A silver ingot weighing 2.1 kg is held by a string so s to be completely immersed in a liquid of relative density `0.8`. The relative density of silver is `0.5`. The tension in the string in `kg-wt` is

Options

A. `1.6`
B. `1.94`
C. `3.1`
D. `5.25`

Answer

To find the tension in the string holding the silver ingot immersed in a liquid, we can follow these steps:

### Step 1: Calculate the Weight of the Silver Ingot
The weight of the silver ingot can be calculated using the formula:
\[ W = m \cdot g \]
where:
- \( m = 2.1 \, \text{kg} \) (mass of the silver ingot)
- \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity)

So,
\[ W = 2.1 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 20.601 \, \text{N} \]

### Step 2: Calculate the Volume of the Silver Ingot
The relative density (RD) of silver is given as 0.5. Relative density is defined as the ratio of the density of a substance to the density of water. Therefore, we can find the density of silver:
\[ \text{Density of silver} = RD \cdot \text{Density of water} \]
Assuming the density of water is \( 1000 \, \text{kg/m}^3 \):
\[ \text{Density of silver} = 0.5 \cdot 1000 \, \text{kg/m}^3 = 500 \, \text{kg/m}^3 \]

Now, using the mass of the silver ingot, we can calculate its volume:
\[ V = \frac{m}{\text{Density}} \]
\[ V = \frac{2.1 \, \text{kg}}{500 \, \text{kg/m}^3} = 0.0042 \, \text{m}^3 \]

### Step 3: Calculate the Buoyant Force Acting on the Silver Ingot
The buoyant force can be calculated using Archimedes' principle:
\[ F_b = V \cdot \text{Density of liquid} \cdot g \]
The density of the liquid can be calculated using its relative density:
\[ \text{Density of liquid} = 0.8 \cdot 1000 \, \text{kg/m}^3 = 800 \, \text{kg/m}^3 \]

Now, substituting the values:
\[ F_b = 0.0042 \, \text{m}^3 \cdot 800 \, \text{kg/m}^3 \cdot 9.81 \, \text{m/s}^2 \]
\[ F_b = 0.0042 \cdot 800 \cdot 9.81 = 32.928 \, \text{N} \]

### Step 4: Calculate the Tension in the String
The tension in the string (T) can be calculated using the equilibrium of forces. The weight of the ingot acts downwards, and the buoyant force acts upwards. Thus, we have:
\[ T + F_b = W \]
Rearranging gives:
\[ T = W - F_b \]
Substituting the values we calculated:
\[ T = 20.601 \, \text{N} - 32.928 \, \text{N} \]
\[ T = -12.327 \, \text{N} \]

Since tension cannot be negative, this indicates that the buoyant force is greater than the weight of the silver ingot, meaning the ingot would rise to the surface. Therefore, the tension in the string is effectively zero when the ingot is fully submerged.

### Final Answer
The tension in the string is \( 0 \, \text{kg-wt} \) (or effectively zero).