A ball of mass m and density `rho` is immersed in a liquid of density `3 rho` at a depth `h` and released. To what height will the ball jump up above the surface of liqud ? (neglect the reistance of water and air).

Options

A. h
B. `h//2`
C. `2h`
D. `3h`

Answer

To solve the problem of how high a ball of mass \( m \) and density \( \rho \) will jump above the surface of a liquid of density \( 3\rho \) when released from a depth \( h \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Ball:**
- The ball experiences two main forces when submerged in the liquid:
- **Weight (W)**: The weight of the ball is given by:
\[ W = \rho V g \]
where \( V \) is the volume of the ball and \( g \) is the acceleration due to gravity.
- **Buoyant Force (F_b)**: The buoyant force acting on the ball is given by Archimedes' principle:
\[ F_b = 3\rho V g \]
since the density of the liquid is \( 3\rho \).

2. **Calculate the Net Upward Force:**
- The net upward force \( F_{net} \) acting on the ball when it is released can be calculated as:
\[ F_{net} = F_b - W = 3\rho V g - \rho V g = 2\rho V g \]

3. **Determine the Acceleration of the Ball:**
- Using Newton's second law, the acceleration \( a \) of the ball can be found:
\[ a = \frac{F_{net}}{m} = \frac{2\rho V g}{\rho V} = 2g \]
- Here, the mass \( m \) of the ball is \( \rho V \), which cancels out.

4. **Calculate the Velocity of the Ball at the Surface:**
- Using the kinematic equation \( v^2 = u^2 + 2as \), where:
- \( u = 0 \) (initial velocity),
- \( a = 2g \) (acceleration),
- \( s = h \) (distance submerged),
- We find:
\[ v^2 = 0 + 2 \cdot 2g \cdot h = 4gh \]
- Therefore, the velocity \( v \) of the ball when it reaches the surface of the liquid is:
\[ v = 2\sqrt{gh} \]

5. **Determine the Height the Ball Will Rise Above the Surface:**
- Once the ball reaches the surface, it will continue to move upward until its velocity becomes zero. We again use the kinematic equation:
\[ 0 = v^2 - 2gs \]
where \( s \) is the height above the surface of the liquid that we want to find. Rearranging gives:
\[ s = \frac{v^2}{2g} = \frac{(2\sqrt{gh})^2}{2g} = \frac{4gh}{2g} = 2h \]

Thus, the height to which the ball will jump above the surface of the liquid is \( 2h \).

### Final Answer:
The ball will jump to a height of \( 2h \) above the surface of the liquid.