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

Answer

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

### Step 1: Understand the Forces Acting on the Block
The forces acting on the block are:
- The weight of the block (W)
- The buoyant force (B) acting upwards
- The tension (T) in the string acting upwards

The weight of the block can be calculated using the formula:
\[ W = mg \]
where:
- \( m \) = mass of the block = 4 kg
- \( g \) = acceleration due to gravity ≈ 10 m/s²

### Step 2: Calculate the Weight of the Block
Using the values given:
\[ W = 4 \, \text{kg} \times 10 \, \text{m/s}^2 = 40 \, \text{N} \]

### Step 3: Calculate the Buoyant Force
The buoyant force can be calculated using the formula:
\[ B = \text{density of fluid} \times g \times V \]
where \( V \) is the volume of the fluid displaced, which is equal to the volume of the block.

The volume of the block can be calculated using its mass and density:
\[ V = \frac{m}{\text{density of block}} \]
Given that the relative density of silver is 10, we can find the density of silver:
\[ \text{Density of silver} = 10 \times 1000 \, \text{kg/m}^3 = 10000 \, \text{kg/m}^3 \]

Now, calculate the volume of the block:
\[ V = \frac{4 \, \text{kg}}{10000 \, \text{kg/m}^3} = 0.0004 \, \text{m}^3 \]

Now we can calculate the buoyant force:
Given that the relative density of the liquid is 0.72, the density of the liquid is:
\[ \text{Density of liquid} = 0.72 \times 1000 \, \text{kg/m}^3 = 720 \, \text{kg/m}^3 \]

Now calculate the buoyant force:
\[ B = 720 \, \text{kg/m}^3 \times 10 \, \text{m/s}^2 \times 0.0004 \, \text{m}^3 = 2.88 \, \text{N} \]

### Step 4: Apply the Equation of Motion
Using the equilibrium of forces, we have:
\[ T + B = W \]
Thus, we can rearrange this to find the tension:
\[ T = W - B \]

### Step 5: Substitute the Values
Now substitute the values we calculated:
\[ T = 40 \, \text{N} - 2.88 \, \text{N} = 37.12 \, \text{N} \]

### Conclusion
The tension in the string is:
\[ T = 37.12 \, \text{N} \]

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