A solid metallic cube having total surface area 24m ^{2} is uniformly heated. If its temperature is

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10-1.Thermometry, Thermal Expansion and Calorimetry

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A solid metallic cube having total surface area $24\;m ^{2}$ is uniformly heated. If its temperature is increased by $10\,^{\circ} C$, calculate the increase in volume of the cube $\left(\right.$ Given $\left.: \alpha=5.0 \times 10^{-4}{ }^{\circ} C ^{-1}\right)$

A

$2.4 \times 10^{6} cm ^{3}$

B

$1.2 \times 10^{5} cm ^{3}$

C

$6.0 \times 10^{4} cm ^{3}$

D

$4.8 \times 10^{5} cm ^{3}$

(JEE MAIN-2022)

Solution

Increase in volume $\Delta V =\gamma V _{0} \Delta T$

$\gamma=3 \alpha$

So $\Delta V =(3 \alpha) V _{0} \Delta T$

Total surface area $=6 a ^{2}$, where $a$ is side length $24=6 a ^{2} \quad a =2 m$

$Volume \,V _{0}=(2)^{3}=8 m ^{3}$

$\Delta V =\left(3 \times 5 \times 10^{-4}\right)(8) \times 10$

$=1.2 \times 10^{5} cm ^{3}$

Standard 11

Physics

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