A block of silver of mass `4 kg` hanging from a string is immersed in a liquid of relative density `0.72`. If relative density of silver is `10`, then tension in the string will be

Options

A. `37.12N`
B. `42N`
C. `73N`
D. `21N`

Explanation

To find the tension in the string when a block of silver is immersed in a liquid, we can follow these steps:

### Step 1: Identify the Given Data
- Mass of the silver block, \( m = 4 \, \text{kg} \)
- Relative density of silver, \( RD_s = 10 \)
- Relative density of the liquid, \( RD_l = 0.72 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the Densities
- Density of silver, \( \rho_s = RD_s \times 1000 \, \text{kg/m}^3 = 10 \times 1000 = 10000 \, \text{kg/m}^3 \)
- Density of the liquid, \( \rho_l = RD_l \times 1000 \, \text{kg/m}^3 = 0.72 \times 1000 = 720 \, \text{kg/m}^3 \)

### Step 3: Calculate the Volume of the Silver Block
Using the formula for density:
\[ \rho = \frac{m}{V} \implies V = \frac{m}{\rho} \]
Substituting the values for the silver block:
\[ V = \frac{4 \, \text{kg}}{10000 \, \text{kg/m}^3} = 0.0004 \, \text{m}^3 \]

### Step 4: Calculate the Weight of the Silver Block
The weight of the silver block (\( W_s \)) is given by:
\[ W_s = m \cdot g = 4 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 40 \, \text{N} \]

### Step 5: Calculate the Buoyant Force
The buoyant force (\( F_B \)) acting on the block is given by Archimedes' principle:
\[ F_B = \rho_l \cdot V \cdot g \]
Substituting the values:
\[ F_B = 720 \, \text{kg/m}^3 \cdot 0.0004 \, \text{m}^3 \cdot 10 \, \text{m/s}^2 = 2.88 \, \text{N} \]

### Step 6: Apply the Force Balance
According to the equilibrium of forces:
\[ T + F_B = W_s \]
Rearranging gives:
\[ T = W_s - F_B \]
Substituting the values:
\[ T = 40 \, \text{N} - 2.88 \, \text{N} = 37.12 \, \text{N} \]

### Conclusion
The tension in the string is \( T = 37.12 \, \text{N} \).

---