NAEST 2018 Screening Test: Buoyancy
By Jitender Singh
A ball is completely immersed in a partially filled glass of water. A string is used to hold the ball at two different positions. In case (A), the ball is at a depth h1 and in case (B), the ball is at a more depth h2 (h2> h1).
Question: As the ball goes from position A to B,
- The pressure at each point of the ball has increased.
- The force by the water on the ball has increased.
- The force by the string on the ball has increased.
- The level of water in the cylinder has increased.
Solution: Let the depth of the centre of the ball be $h_1$ in case (A) and it is $h_2$ in case (B). Thus, the depth of each point of the ball increases by $h=h_2-h_1$ when it moves from A to B. Thus, hydrostatic pressure at each of the ball increase by \begin{align} p_B-p_A=\rho g h=\rho g (h_2-h_1) \end{align}
The force on the ball by the water is equal to buoyant force ($F_b$) on it. By Archimedes principle, the buoyant force on the ball is equal to the weight of displaced water. This remains the same as the ball moves from A to B.
The force by the string on the ball balances the difference between the ball's weight (W) and the buoyant force ($F_b$) on it i.e., $T=W-F_b$. The forces $W$ and $F_b$ remains constant as the ball moves from A to B. Thus, the tension in the string also remains constant.
The ball is completely immersed in both cases. Thus, the water level in the glass remains the same as ball moves from A to B.
See a detailed explanation of this question by Dr HC Verma.
