Published by:
CGP EDU Academic Team
Published on: September 12, 2026
The figure shows a hollow cube of side 'a' of volume V. There is a small chamber of volume
in the cube as shown. This chamber is completely filled by m kg of water. Water leaks through a hole H and spreads in the whole cube. Then the work done by gravity in this process assuming that the complete water finally lies at the bottom of the cube is:

Text Solution
Verified by ExpertsThe correct answer is:
B
Step 1: Determine the height from which the water falls. The height at which the water is initially located is \( \frac{a}{2} \).
Step 2: The work done by gravity is calculated using the formula: \( W = mgh \), where \( m \) is the mass of water, \( g \) is the acceleration due to gravity (\( 9.81 \, \text{m/s}^2 \)), and \( h \) is the falling height (\( \frac{a}{2} \)).
Step 3: Substitute the values into the equation:
\( W = m \cdot g \cdot \frac{a}{2} \).
Step 4: The work done by gravity is thus: \( W = \frac{1}{2} \, m \, g \, a \).
Therefore, B.
Step 2: The work done by gravity is calculated using the formula: \( W = mgh \), where \( m \) is the mass of water, \( g \) is the acceleration due to gravity (\( 9.81 \, \text{m/s}^2 \)), and \( h \) is the falling height (\( \frac{a}{2} \)).
Step 3: Substitute the values into the equation:
\( W = m \cdot g \cdot \frac{a}{2} \).
Step 4: The work done by gravity is thus: \( W = \frac{1}{2} \, m \, g \, a \).
Therefore, B.
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