Q. 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) $\rho = \rho_{0}/\gamma d\theta$
Explanation
(b) Since, with increase in temperature, volume of given body increases, while mass remains constant so that density will decrease
$i.e.\frac{\rho}{\rho_{0}} = \frac{m/V}{m/V_{0}} = \frac{V_{0}}{V} = \frac{V_{0}}{V_{0}(1 + r\mathrm{\Delta}\theta)} = (1 - \gamma\mathrm{\Delta}\theta)$
$\therefore\rho = \rho_{0}(1 - \gamma\mathrm{\Delta}\theta)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.