Question

A wooden cube is floating in a bucket of water with \(\frac{3}{4}\) of its volume immersed. If this bucket with the wooden block is now placed in a lift moving down with an acceleration of \(\frac{g}{2}\), the fraction of volume of the wooden cube immersed in water is:

• A. \(\frac{3}{4}\)
• B. \(\frac{3}{8}\)
• C. \(\frac{3}{2}\)
• D. \(\frac{1}{2}\)

Hint:

When the lift is accelerating down, the apparent weight of the object decreases, but the fraction of the object immersed in a fluid may remain unchanged in some conditions.

Answer 1

The Correct Option is A

- Let density of cube = \(\rho_c\), density of water = \(\rho_w\).
- Fraction immersed is given by: \[ \frac{V_{\text{immersed}}}{V} = \frac{\rho_c}{\rho_w} \]
- In a lift moving downward with acceleration \(a = \frac{g}{2}\), effective gravity becomes: \[ g' = g - a = \frac{g}{2} \]
- Weight of cube: \[ W' = \rho_c V g' \]
- Buoyant force: \[ B' = \rho_w V_{\text{immersed}} g' \]
- For floating condition \(B' = W'\): \[ \rho_w V_{\text{immersed}} g' = \rho_c V g' \]
- Cancelling \(g'\): \[ \frac{V_{\text{immersed}}}{V} = \frac{\rho_c}{\rho_w} \]
- Hence, fraction immersed remains unchanged \(= \frac{3}{4}\).