If the temperature of body is increased by 100°C then the percentage decrease in it's density is [if `gamma` =`75 ×10^-5°C^-1`]

Options

A. 0.75
B. 0.075
C. 0.0075
D. 0.0008

Answer

To solve the problem of finding the percentage decrease in density when the temperature of a body is increased by 100°C, given that the coefficient of volumetric expansion (γ) is \(75 \times 10^{-5} °C^{-1}\), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between volume and temperature**:
The volume \(V\) of a body at a temperature \(T\) can be expressed in terms of its initial volume \(V_0\) at temperature \(T_0\) using the formula:
\[ V = V_0 (1 + \gamma \Delta T) \]
where \(\Delta T = T - T_0\).

2. **Define the change in temperature**:
Here, the change in temperature \(\Delta T\) is given as:
\[ \Delta T = 100 °C \]

3. **Substituting values into the volume formula**:
The new volume \(V\) becomes:
\[ V = V_0 (1 + \gamma \cdot 100) \]

4. **Express density in terms of volume**:
Density \(\rho\) is defined as mass \(M\) divided by volume \(V\):
\[ \rho = \frac{M}{V} \]
Initially, the density is:
\[ \rho_0 = \frac{M}{V_0} \]
After the temperature change, the new density \(\rho\) can be expressed as:
\[ \rho = \frac{M}{V_0 (1 + \gamma \cdot 100)} \]

5. **Relate the new density to the initial density**:
By substituting the expression for \(V\) into the density formula, we have:
\[ \frac{\rho}{\rho_0} = \frac{V_0}{V_0 (1 + \gamma \cdot 100)} = \frac{1}{1 + \gamma \cdot 100} \]

6. **Using the approximation for small changes**:
For small values of \(\gamma \cdot 100\), we can use the binomial approximation:
\[ \frac{1}{1 + x} \approx 1 - x \quad \text{for small } x \]
Thus:
\[ \frac{\rho}{\rho_0} \approx 1 - \gamma \cdot 100 \]

7. **Calculate the percentage decrease in density**:
The percentage decrease in density is given by:
\[ \text{Percentage Decrease} = \left( \frac{\rho_0 - \rho}{\rho_0} \right) \times 100 = \left( 1 - \frac{\rho}{\rho_0} \right) \times 100 \]
Substituting the earlier result:
\[ \text{Percentage Decrease} \approx \left(1 - (1 - \gamma \cdot 100)\right) \times 100 = \gamma \cdot 100 \times 100 \]

8. **Substituting the value of \(\gamma\)**:
Now substituting \(\gamma = 75 \times 10^{-5}\):
\[ \text{Percentage Decrease} = 75 \times 10^{-5} \times 100 \times 100 = 75 \times 10^{-5} \times 10000 = 7.5\% \]

### Final Answer:
The percentage decrease in density is **7.5%**.