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

Answer

`gamma=2 xx 10^(-4)//"^(@)C`

Explanation

Density of a liquid varies with temperature as-
`rho_(t^(@)c) = ((rho_(0^(@)c))/(1+gammat))`
Here `gamma` is the coefficient of volume expansion of liquid.
ltbr. In the figure-
`h_(1) = 52.8 cm, h_(2) = 51 cm` and `h = 49 cm`
Now pressure at `B ="pressure at "C`
Therefore
`P_(0) +h_(1) rho_(95^(@))g-h rho_(5^(@))g =P_(0)+h_(2) rho_(5^(@))g-h rho_(95^(@))g`
`rArr rho_(95^(@)) (h_(1) +h) = rho_(5^(@)) (h_(2) +h)`
`rArr (rho_(95^(@)))/(rho_(5^(@))) = (h_(2)+h)/(h_(1)+h)`
`rArr (rho_(0^(@)))/(((1+95gamma))/(rho_(0^(@))/((1+5gamma)))) =(h_(2)+h)/(h_(1)+h)`
`rArr (1+5 gamma)/(1+95 gamma) = (51+49)/(52.8 +49) = (100)/(101.8)`
Solving this equaiton, we get
`gamma = 2 xx 10^(-4) per.^(@)C`