Question

Density of mercury is $13.6gmc{m}^{-3}$ at ${15}^{o}C$. Coefficient of real expansion of mercury is $182\times {10}^{-6}{\left(\right)}^{-1}$. Calculate its density at ${65}^{o}C$:

• A. $13.48\times {10}^{-3}kgc{m}^{-3}$
• B. $130.48\times {10}^{3}kgc{m}^{-3}$
• C. $133.40\times {10}^{3}kgc{m}^{-3}$
• D. $13.28\times {10}^{3}kgc{m}^{-3}$

Answer 1

The correct option is A $13.48\times {10}^{-3}kgc{m}^{-3}$

Given,

$\text{initial}\text{density,}\rho =13.6gmc{m}^{-3}$

$\text{coefficent}\text{of}\text{expansion,}\alpha =182\times {10}^{-6}{}^o{C}^{-1}$

$\text{change}\text{in}\text{temperature},\Delta T=65-15=50{}^{o}C$

$\rho =\frac{{\rho }_o}{(1+\alpha \Delta T)}=\frac{13.6}{[1+182\times {10}^{-6}(65-15\left)\right]}$

$\rho =13.48gm/c{m}^{-3}=13.48\times {10}^{-3}kg/c{m}^{3.}$