Q 11-08-006NEETJEE MainHard
Two rods of equal length and equal cross-section, with Young's moduli $Y_1$ and $Y_2$, are joined end to end. The combination is stretched by a force. The effective Young's modulus of the combined rod is
Answer: (C) $\dfrac{2Y_1Y_2}{Y_1 + Y_2}$
Concept. Rods joined end to end carry the same force, and their extensions add.
Step 1. Extensions. Each rod has length $L$: $\Delta L_1 = \dfrac{FL}{AY_1}$, $\Delta L_2 = \dfrac{FL}{AY_2}$.
Step 2. Equivalent rod. Length $2L$, same area, same force:
$$\frac{F(2L)}{AY} = \frac{FL}{AY_1} + \frac{FL}{AY_2} \;\Rightarrow\; \frac{2}{Y} = \frac{1}{Y_1} + \frac{1}{Y_2}$$
Step 3. Solve.
$$Y = \frac{2Y_1Y_2}{Y_1 + Y_2}$$
Answer. $\dfrac{2Y_1Y_2}{Y_1 + Y_2}$ (Option C).
Solution by Sreeraj P, M.Sc Physics