Question

Three steel rods having lengths 1L, 3L, 3L and cross-sectional areas A, 2A, 3A respectively are joined in the form of a single rod. If the cross rod is subjected to a stretching force F, then the increase in length is: (Y = Young's modulus of steel)

• A. $\frac{7FL}{2AY}$
• B. $\frac{FL}{AY}$
• C. $\frac{FL}{3AY}$
• D. $\frac{AY}{3FL}$

Hint:

For rods in series, add the elongations of each segment. Use the formula $\Delta L = \frac{FL}{AY}$ for each rod and sum them up.

Answer 1

The Correct Option is A

The rods are in series, so the total elongation is the sum of elongations of each rod. The formula for elongation is $\Delta L = \frac{FL}{AY}$, where $L$ is the length, $A$ is the area, and $Y$ is Young's modulus. 
For rod 1: $L_1 = L$, $A_1 = A$, elongation $\Delta L_1 = \frac{F \cdot L}{A \cdot Y}$. 
For rod 2: $L_2 = 3L$, $A_2 = 2A$, elongation $\Delta L_2 = \frac{F \cdot 3L}{2A \cdot Y}$. 
For rod 3: $L_3 = 3L$, $A_3 = 3A$, elongation $\Delta L_3 = \frac{F \cdot 3L}{3A \cdot Y} = \frac{F \cdot L}{A \cdot Y}$. 
Total elongation $\Delta L = \Delta L_1 + \Delta L_2 + \Delta L_3 = \frac{FL}{AY} + \frac{3FL}{2AY} + \frac{FL}{AY}$. 
Combine: $\Delta L = \frac{FL}{AY} (1 + \frac{3}{2} + 1) = \frac{FL}{AY} \left(1 + 1.5 + 1\right) = \frac{FL}{AY} \times \frac{7}{2} = \frac{7FL}{2AY}$.