Two wires of equal length and equal cross sectional areas are suspended as shown in the figure. Their Young’s modulii are $Y_{1}$ and $Y_{2},$ respectively. The equivalent Young’s modulus is

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$Y_{1} + Y_{2}$

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

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

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

answer is B.

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Detailed Solution

Given, two wires have same length and equal cross – sectional area
Question Image
i.e., $l_{1} = l_{2} = l a n d A_{1} = A_{2} = A$
Suppose, m is the mass of the hanging load.
From figure, $2 T = m g$
$i . e ., T = \frac{m g}{2}$
Hence, stress on each wire= $\frac{T}{A} = \frac{m g}{2 A}$
$∴ Y_{1} = \frac{s t r e s s}{s t r a i n} = \frac{\frac{m g}{2 A}}{\frac{Δ l}{l}} = \frac{m g l}{2 A Δ l} ...... (i)$
$Y_{2} = \frac{s t r e s s}{s t r a i n} = \frac{\frac{m g}{2 A}}{\frac{Δ l}{l}} = \frac{m g l}{2 A Δ l} ...... (i i)$
If Y be the equivalent Young’s modulus of the combination.
Then, $Y = \frac{m g l}{A Δ l} ...... (i i i)$
From Eqs. (i), (ii) amd (iii), we get
$Y = \frac{Y_{1} + Y_{2}}{2}$