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Q. A thin copper wire of length $L$ increases in length by $1\%$ when heated from temperature $T_{1}$ to $T_{2}$ . The percentage change in the area when a thin copper plate having dimensions $2L\times L$ is heated from $T_{1}$ to $T_{2}$ is

NTA AbhyasNTA Abhyas 2020

Solution:

Fractional charge in length
$\frac{\Delta l}{l}=\alpha .\Delta \theta $
Fractional change in area
$\frac{\Delta A}{A}=\beta .\Delta \theta $
But $\beta =2\alpha $
$\therefore \, \, \frac{\Delta A}{A}=\left(2 \alpha \right)\Delta \theta $
$\frac{\Delta A}{A}=2\left(\frac{\Delta l}{l}\right)$
$\frac{\Delta A}{A}=2\left(1 \%\right)=2\%$