A ball submerges completely and floats inside a liquid of density p If same ball is dropped in a liquid of density 4p, then the percentage volume of the ball immersed inside this new liquid will be?

Options

A. 0.25
B. 0.5
C. 0.75
D. 0.2

Answer

To solve the problem step by step, we can follow these logical steps:

### Step 1: Understand the Initial Condition
We have a ball that is completely submerged in a liquid of density \( \rho \) and is floating. This indicates that the buoyant force acting on the ball is equal to the weight of the ball.

### Step 2: Write the Equilibrium Condition
For the ball to float, the buoyant force \( F_B \) must equal the weight \( W \) of the ball. Mathematically, this can be expressed as:
\[ F_B = W \]
Where:
- \( F_B = \rho \cdot V \cdot g \) (buoyant force)
- \( W = \rho_b \cdot V \cdot g \) (weight of the ball)

Here, \( V \) is the volume of the ball, \( g \) is the acceleration due to gravity, and \( \rho_b \) is the density of the ball.

### Step 3: Set Up the Equation
Since the ball is completely submerged in the first liquid:
\[ \rho \cdot V \cdot g = \rho_b \cdot V \cdot g \]
Cancelling \( V \cdot g \) from both sides gives:
\[ \rho = \rho_b \]
This means the density of the ball is equal to the density of the first liquid.

### Step 4: Analyze the New Condition
Now, we drop the same ball into a new liquid with a density of \( 4\rho \). In this case, only a portion of the ball will be submerged.

### Step 5: Write the New Equilibrium Condition
Let \( V' \) be the volume of the ball that is submerged in the new liquid. The equilibrium condition now becomes:
\[ F_B = W \]
Where:
\[ F_B = 4\rho \cdot V' \cdot g \]
And the weight remains the same:
\[ W = \rho_b \cdot V \cdot g \]
Substituting \( \rho_b \) with \( \rho \) (from Step 3):
\[ W = \rho \cdot V \cdot g \]

### Step 6: Set Up the New Equation
Setting the two forces equal gives:
\[ 4\rho \cdot V' \cdot g = \rho \cdot V \cdot g \]
Cancelling \( g \) from both sides and \( \rho \) gives:
\[ 4V' = V \]
Thus, we find:
\[ V' = \frac{V}{4} \]

### Step 7: Calculate the Percentage Volume Immersed
To find the percentage of the volume of the ball that is submerged in the new liquid:
\[ \text{Percentage Volume Immersed} = \left(\frac{V'}{V}\right) \times 100 = \left(\frac{V/4}{V}\right) \times 100 = 25\% \]

### Final Answer
The percentage volume of the ball immersed inside the new liquid is **25%**.

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