JEE MAIN - Physics (2021 - 24th February Morning Shift - No. 6)
Each side of a box made of metal sheet in cubic shape is 'a' at room temperature 'T', the coefficient of linear expansion of the metal sheet is '$$\alpha$$'. The metal sheet is heated uniformly, by a small temperature $$\Delta$$T, so that its new temperature is T + $$\Delta$$T. Calculate the increase in the volume of the metal box.
3a3$$\alpha$$$$\Delta$$T
4$$\pi$$a3$$\alpha$$$$\Delta$$T
$${{4 \over 3}}$$$$\pi$$a3$$\alpha$$$$\Delta$$T
4a3$$\alpha$$$$\Delta$$T
Explanation
We know that, $$\gamma = 3\alpha $$ .... (i)
where, $$\alpha$$ is the coefficient of linear expansion and $$\gamma$$ is the coefficient of volume expansion.
We know that,
$${{\Delta V} \over V} = \gamma \Delta T$$
$$ \Rightarrow {{\Delta V} \over V} = 3\alpha \Delta T$$ [from Eq. (i)]
$$\Delta V = 3{a^3}\alpha \Delta T$$ [ $$\because$$ volume of cube = a3 ]
where, $$\alpha$$ is the coefficient of linear expansion and $$\gamma$$ is the coefficient of volume expansion.
We know that,
$${{\Delta V} \over V} = \gamma \Delta T$$
$$ \Rightarrow {{\Delta V} \over V} = 3\alpha \Delta T$$ [from Eq. (i)]
$$\Delta V = 3{a^3}\alpha \Delta T$$ [ $$\because$$ volume of cube = a3 ]
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