The volume of a metal block increases by 0.22 | vedclass

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Question

(C) The fractional change in volume is given by $\frac{\Delta V}{V} = \frac{0.225}{100} = 0.00225$.The change in temperature is $\Delta T = 30^{\circ}

Vedclass · Accepted Answer

$2.5 	imes 10^{-5} /{ }^{\circ} C$.

Vedclass · Answer

(C) The fractional change in volume is given by $\frac{\Delta V}{V} = \frac{0.225}{100} = 0.00225$.The change in temperature is $\Delta T = 30^{\circ} C$.We know that the coefficient of volume expansion $\gamma$ is related to the coefficient of linear expansion $\alpha$ by the relation $\gamma = 3\alpha$.The formula for volume expansion is $\Delta V = V \gamma \Delta T$,which can be rewritten as $\frac{\Delta V}{V} = \gamma \Delta T$.Substituting $\gamma = 3\alpha$,we get $\frac{\Delta V}{V} = 3\alpha \Delta T$.Substituting the given values: $0.00225 = 3 	imes \alpha 	imes 30$.Solving for $\alpha$: $\alpha = \frac{0.00225}{90} = 2.5 	imes 10^{-5} /{ }^{\circ} C$.

The volume of a metal block increases by $0.225 \%$ when its temperature is increased by $30^{\circ} C$. Hence,the coefficient of linear expansion of the material of the metal block is

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