A sphete of deamrter 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 liqued is `1.527 g cm^(-3)` at` 0^@C`, find the coeffiecient of cubical expamsion of the liquid. Neglect the expansion of the sphere.

Explanation

To solve the problem step by step, we will follow the given information and apply the relevant formulas.

### Step 1: Calculate the radius of the sphere
The diameter of the sphere is given as 7.0 cm. To find the radius, we use the formula:
\[ r = \frac{d}{2} = \frac{7.0 \, \text{cm}}{2} = 3.5 \, \text{cm} \]

**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 value of \( r \):
\[ V = \frac{4}{3} \pi (3.5 \, \text{cm})^3 \]
Calculating \( (3.5)^3 \):
\[ (3.5)^3 = 42.875 \, \text{cm}^3 \]
Now substituting back:
\[ V = \frac{4}{3} \pi (42.875) \approx \frac{4}{3} \times 3.14 \times 42.875 \approx 179.594 \, \text{cm}^3 \]

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

### Step 3: Calculate the density of the sphere at 35°C
The mass \( m \) of the sphere is given as 266.5 g. The density \( \rho \) can be calculated using the formula:
\[ \rho = \frac{m}{V} \]
Substituting the values:
\[ \rho = \frac{266.5 \, \text{g}}{179.594 \, \text{cm}^3} \approx 1.484 \, \text{g/cm}^3 \]

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

### Step 4: Use the density formula to find the coefficient of cubical expansion
The density of the liquid at 0°C is given as \( \rho_0 = 1.527 \, \text{g/cm}^3 \). The density of the liquid at temperature \( t \) is given by:
\[ \rho_t = \frac{\rho_0}{1 + \gamma t} \]
Where \( \gamma \) is the coefficient of cubical expansion. We know that at \( t = 35°C \), the sphere begins to sink, meaning the density of the liquid at this temperature is equal to the density of the sphere:
\[ 1.484 \, \text{g/cm}^3 = \frac{1.527 \, \text{g/cm}^3}{1 + \gamma \cdot 35} \]

**Hint:** Set the densities equal to each other since the sphere sinks when they are equal.

### Step 5: Rearranging the equation to solve for \( \gamma \)
Rearranging the equation gives:
\[ 1 + \gamma \cdot 35 = \frac{1.527}{1.484} \]
Calculating the right side:
\[ \frac{1.527}{1.484} \approx 1.029 \]
Now, we can solve for \( \gamma \):
\[ 1 + 35\gamma = 1.029 \]
\[ 35\gamma = 1.029 - 1 \]
\[ 35\gamma = 0.029 \]
\[ \gamma = \frac{0.029}{35} \approx 0.00082857 \, \text{°C}^{-1} \]

**Hint:** To isolate \( \gamma \), subtract 1 from both sides and then divide by 35.

### Final Answer
The coefficient of cubical expansion of the liquid is approximately:
\[ \gamma \approx 8.3 \times 10^{-4} \, \text{°C}^{-1} \]