In the figure shown, `A` and `B` are two short steel rods each of cross-sectional area `5 cm^(2)`. The lower ends of `A` and `B` are welded to a fixed plate `CD`. The upper end of `A` is welded to the `L`-shaped piece `EFG`, which can slide without friction on upper end of `B`. A horizontal pull of `1200 N` is exerted at `G` as shown. Neglect the weight of `EFG`.
Longitudinal stress in `A` is

Options

A. (a)tensile in nature and having magnitude `180 N//m^(2)`
B. (b)tensile in nature and having magnitude `240 N//m^(2)`
C. (c)compressive in nature and having magnitude `180 N//m^(2)`
D. (d)compressive in nature and having magnitude `240 N//m^(2)`

Explanation

To solve the problem, we need to determine the longitudinal stress in rod A when a horizontal pull of 1200 N is exerted at point G.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the System**:
- There is a horizontal pull of 1200 N at point G.
- Let \( N_1 \) be the force in rod A and \( N_2 \) be the force in rod B.

2. **Establish Equilibrium Conditions**:
- Since the system is in equilibrium, the sum of vertical forces must equal zero:
\[ N_1 = N_2 \]

3. **Torque Equilibrium**:
- We can analyze the torques about point E to find the relationship between \( N_2 \) and the applied force.
- The distance from E to C (where rod B is fixed) is 4 cm, and the distance from G to F is 3 m (or 300 cm).
- The torque due to \( N_2 \) is:
\[ \text{Torque due to } N_2 = N_2 \times 4 \text{ cm} \]
- The torque due to the applied force (F) is:
\[ \text{Torque due to } F = 1200 \text{ N} \times 300 \text{ cm} \]
- Setting these equal for rotational equilibrium:
\[ N_2 \times 4 = 1200 \times 300 \]

4. **Solve for \( N_2 \)**:
- Rearranging gives:
\[ N_2 = \frac{1200 \times 300}{4} \]
- Calculate \( N_2 \):
\[ N_2 = \frac{360000}{4} = 90000 \text{ N} \]

5. **Determine the Longitudinal Stress in Rod A**:
- The longitudinal stress \( \sigma_A \) in rod A is given by:
\[ \sigma_A = \frac{N_1}{A} \]
- Since \( N_1 = N_2 \), we have:
\[ \sigma_A = \frac{90000 \text{ N}}{A} \]
- The cross-sectional area \( A \) is given as \( 5 \text{ cm}^2 \) (which is \( 5 \times 10^{-4} \text{ m}^2 \)).
- Therefore:
\[ \sigma_A = \frac{90000}{5 \times 10^{-4}} = 180000000 \text{ N/m}^2 = 180 \text{ N/cm}^2 \]

6. **Conclusion**:
- The longitudinal stress in rod A is \( 180 \text{ N/cm}^2 \) and is tensile in nature.

### Final Answer:
The longitudinal stress in rod A is \( 180 \text{ N/cm}^2 \) (tensile).