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A wire 3m in length and 1mm in diameter at $30^{0} C$ and is kept at temperature $170^{0} C$ and is stretched by hanging a weight of 10kg at one end calculate the change in the length of the wire. Given $α = 1.2 \times 10^{- 5} /^{0} C$ $Y = 2 \times 10^{11} N m^{- 2}$ $g = 10 m s^{- 2}$

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$2 \times 10^{- 3} m$

$7.2 \times 10^{- 3} m$

$9.2 \times 10^{- 3} m$

5.2 mm

answer is C.

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Detailed Solution

$e_{1} = ℓ α Δ t = 3 \times (1.2 \times 10^{- 5}) (170 - 30)$

$= 7.2 \times 10^{- 3} m$

$e_{2} = \frac{F ℓ}{A y} = \frac{20}{11} \times 10^{- 3} m \approx 2 \times 10^{- 3} m$

$Δ l = e_{2} + e_{1} = (2 + 7.2) \times 10^{- 3} = 9.2 \times 10^{- 3} m$ $= 9.2 m m$

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