A sphere of mass `180g` and diameter `6 cm` floats on the surface of a liquid. When the liquid is heated to `35^(0)C`, the sphere sinks in the liquids. If the density of liquid at `0^(0)C` is `2 gcm^(-5)`. The coefficient of real expansion of liquid is

Options

A. `71.4 xx 10^(-4)//^(@)C`
B. `81.4 xx 10^(-4)//^(@)C`
C. `91.4 xx 10^(-4)//^(@)C`
D. `61.4 xx 10^(-4)//^(@)C`

Explanation

To find the coefficient of real expansion of the liquid, we can follow these steps:

### Step 1: Understand the problem
We have a sphere of mass 180 g and diameter 6 cm that floats on a liquid. When the liquid is heated to 35°C, the sphere sinks. We know the density of the liquid at 0°C is 2 g/cm³. We need to find the coefficient of real expansion (γ) of the liquid.

### Step 2: Calculate the volume of the sphere
The volume (V) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
First, we need to find the radius (r) of the sphere:
- Diameter = 6 cm, so radius \( r = \frac{6}{2} = 3 \) cm.

Now, substituting the radius into the volume formula:
\[ V = \frac{4}{3} \pi (3)^3 = \frac{4}{3} \pi (27) = 36\pi \text{ cm}^3 \]

### Step 3: Calculate the density of the sphere
Density (d) is defined as mass (m) divided by volume (V):
\[ d = \frac{m}{V} \]
Substituting the values:
- Mass of the sphere \( m = 180 \) g,
- Volume of the sphere \( V = 36\pi \text{ cm}^3 \).

Calculating the density:
\[ d = \frac{180}{36\pi} \approx \frac{180}{113.1} \approx 1.59 \text{ g/cm}^3 \]

### Step 4: Use the formula for the coefficient of real expansion
The relationship between the densities and the coefficient of real expansion is given by:
\[ \gamma = \frac{d_0}{d_T} - 1 \times \frac{1}{\Delta T} \]
Where:
- \( d_0 \) = density at 0°C = 2 g/cm³,
- \( d_T \) = density at 35°C (which we calculated as 1.59 g/cm³),
- \( \Delta T = 35 - 0 = 35°C \).

### Step 5: Substitute the values into the formula
Now substituting the values into the formula:
\[ \gamma = \frac{2}{1.59} - 1 \times \frac{1}{35} \]

Calculating \( \frac{2}{1.59} \):
\[ \frac{2}{1.59} \approx 1.26 \]

Now substituting back:
\[ \gamma = (1.26 - 1) \times \frac{1}{35} \]
\[ \gamma = 0.26 \times \frac{1}{35} \]
\[ \gamma \approx \frac{0.26}{35} \approx 0.00743 \text{ per °C} \]
\[ \gamma \approx 73.6 \times 10^{-4} \text{ per °C} \]

### Final Answer
The coefficient of real expansion of the liquid is approximately \( 73.6 \times 10^{-4} \text{ per °C} \).

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