Find density of ethyl alcohol

Pressure at left hand of U-tube, $p_1=p_0+\rho_1 g h_1=p_0+\rho_1 g$\left(10 \times 10^{-2}\right)$or$\quad \mathrm{p}_1=\mathrm{p}_0+\rho_1$Pressure at right hand of u-tube,$$\begin{aligned} & p_2=p_0+\rho_2 g h_2=p_0+\rho_2 g\left(12 \times 10^{-2}\right) \\ & \text{ or } \quad \mathrm{p}_2=\mathrm{p}_0+1.2 \rho_2 \\ & \end{aligned}$$The mercury column in both arms of$U$-tube are at same level, therefore pressure in both arms will be same. or$\quad \mathrm{p}_1=\mathrm{p}_2 \Rightarrow \quad \mathrm{p}_0+\rho_1=\mathrm{p}_0+1.2 \rho_2 \therefore$Density of ethyl alcohol,$\rho_2=\frac{\rho_1}{1.2}=\frac{1000}{1.2}=833.3 \mathrm{~kg} / \mathrm{m}^3=0.83 \mathrm{gm} / \mathrm{cm}^3