The liquid is in hydrostatic equilibrium ` implies rho_(t) gh_(t) = rho_(0)gh_(0)`
Where, ` rho_(t)` is density of liquid in hot bath, `rho_(0)` is density of liquid in cold bath.
Volumes of a given mass M of liquid at temperature t and `0^(@)C`
are ralated by `V_(t) = V_(0) (1+gammat) "Since" rho_(t)V_(t) = rho_(0) V_(0) implies rho_(t) = (rho_(0)V_(0))/(V_(t)) = (rho_(0))/((1+gammat))`
Since ` h_(t) = (rho_(0)h_(0))/(rho_(t)) = h_(0)(1+gammat)` which on solving for ` gamma `, yields `gamma = ((h_(t)-h_(0)))/(h_(0)t) `