The Young's modulus of a metal is `2 xx 10^(12) " dyne/cm^(2))` and its breaking stress is `11000 kg//cm^(2)`. In case of longitudinal strain the maximum energy that can be stored per cubic metre of this metal is approximately (Assume `g= 10 m//s^(2)`)

Options

A. `58.28 xx 10^(5) J`
B. `30.25 xx 10^(5) J`
C. `37.15 xx 10^(5) J`
D. `15.15 xx 10^(5) J`

Explanation

To solve the problem, we need to find the maximum energy that can be stored per cubic meter of the metal given its Young's modulus and breaking stress.

### Step-by-Step Solution:

1. **Understanding the Given Values:**
- Young's modulus (E) = \(2 \times 10^{12} \, \text{dyne/cm}^2\)
- Breaking stress (\(\sigma_{max}\)) = \(11000 \, \text{kg/cm}^2\)

2. **Convert Units:**
- Convert Young's modulus from dyne/cm² to N/m² (Pascal):
\[ 1 \, \text{dyne} = 10^{-5} \, \text{N} \]
\[ 1 \, \text{cm}^2 = 10^{-4} \, \text{m}^2 \]
Therefore,
\[ E = 2 \times 10^{12} \, \text{dyne/cm}^2 = 2 \times 10^{12} \times 10^{-5} \, \text{N} / 10^{-4} \, \text{m}^2 = 2 \times 10^{12} \times 10^{4} \, \text{N/m}^2 = 2 \times 10^{16} \, \text{N/m}^2 \]

- Convert breaking stress from kg/cm² to N/m²:
\[ \sigma_{max} = 11000 \, \text{kg/cm}^2 = 11000 \times 10^{3} \, \text{N} / 10^{-4} \, \text{m}^2 = 11000 \times 10^{3} \times 10^{4} \, \text{N/m}^2 = 1.1 \times 10^{8} \, \text{N/m}^2 \]

3. **Calculate Maximum Energy per Unit Volume:**
- The maximum energy (U) that can be stored per unit volume in a material under elastic deformation is given by the formula:
\[ U = \frac{\sigma_{max}^2}{2E} \]
- Substitute the values:
\[ U = \frac{(1.1 \times 10^{8})^2}{2 \times (2 \times 10^{16})} \]

4. **Calculate the Values:**
- First, calculate \((1.1 \times 10^{8})^2\):
\[ (1.1 \times 10^{8})^2 = 1.21 \times 10^{16} \]
- Now substitute this back into the energy formula:
\[ U = \frac{1.21 \times 10^{16}}{4 \times 10^{16}} = \frac{1.21}{4} \times 10^{0} = 0.3025 \, \text{J/m}^3 \]

5. **Final Result:**
- The maximum energy that can be stored per cubic meter of this metal is approximately \(0.3025 \, \text{J/m}^3\).