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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 wires

b) mass m should be suspended close to B to have equal stresses in both the wires

c) mass m should be suspended at the middle of the wires to have equal stresses in both the wires

d) mass m should be suspended close to wire A to have equal strain in both wires

Answers (1)

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

$\frac{T_{b}}{A_{b}}=\frac{T_{b}}{2A_{a}}$

Since stress on steel = stress on Al

$\frac{T_{a}}{A_{a}}=\frac{T_{b}}{2A_{a}}\rightarrow T_{a}=\frac{T_{b}}{2}$

$\frac{T_{b}}{T_{a}}=2$

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

$x = \frac{l}{3}$ from point b

so, distance from A = $l-x = l-\frac{l}{3} = \frac{2l}{3}$

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)

$\frac{S_{a}}{Y_{a}} =\frac{S_{b}}{Y_{b}}$

$Y steel / Y Al =$ $(\frac{T_a}{T_{b}}) (\frac{A_{b}}{A_{a}}) = (\frac{x}{l}-x) (\frac{2Aa}{Aa})$

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

$14 x = 20l - 20x$

$34x = 20l$

$x = \frac{20l}{34} = \frac{10l}{17}$

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

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

Hence option d is correct.

Posted by

infoexpert24

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