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Physics Question on thermal properties of matter

A metallic bar of coefficient of linear expansion 10−5K−110^{-5} K^{-1} is heated from 0∘C0^{\circ}C to 100∘C100^{\circ}C. The percentage increase in its length is

A

0.10%

B

0.01

C

0.1

D

0.01%

Answer

0.10%

Explanation

Solution

Given, α=10−5K−1\alpha=10^{-5} K ^{-1}
and ΔT=100∘−0∘=100∘C\Delta T=100^{\circ}-0^{\circ}=100^{\circ} C
We know that, percentage increase in its length,
Δll×100=αΔT×100\frac{\Delta l}{l} \times 100=\alpha \Delta T \times 100
Δll=10−5×100=10−3\frac{\Delta l}{l}=10^{-5} \times 100=10^{-3}
∴Δll×100=0.1%\therefore \frac{\Delta l}{l} \times 100=0.1 \%