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A 30 kg load is suspended from the lower end of a wire 10 cm and $2 m m^{2}$ in cross sectional area. The upper half of the wire is iron and the lower half is copper. Find the elongation in the wire(Nearly) $(Y_{iron} = 20 \times 10^{10} N m^{- 2}, Y_{Cu} = 1.1 \times 10^{11} N m^{- 2})$

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$20 \times 10^{- 5} m$

$15 \times 10^{- 5} m$

$10 \times 10^{- 5} m$

$5 \times 10^{- 5} m$

answer is C.

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Detailed Solution

$Y = \frac{2 Y_{1} Y_{2}}{Y_{1} + Y_{2}} = \frac{2 (20 \times 10^{10}) (1.1 \times 10^{11})}{20 \times 10^{10} + 1.1 \times 10^{11}} = 14.66 \times 10^{10}$

$= 15 \times 10^{10} N m^{- 2}$

$e = \frac{F L}{A y} = \frac{30 \times 10 \times 10 \times 10^{- 2}}{2 \times 10^{- 6} \times 15 \times 10^{10}}$

$= 10 \times 10^{- 5} m$

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