The loss of weight of a solid when immersed in a liquid at `0^(@)C` is `W_(0)` and at `t^(@)C` is `'W'`. If cubical coefficient of expansion of the solid and the liquid are `gamma_(s)` and `gamma_(1)` then `W =`

Options

A. `W_(0)[1+(gamma_(s)-gamma_(L))t]`
B. `W_(0)[1-(gamma_(s)-gamma_(L)t]`
C. `W_(0)[(gamma_(s)-gamma_(L))t]`
D. `W_(0)t//(gamma_(s)-gamma_(L))`

Explanation

To solve the problem, we need to find the relationship between the loss of weight of a solid when immersed in a liquid at two different temperatures, given the cubical coefficients of expansion for both the solid and the liquid.

### Step-by-Step Solution:

1. **Understanding the Loss of Weight**:
The loss of weight of a solid when immersed in a liquid is equal to the buoyant force acting on it. This buoyant force depends on the density of the liquid and the volume of the solid.

2. **Initial Conditions**:
At `0°C`, the loss of weight is given as `W₀`. At a temperature of `t°C`, we denote the loss of weight as `W`.

3. **Buoyant Force Expression**:
The buoyant force can be expressed as:
\[ F_b = \rho_{liquid} \cdot V_{solid} \cdot g \]
where \( \rho_{liquid} \) is the density of the liquid, \( V_{solid} \) is the volume of the solid, and \( g \) is the acceleration due to gravity.

4. **Density Changes with Temperature**:
The density of the liquid changes with temperature according to its cubical coefficient of expansion \( \gamma_{l} \):
\[ \rho_{liquid}(t) = \frac{\rho_{liquid}(0)}{1 + \gamma_{l} \cdot t} \]
where \( \rho_{liquid}(0) \) is the density of the liquid at `0°C`.

5. **Volume Changes with Temperature**:
The volume of the solid also changes with temperature according to its cubical coefficient of expansion \( \gamma_{s} \):
\[ V_{solid}(t) = V_{solid}(0) \cdot (1 + \gamma_{s} \cdot t) \]
where \( V_{solid}(0) \) is the volume of the solid at `0°C`.

6. **Expressing W in Terms of W₀**:
At `0°C`, the loss of weight is:
\[ W₀ = \rho_{liquid}(0) \cdot V_{solid}(0) \cdot g \]
At `t°C`, the loss of weight is:
\[ W = \rho_{liquid}(t) \cdot V_{solid}(t) \cdot g \]
Substituting the expressions for \( \rho_{liquid}(t) \) and \( V_{solid}(t) \):
\[ W = \left(\frac{\rho_{liquid}(0)}{1 + \gamma_{l} \cdot t}\right) \cdot \left(V_{solid}(0) \cdot (1 + \gamma_{s} \cdot t)\right) \cdot g \]

7. **Relating W to W₀**:
By substituting \( W₀ \) into the equation:
\[ W = W₀ \cdot \frac{(1 + \gamma_{s} \cdot t)}{(1 + \gamma_{l} \cdot t)} \]
For small values of \( \gamma_{s} \cdot t \) and \( \gamma_{l} \cdot t \), we can use the approximation:
\[ \frac{1 + \gamma_{s} \cdot t}{1 + \gamma_{l} \cdot t} \approx 1 + (\gamma_{s} - \gamma_{l}) \cdot t \]
Therefore, we can express \( W \) as:
\[ W \approx W₀ \cdot (1 + (\gamma_{s} - \gamma_{l}) \cdot t) \]

### Final Expression:
Thus, the final expression for the loss of weight at temperature `t°C` is:
\[ W = W₀ \cdot (1 + (\gamma_{s} - \gamma_{l}) \cdot t) \]