Sol. $$ \begin{aligned} h_1 & +h_2=0.29 \times 2+0.1 \\ h_1 & +h_2=0.68 \\ \Rightarrow & P_0+\rho_k g(0.1)+\rho_w g\left(h_1-0.1\right)\left[\rho_k=\text { density of kerosene } \& \rho_w=\text { density of water }\right] \\ & -\rho_w g h_2=P_0 \\ \Rightarrow & \rho_k g(0.1)+\rho_w g h_1-\rho_w g \times(0.1) \\ & =\rho_w g h_2 \\ \Rightarrow & 800 \times 10 \times 0.1+1000 \times 10 \times h_1 \\ & -1000 \times 10 \times 0.1=1000 \times 10 \times h_2 \\ \Rightarrow & 10000\left(h_1-h_2\right)=200 \\ \Rightarrow & h_1-h_2=0.02 \\ \Rightarrow & h_1=0.35 \\ \Rightarrow & h_2=0.33 \end{aligned} $$ So, $\frac{\mathrm{h}_1}{\mathrm{~h}_2}=\frac{35}{33}$