Question

Coefficient of real expansion mercury is $0.18\times {10}^{-3}{/}^{0}C.$ if the density of mercury at $0^{0}C$ is $13.6gm/C.C,$ then is density at $473K$ will be$:$

• A. $13.11\%$
• B. $3.6\%$
• C. $36\%$
• D. $24.4\%$

Answer 1

The correct option is A $13.11\%$

Coefficient of real expansion of mercury, $r_0=0.18\times {10}^{-3}{/}^{0}C$

Initial temperature $0^{0}C=273K$

Density of mercury $13.6gm/c^3$

Final temperature $473K$

We need to calculate the density at $473K$ or ${200}^{0}C$

We have,

$\rho =\frac{{\rho }_0}{1+r_{0}T}$

where,

${\rho }_0=$ Density at $0^{0}C$

$r_0=$ Coefficient of volume expansion

$T=$ Temperature at which density is to be calculated

Putting the values in the formula,

$\rho =\frac{13.6}{1+0.18\times 200}$

$=\frac{13.6}{1.036}=13.11g/cc$