JEE MainPhysicsToughHeat

When the temperature of copper coin is raised by 80ºC, its
diameter increases by 0.2.

Show answer and solution

100 \times \frac{\Delta D}{D} = \alpha\Delta T \times 100

100 \times \left( \frac{\Delta A}{A} \right) = \beta\Delta T \times 100

= 2\alpha\Delta T \times 100

= 2(0.2)

= 0.4\%

100 \times \frac{\Delta t}{t} = \alpha\Delta T \times 100

\alpha = \frac{\frac{\Delta d}{d}}{\Delta T} = \frac{0.2 \times 10^{- 2}}{80}\left( \frac{\Delta d}{d} \times 100 = 0.2 \right)

\alpha = 0.25 \times 10^{- 4}{}^{\circ}C^{- 1}

100 \times \frac{\Delta V}{V} = \gamma\Delta T \times 100

= 3(0.2) = 0.6\%

Source: Grass-G / P-08 Physics paper 23 October.docx

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