# Rod of length l and negligible mass is suspended at its two ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths. The cross-sectional areas of wires A and B are 1.0 mm2 and 2.0 mm2 respectively.a) mass m should be suspended close to wire A to have equal stresses in both the wiresb) mass m should be suspended close to B to have equal stresses in both the wiresc) mass m should be suspended at the middle of the wires to have equal stresses in both the wiresd) mass m should be suspended close to wire A to have equal strain in both wires

The answer is the option (b) and (d)

$T_b x = T_a (l-x)$

$\frac{T_{b}}{T_{a}} = (\frac{l}{x}-1)$ ------------------(1)

For wire A, stress

$=\frac{T_{a}}{T_{b}}$

For wire B, stress

Since stress on steel = stress on Al

$\frac{l}{x}-1 = \frac{2}{1}$

from point b

so, distance from A =

hence m is closer to B than A. so, option b is correct.

the rod remains in a horizontal and balanced. so, the strain will also be equal.

Strain (A) = Strain (B)

$Y steel / Y Al =$

$\frac{200\times10^{9}}{70\times10^{9}}=\frac{2x}{l-x}$

$(l-x) = \frac{7}{17}$

So, for an equal amount of strain, mass m will be closer to A

Hence option d is correct.

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