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(D) $1.$ Using the equation of continuity,$A_1 v_1 = A_2 v_2$,where $A_1$ and $A_2$ are the cross-sectional areas of the piston and nozzle respectivel

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(D) $1.$ Using the equation of continuity,$A_1 v_1 = A_2 v_2$,where $A_1$ and $A_2$ are the cross-sectional areas of the piston and nozzle respectively.Given $r_1 = 20 \ mm$ and $r_2 = 1 \ mm$.$A_1 = \pi r_1^2 = \pi (20)^2 = 400 \pi \ mm^2$ and $A_2 = \pi r_2^2 = \pi (1)^2 = 1 \pi \ mm^2$.Thus,$A_1 = 400 A_2$.Given $v_1 = 5 \ mm \ s^{-1}$.$400 	imes 5 = 1 	imes v_2 \Rightarrow v_2 = 2000 \ mm \ s^{-1} = 2 \ m \ s^{-1}$.Therefore,the correct option is $(C)$.$2.$ Applying Bernoulli's principle at the nozzle opening and the liquid surface in the container:$P_0 = P_{nozzle} + \frac{1}{2} ho_a v_a^2$ (where $P_{nozzle}$ is the pressure at the nozzle).Also,for the liquid to rise to height $h$,$P_0 = P_{nozzle} + ho_{\ell} g h$.Equating the pressure drop,$\frac{1}{2} ho_a v_a^2 = ho_{\ell} g h$.For a given height $h$,the velocity of the liquid $v_{\ell}$ is related to the pressure difference. The rate of flow is proportional to the velocity of the air stream. From the pressure balance,$v_a \propto \sqrt{\frac{ho_{\ell}}{ho_a}}$. However,the question asks for the rate of liquid sprayed,which depends on the pressure drop created by the air flow. The pressure drop $\Delta P = \frac{1}{2} ho_a v_a^2$. The liquid flow rate is proportional to $\sqrt{\Delta P / ho_{\ell}} = \sqrt{\frac{ho_a v_a^2}{2 ho_{\ell}}} \propto \sqrt{\frac{ho_a}{ho_{\ell}}}$.Thus,the correct option is $(A)$.

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