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A steel tape measures the length of a copper rod as 90.0cm when both are at $10 ° C,$ the calibration temperature for the tape. If the change in the tape read for the length of the rod when both are at $30 ° C$ is $x \times 10^{- 3}$ cm. Find the value of ' $x$ ' $(Given α_{steel} = 1.2 \times 10^{- 5} / ° C, α_{copper} = 1.7 \times 10^{- 5} / ° C)$ (round off to nearest integer)

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answer is 9.

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Detailed Solution

$\frac{L_{0} (1 + α_{c} Δ θ)}{1 (1 + α_{s} Δ θ)} = \frac{90 (1 + 1.7 \times 10^{- 5} \times 20)}{1 (1 + 1.2 \times 10^{- 5} \times 20)} \approx 90.089$

Change in length is 8.9x10^-3cm

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