Question

A vertical column $50cm$ long at ${50}^{\circ }C$ balances another column of same liquid $60cm$ long at ${100}^{\circ }C$. The coefficient of absolute expansion of the liquid is:

• A. $0.005{/}^{\circ }C$
• B. $0.0005{/}^{\circ }C$
• C. $0.002{/}^{\circ }C$
• D. $0.0002{/}^{\circ }C$

Answer 1

Answer: (A)

$0.005{/}^{\circ }C$

Given vertical column of liquid $h_1=50cm$

Temperature $T_1={50}^{o}C$

given another vertical column of same liquid

$h_2=60cm$

$T_2={100}^{o}C$

Let the density of gives liquid at $STP={\rho }_0$

if both vertical columns balances each other then their pressure should be equal

$P=\rho gH$

$\rho =$ density

$g=$ acceleration due gravity

$H=$ height

${\rho }_{1}g{h}_1={\rho }_{2}g{h}_2\Rightarrow \frac{{\rho }_1}{{\rho }_2}=\frac{h_2}{h_1}....\left(1\right)$

let $r$ be coefficient of expansion of liquid

$={\rho }_1={\rho }_0/(1+r{T}_1)$ ${\rho }_2={\rho }_0/(1+r{T}_2)$

Putting this in $\left(1\right)$

$\frac{{\rho }_1}{{\rho }_2}\frac{\rho /(1+r{T}_1)}{{\rho }_0(1+r{T}_2)}=\frac{h_2}{h_1}=\frac{60}{50}$

$=\frac{1+r{T}_2}{1+r{T}_1}=\frac{6}{5}\Rightarrow 5+5r{T}_2+6+6r{T}_1$

$\Rightarrow 5r{T}_2-6r{T}_1=1$

$=5\times r\times 100-6\times r\times 30=1$

$=200r=1\Rightarrow r=0.005/{}^{o}C$

Hence coefficient of absolute expansion of liquid

is $0.005/{}^{o}C$