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(B) As the temperature increases,the volume $V$ of a body increases due to thermal expansion,while its mass $m$ remains constant. Since density $\rho

Vedclass · Accepted Answer

$ho = ho_0(1 - \gamma \Delta 	heta)$

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(B) As the temperature increases,the volume $V$ of a body increases due to thermal expansion,while its mass $m$ remains constant. Since density $ho = m/V$,the density decreases.Given the relation for volume expansion: $V = V_0(1 + \gamma \Delta 	heta)$,where $\gamma$ is the coefficient of volume expansion.The density at temperature $	heta$ is $ho = m/V$ and at initial temperature $	heta_0$ is $ho_0 = m/V_0$.Therefore,$\frac{ho}{ho_0} = \frac{V_0}{V} = \frac{V_0}{V_0(1 + \gamma \Delta 	heta)} = (1 + \gamma \Delta 	heta)^{-1}$.Using the binomial approximation for small $\gamma \Delta 	heta$,we get $(1 + \gamma \Delta 	heta)^{-1} \approx (1 - \gamma \Delta 	heta)$.Thus,$ho = ho_0(1 - \gamma \Delta 	heta)$.

With a rise in temperature,the density of a given body changes according to which of the following relations?

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