A sphere of diameter 7.0 cm and mass 266.5 g float in a bath of liquid. As the temperature is raised, the sphere begins to sink at a temperature of `35^@ C`. If the density of liquid is `1.527 g cm^(-3)` at` 0^@C`, find the coefficient of cubical expansion of the liquid. Neglect the expansion of the sphere.

Options

A. `0.0043//^(@)C`
B. `0.00083//^(@)C`
C. `0.00025//^(@)C`
D. `0.00010//^(@)C`

Explanation

To solve the problem, we need to find the coefficient of cubical expansion of the liquid based on the given information. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the radius of the sphere
The diameter of the sphere is given as 7.0 cm. Therefore, the radius \( R \) can be calculated as:
\[ R = \frac{\text{Diameter}}{2} = \frac{7.0 \, \text{cm}}{2} = 3.5 \, \text{cm} \]

**Hint:** Remember that the radius is half of the diameter.

### Step 2: Calculate the volume of the sphere
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi R^3 \]
Substituting the radius:
\[ V = \frac{4}{3} \pi (3.5 \, \text{cm})^3 = \frac{4}{3} \pi (42.875 \, \text{cm}^3) \approx 179.594 \, \text{cm}^3 \]

**Hint:** Use \( \pi \approx 3.14 \) for calculations.

### Step 3: Calculate the density of the sphere
The density \( \rho_s \) of the sphere can be calculated using the formula:
\[ \rho_s = \frac{\text{mass}}{\text{volume}} = \frac{266.5 \, \text{g}}{179.594 \, \text{cm}^3} \approx 1.483 \, \text{g/cm}^3 \]

**Hint:** Density is mass divided by volume.

### Step 4: Set up the equation for the density of the liquid at 35°C
At the temperature of 35°C, the density of the liquid \( \rho_l \) must equal the density of the sphere for it to sink:
\[ \rho_l(35°C) = 1.483 \, \text{g/cm}^3 \]

### Step 5: Use the formula for density change with temperature
The relationship between the initial density \( \rho_0 \) at 0°C and the density at a higher temperature is given by:
\[ \rho_l(T) = \rho_0 \left(1 - \gamma \Delta T\right) \]
Where:
- \( \rho_0 = 1.527 \, \text{g/cm}^3 \) (density at 0°C)
- \( \Delta T = 35°C - 0°C = 35°C \)
- \( \gamma \) is the coefficient of cubical expansion.

Setting the two densities equal gives:
\[ 1.483 = 1.527 \left(1 - \gamma \times 35\right) \]

**Hint:** Rearranging the equation helps isolate the variable you want to solve for.

### Step 6: Solve for \( \gamma \)
Rearranging the equation:
\[ 1 - \gamma \times 35 = \frac{1.483}{1.527} \]
Calculating the right side:
\[ 1 - \gamma \times 35 \approx 0.971 \]
Thus:
\[ \gamma \times 35 = 1 - 0.971 = 0.029 \]
\[ \gamma = \frac{0.029}{35} \approx 0.00083 \, \text{per °C} \]

**Hint:** Make sure to divide correctly to find the coefficient.

### Final Answer
The coefficient of cubical expansion of the liquid is approximately:
\[ \gamma \approx 0.00083 \, \text{per °C} \]