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Two bars of steel $(Y = 2 \times 10^{11} N / m^{2})$ are joined together as shown. The area of cross-section of the left bar is $15 c m^{2}$ and the area of right bar is unknown. The extension in both bars is the same.

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The decrease in thickness of bar is more for the left

The area of right bar is $10 c m^{2}$

The stresses in left and right bar are in ratio $3 : 2$

The decrease in thickness of bar is more for right bar

answer is A, D.

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Detailed Solution

$∆ l = \frac{F l_{0}}{A y} \Rightarrow \frac{l_{0}}{A} = \frac{l_{0}^{'}}{A'} \frac{0.6}{15} = \frac{0.4}{A'} \Rightarrow A' = 10 c m^{2}$

stress $= \frac{F}{A} \Rightarrow \frac{σ_{1}}{σ_{2}} = \frac{A_{2}}{A_{1}} = \frac{10}{15}$

$\frac{∆ r}{r} = - σ \frac{∆ l}{l} ∆ r = - σ \frac{∆ l}{l} \times r = - σ ∆ l \frac{\sqrt{A}}{l} ∆ r_{l e f t} = \frac{\sqrt{15}}{0.6} = \sqrt{\frac{15}{0.36}} ∆ r_{r i g h t} = \frac{\sqrt{10}}{0.4} = \sqrt{\frac{10}{0.16}}$

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