A solid metallic cube having total surface area 24 m2 is uniformly heated. If its temperature is increased by 10$^\circ$C, calculate the increase in volume of the cube. (Given $\alpha$ = 5.0 $\times$ 10$-$4 $^\circ$C$-$1).
Solution
<p>$6 \times {l^2} = 24$</p>
<p>$\Rightarrow l = 2$ m</p>
<p>$\therefore$ ${{\Delta V} \over V} = 3 \times {{\Delta l} \over l}$</p>
<p>$\Rightarrow \Delta V = 3 \times (\alpha \Delta T) \times V$</p>
<p>$= 3 \times 5 \times {10^{ - 4}} \times 10 \times (8)$</p>
<p>$= 120 \times {10^{ - 3}}$ m<sup>3</sup></p>
<p>$= 120 \times {10^{ - 3}} \times {10^6}$ cm<sup>3</sup></p>
<p>$= 1.2 \times {10^5}$ cm<sup>3</sup></p>
About this question
Subject: Physics · Chapter: Thermodynamics · Topic: Zeroth and First Law
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