JEE MAIN - Physics (2022 - 25th June Evening Shift - No. 4)
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).
2.4 $$\times$$ 106 cm3
1.2 $$\times$$ 105 cm3
6.0 $$\times$$ 104 cm3
4.8 $$\times$$ 105 cm3
Explanation
$$6 \times {l^2} = 24$$
$$ \Rightarrow l = 2$$ m
$$\therefore$$ $${{\Delta V} \over V} = 3 \times {{\Delta l} \over l}$$
$$ \Rightarrow \Delta V = 3 \times (\alpha \Delta T) \times V$$
$$ = 3 \times 5 \times {10^{ - 4}} \times 10 \times (8)$$
$$ = 120 \times {10^{ - 3}}$$ m3
$$ = 120 \times {10^{ - 3}} \times {10^6}$$ cm3
$$ = 1.2 \times {10^5}$$ cm3
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