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-gamme_L)`

Explanation

To solve the problem, we will use the principles of buoyancy and the concept of thermal expansion. The loss of weight of a solid when immersed in a liquid can be determined by considering the change in volume and density of both the solid and the liquid due to temperature changes.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- Let the loss of weight of the solid when immersed in the liquid at \(0^\circ C\) be \(W_0\).
- Let the loss of weight of the solid when immersed in the liquid at \(t^\circ C\) be \(W\).
- The cubical coefficient of expansion for the solid is \(\gamma_s\) and for the liquid is \(\gamma_l\).

2. **Applying Archimedes' Principle**:
- According to Archimedes' principle, the loss of weight \(W\) is equal to the weight of the liquid displaced by the solid.
- This can be expressed as:
\[ W = V_t \cdot \rho_l \cdot g \]
where \(V_t\) is the volume of the solid at temperature \(t\), \(\rho_l\) is the density of the liquid at temperature \(t\), and \(g\) is the acceleration due to gravity.

3. **Volume and Density Changes**:
- The volume of the solid at temperature \(t\) can be expressed as:
\[ V_t = V_0 (1 + \gamma_s \cdot t) \]
where \(V_0\) is the original volume of the solid at \(0^\circ C\).
- The density of the liquid at temperature \(t\) can be expressed as:
\[ \rho_l = \frac{\rho_0}{1 + \gamma_l \cdot t} \]
where \(\rho_0\) is the original density of the liquid at \(0^\circ C\).

4. **Substituting Values**:
- Substitute \(V_t\) and \(\rho_l\) into the equation for \(W\):
\[ W = V_0 (1 + \gamma_s \cdot t) \cdot \left(\frac{\rho_0}{1 + \gamma_l \cdot t}\right) \cdot g \]

5. **Expressing \(W_0\)**:
- At \(0^\circ C\), the loss of weight \(W_0\) can be expressed as:
\[ W_0 = V_0 \cdot \rho_0 \cdot g \]

6. **Finding the Ratio \(\frac{W}{W_0}\)**:
- Now, we can find the ratio \(\frac{W}{W_0}\):
\[ \frac{W}{W_0} = \frac{V_0 (1 + \gamma_s \cdot t) \cdot \left(\frac{\rho_0}{1 + \gamma_l \cdot t}\right) \cdot g}{V_0 \cdot \rho_0 \cdot g} \]
- This simplifies to:
\[ \frac{W}{W_0} = \frac{(1 + \gamma_s \cdot t)}{(1 + \gamma_l \cdot t)} \]

7. **Final Expression for \(W\)**:
- Rearranging gives:
\[ W = W_0 \cdot \frac{(1 + \gamma_s \cdot t)}{(1 + \gamma_l \cdot t)} \]
- This can be further simplified to:
\[ W = W_0 \cdot \left(1 + \gamma_s \cdot t - \gamma_l \cdot t\right) \]

### Final Result:
\[ W = W_0 \cdot \left(1 + (\gamma_s - \gamma_l) \cdot t\right) \]