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) $\dfrac{2Y_1Y_2}{Y_1 + Y_2}$
Both rods carry the same force $F$. Extensions: $\dfrac{FL}{AY_1}$ and $\dfrac{FL}{AY_2}$.
For the combined rod of length $2L$: $\dfrac{F(2L)}{AY} = \dfrac{FL}{AY_1} + \dfrac{FL}{AY_2}$, so $\dfrac{2}{Y} = \dfrac{1}{Y_1} + \dfrac{1}{Y_2}$ and
$$Y = \frac{2Y_1Y_2}{Y_1 + Y_2}$$
Solution by Sreeraj P, M.Sc Physics