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A two metre long rod is suspended with the help of two wires of equal length. One wire is of steel and its cross-sectional area is 0.1 cm² and another wire is of brass and its cross-sectional area is 0.2 cm². If a load W is suspended from the rod and stress produced in both the wires is same then the ratio of tensions in them will be:

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Will depend on the position of W

$\frac{T_{1}}{T_{2}} = 0.5$

$\frac{T_{1}}{T_{2}} = 1$

$\frac{T_{1}}{T_{2}} = 2$

answer is D.

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Detailed Solution

$S t r e s s = \frac{T e n s i o n}{A r e a o f c r o s s - s e c t i o n} = c o n s \tan t ∴ \frac{T_{1}}{A_{1}} = \frac{T_{2}}{A_{2}} \Rightarrow \frac{T_{1}}{T_{2}} = \frac{A_{1}}{A_{2}} = \frac{0.1}{0.2} = \frac{1}{2} = 0.5$

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