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.

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