A metallic bob weights `50 g` in air. If it is immersed in a liquid at a temperature of `25^@ C`, it weights `45 g` . When the temperature of the liquid is raised to `100^@ C`, it weights `45.1 g`. Calculate the coefficient of cubical expansion of the liquid. Given that coefficient of cubical expansion of the metal is `12 xx 10^(-6) .^@ C^-1`.

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the problem
We have a metallic bob that weighs 50 g in air and 45 g when immersed in a liquid at 25°C. When the temperature of the liquid is raised to 100°C, the weight of the bob becomes 45.1 g. We need to calculate the coefficient of cubical expansion of the liquid.

### Step 2: Identify the relevant equations
The change in weight of the bob when immersed in the liquid can be related to the buoyant force acting on it. The relationship can be expressed as:
\[ \frac{\Delta W'}{\Delta W} = \frac{F'}{F} = \frac{1 + \gamma_s \Delta \theta}{1 + \gamma_l \Delta \theta} \]
where:
- \(\Delta W' = 50 - 45.1\) (weight in air - weight in liquid at 100°C)
- \(\Delta W = 50 - 45\) (weight in air - weight in liquid at 25°C)
- \(\gamma_s\) = coefficient of cubical expansion of the metal = \(12 \times 10^{-6} \, ^\circ C^{-1}\)
- \(\Delta \theta = 100 - 25 = 75^\circ C\) (temperature change)

### Step 3: Calculate the changes in weight
Calculate \(\Delta W'\) and \(\Delta W\):
\[ \Delta W' = 50 - 45.1 = 4.9 \, \text{g} \]
\[ \Delta W = 50 - 45 = 5 \, \text{g} \]

### Step 4: Substitute the values into the equation
Substituting the values into the equation:
\[ \frac{4.9}{5} = \frac{1 + (12 \times 10^{-6} \times 75)}{1 + \gamma_l \times 75} \]

### Step 5: Simplify the equation
Calculate \(12 \times 10^{-6} \times 75\):
\[ 12 \times 10^{-6} \times 75 = 9 \times 10^{-4} \]
Thus, the equation becomes:
\[ \frac{4.9}{5} = \frac{1 + 9 \times 10^{-4}}{1 + \gamma_l \times 75} \]

### Step 6: Cross-multiply to solve for \(\gamma_l\)
Cross-multiplying gives:
\[ 4.9(1 + \gamma_l \times 75) = 5(1 + 9 \times 10^{-4}) \]
Expanding both sides:
\[ 4.9 + 4.9 \gamma_l \times 75 = 5 + 4.5 \times 10^{-3} \]

### Step 7: Rearranging the equation
Rearranging to isolate \(\gamma_l\):
\[ 4.9 \gamma_l \times 75 = 5 + 4.5 \times 10^{-3} - 4.9 \]
\[ 4.9 \gamma_l \times 75 = 0.1 + 4.5 \times 10^{-3} \]
\[ 4.9 \gamma_l \times 75 = 0.1045 \]

### Step 8: Solve for \(\gamma_l\)
Now, divide both sides by \(4.9 \times 75\):
\[ \gamma_l = \frac{0.1045}{4.9 \times 75} \]
Calculating the denominator:
\[ 4.9 \times 75 = 367.5 \]
Thus:
\[ \gamma_l = \frac{0.1045}{367.5} \approx 2.84 \times 10^{-4} \, ^\circ C^{-1} \]

### Final Result
The coefficient of cubical expansion of the liquid is approximately:
\[ \gamma_l \approx 2.84 \times 10^{-4} \, ^\circ C^{-1} \]