Physics - Fluid Question with Solution | TestHub

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