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Question: Length of a wire at room temperature is \(4.55\) m, when the temperature increases upto \(100^{o}C\)...

Length of a wire at room temperature is 4.554.55 m, when the temperature increases upto 100oC100^{o}C then its length becomes 4.574.57 m. the coefficient of linear expansion. (α)(\alpha) of the given wire is.

A

5.021×10−5K−15.021 \times 10^{- 5}K^{- 1}

B

6.021×10−5K−16.021 \times 10^{- 5}K^{- 1}

C

7.021×10−5K−17.021 \times 10^{- 5}K^{- 1}

D

8.021×10−5K−18.021 \times 10^{- 5}K^{- 1}

Answer

6.021×10−5K−16.021 \times 10^{- 5}K^{- 1}

Explanation

Solution

Here, T1=27∘C=273+27=300lT_{1} = 27{^\circ}C = 273 + 27 = 300l

L1=4.55m,L2=4.57mL_{1} = 4.55m,L_{2} = 4.57m

As, L2=L1[1+α(T2−T1)]L_{2} = L_{1}\lbrack 1 + \alpha(T_{2} - T_{1})\rbrack

∴4.57=4.55[1+α(373−300)]\therefore 4.57 = 4.55\lbrack 1 + \alpha(373 - 300)\rbrack

4.57=4.55[1+α×73]4.57 = 4.55\lbrack 1 + \alpha \times 73\rbrack

Or α=[4.574.55−1]×173=[4.57−4.554.55]×173\alpha = \left\lbrack \frac{4.57}{4.55} - 1 \right\rbrack \times \frac{1}{73} = \left\lbrack \frac{4.57 - 4.55}{4.55} \right\rbrack \times \frac{1}{73}

α=6.021×10−5K−1\alpha = 6.021 \times 10^{- 5}K^{- 1}