Two wires of equal length and cross section area suspended as shown in figure. Their Youngs modulus are $Y_{1} a n d Y_{2}$ respectively The equivalent Youngs modulus will be

Y_{1} + Y_{2}

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

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

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\sqrt{Y_{1} Y_{2}}

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Solution

The correct option is B $\frac{Y_{1} + Y_{2}}{2}$
$k = \frac{Y A}{L}$
$k_{1} = \frac{Y_{1} A}{L}$
$k_{2} = \frac{Y_{2} A}{L}$
$k_{1}$ and $k_{2}$ are in parallel.
$k_{e q} = k_{1} + k_{2}$
$\frac{Y_{e q} (2 A)}{L} = (Y_{1} + Y_{2}) \frac{A}{L}$
$Y_{e q} = \frac{Y_{1} + Y_{2}}{2}$

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The correct option is B $\frac{Y_{1} + Y_{2}}{2}$
$k = \frac{Y A}{L}$
$k_{1} = \frac{Y_{1} A}{L}$
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$k_{1}$ and $k_{2}$ are in parallel.
$k_{e q} = k_{1} + k_{2}$
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$Y_{e q} = \frac{Y_{1} + Y_{2}}{2}$