifting in a liquid: \[ \boxed{ W=(\text{Weight}-\text{Buoyant force})\times \text{distance} } \]

Answer 1

The Correct Option is D

Step 1: Calculate effective gravity. Since the lift moves downward with acceleration \(2\,\text{m s}^{-2}\), \[ g_{\text{eff}}=g-a=10-2=8\,\text{m s}^{-2}. \]

Step 2: Calculate the weight and buoyant force. Volume of the cube, \[ V=(0.1)^3=10^{-3}\,\text{m}^3. \] Density of cube, \[ \rho_c=1.5\,\text{g cm}^{-3}=1500\,\text{kg m}^{-3}. \] Weight, \[ W=\rho_c V g_{\text{eff}} =1500\times10^{-3}\times8 =12\,\text{N}. \] Buoyant force, \[ F_B=\rho_w V g_{\text{eff}} =1000\times10^{-3}\times8 =8\,\text{N}. \]

Step 3: Find the work done. Required pulling force, \[ F=W-F_B=12-8=4\,\text{N}. \] Distance moved, \[ s=1\,\text{m}. \] Therefore, \[ W=Fs=4\times1=4\,\text{J}. \] Hence, \[ \boxed{4\,\text{J}} \] Therefore, \[ \boxed{(D)} \] is the correct answer.