Question

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

• A. $\rho = \rho_{0}\lbrack 1 + \gamma d\theta\rbrack$
• B. $\rho = \rho_{0}\lbrack 1 - \gamma d\theta\rbrack$
• C. $\rho = \rho_{0}\gamma d\theta$
• D. $\rho = \rho_{0}/\gamma d\theta$

Answer 1

Answer: (B)

$\rho = \rho_{0}\lbrack 1 - \gamma d\theta\rbrack$

Answer 2

Since, with increase in temperature, volume of given body increases, while mass remains constant so that density will decrease.

i.e. $\frac{\rho}{\rho_{0}} = \frac{m ⥂ / ⥂ V}{m ⥂ / ⥂ V_{0}} = \frac{V_{0}}{V} = \frac{V_{0}}{V_{0}(1 + r\Delta\theta)} = (1 - \gamma\Delta\theta)$

∴ $\rho = \rho_{0}(1 - \gamma\Delta\theta)$