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

A, C

Explanation

Let the pressure at point O be `P_0.` Since the liquid is at
equilibrium at M
`P_A+h_1rho_(95^@)g=P_0+hrho_(5^@)g`
`rArrP_0=P_A+h_1rho_(95^@)g-hrho_(5^@)g…(i)`
Since the liquid is at equilibrium at N
`rArrP_A+h_2rho_(5^@)g=P_0+hrho_(95^@)g`
`rArrP_0=P_A+h_2rho_(5^@)g-hrho_(95^@)g…(ii)`
From (i) and (ii)
`P_A+h_1rho_(95^@)g-hrho_(5^@)g`
`=P_A+h_2rho_(5^@)g-hrho_(95^@)g`
`rArr(rho_(5^@))/(rho_(95^@))=1.018....(i)`

we know that
`rho_0=rho_1(1+gammaDeltaT)`
Applying the above formula,we get
`rho_0=rho_(95^@)(1+gammaxx95)`
rho_0=rho_(5^@)(1+gammaxx5)`
`:. (rho_(5^@))/(rho_(95^@))=(1+95gamma)/(1+5gamma).....(ii)`
But `gamma=3alpha`
`rArralpha=gamma/3=(2.002xx10^-4)/3=6.67xx10^(-5)@c^-1`