Two identical rods of identical dimensions and Young's moduli $Y_{1}$ and $Y_{2}$ are joined end to end. The equivalent young's modulus for the composite rod is :

\frac{Y_{1} + Y_{2}}{2}

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\frac{2 Y_{1} Y_{2}}{Y_{1} + Y_{2}}

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Y_{1} + Y_{2}

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\frac{Y_{1} Y_{2}}{Y_{1} + Y_{2}}

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Solution

The correct option is B $\frac{2 Y_{1} Y_{2}}{Y_{1} + Y_{2}}$
Since they are joined end to end, the total extension will be sum of individual ones. Also, for combined material, area will be same as each of these, say A; and force will also be same.
Thus, $Δ l = Δ l_{1} + Δ l_{2} o r, \frac{F .2 L}{A . Y} = \frac{F . L}{A Y_{1}} + \frac{F . L}{A Y_{2}} \Rightarrow \frac{2}{Y} = \frac{1}{Y_{1}} + \frac{1}{Y_{2}} o r Y = \frac{2 Y_{1} Y_{2}}{Y_{1} {+ Y}_{2}}$

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Two identical rods of identical dimensions and Young's moduli Y1 and Y2 are joined end to end. The equivalent young's modulus for the composite rod is :

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