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Q.

A $20 k g$ load is suspended from the lower end of a wire $20 c m$ and $1 m m^{2}$ in cross sectional area. The upper half of the wire is iron and the lower half is aluminum Find the elongation in the wire $(Y_{i r o n} = 20 \times 10^{10} N m^{- 2}, Y_{A l} = 7 \times 10^{10} N m^{- 2} a n d g = 10 m s^{- 2})$

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a

None

b

$1.78 \times 10^{- 3} m$

c

$38.5 \times 10^{- 3} m$

d

$18.9 \times 10^{- 5} m$

answer is A.

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Detailed Solution

$e = e_{1} + e_{2}$ $(∵ Y = \frac{F L}{A e})$

$= \frac{m g ℓ}{A} (\frac{1}{y_{1}} + \frac{1}{y_{2}}) = \frac{20 \times 10 \times (10 \times 10^{- 2})}{10^{- 6}} (\frac{1}{20 \times 10^{10}} + \frac{1}{7 \times 10^{10}}) = \frac{20}{10^{- 6 + 10}} (\frac{1}{20} + \frac{1}{7}) = 20 \times 10^{- 4} (\frac{7 + 20}{140}) = 3.85 \times 10^{- 4} = 38.5 \times 10^{- 3}$

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