The apparatus shown in the figure consists of four glass columns connected by horizontal section. The height of two central column B and C are 49 cm each. The two outer columns A and D are open to the temperature. A and C are maintained at a temperature of `95^@C` while the columns B and D are maintained at `5^@C`. The height of the liquid in A and D measured from the base the are 52.8 cm and 51cm respectively. Determine the coefficient of thermal expansion of the liquid

Explanation

Density of a liquid varies with temperature as
`rho_(t^@C)=((rho_(0^@C))/(1+gammat))`
here `gamma` is the corfficiecnt of volume expansion of temperature.
In the figure,
`h_1=52.8cm`,`h_2cm` and `h=49cm`
Now pressure at `B=`pressure at C. Therefore
`P_0+h_1rho_(95^@)g-hrho_(5^@)g=P_0+h_2rho_(5^@)g-hrho_(95^@)`
`grho_(95^@)(h_1+h)=rho_(5^@)(h_2+h)`
`implies(rho_(95^@))/(rho_(5^@))=(h_2+h)/(h_1+h)implies((rho_(0^@))/((1+95gamma)))/((rho_(0^@))/((1+5gamma)))=(h_2+h)/(h_1+h)`
`implies(1+5gamma)/(1+95gamma)=(51+49)/(52.8+49)=(100)/(101.8)`
Solving this equation we get
`gamma=2xx10^-4per^@C`