Here, for water column in one arm of U-tube : $h_{1}=10.0\ \text{cm}$ $\rho_{1}=1\ \mathrm{g\ cm}^{-3}$ For spirit column in other arm of U-tube, $h_{2}=12.5\ \text{cm}$ $\rho_{2}=?$ If $P_1\ \text{and}\ P_2$ be the pressures exerted by water column and spirit column respectively, $\text{Then }P_{1}=h_{1}\rho_{1}g\ldots(i)$ $\text{and }P_{2}=h_{2}\rho_{2}g\ldots(ii)$ Also as the mercury columns in the two arms of U-tube are at the same level, thus in equilibrium, $P_{1}=P_{2}$ or $h_{1}\rho_{1}g=h_{2}\rho_{2}g$ Now specific gravity of spirit $= \text{density of spirit}\text{ density of water }$ $= \frac{0.8\ \mathrm{g\ cm}^{-3}}{1\ \mathrm{g\ cm}^{-3}}=0.800$.