Water flows through a horizontal pipe at a speed ' \(\mathrm{V}\) '. Internal diameter of the pipe is ' \(\mathrm{d}\) '. If the water is coming out at a speed ' \(V_1\) ' then the diameter of the nozzle is

From equation of continuity,
\(\begin{aligned}
& \mathrm{A}_1 \mathrm{~V}_1=\mathrm{A}_2 \mathrm{~V}_2 \\
& \text { Given: } \mathrm{V}_1=\mathrm{V} \text { and } \mathrm{V}_2=\mathrm{V}_1 \\
& \Rightarrow \mathrm{A}_1 \mathrm{~V}=\mathrm{A}_2 \mathrm{~V}_1 \\
& \frac{\pi \mathrm{d}^2}{4} \mathrm{~V}=\frac{\pi \mathrm{d}_{\mathrm{n}}^2 \mathrm{n}}{4} \mathrm{~V}_1 \quad \ldots\left(\because \mathrm{A}=\frac{\pi \mathrm{d}^2}{4}\right)
\end{aligned}\)
where \(d_n\) is the diameter of the nozzle.
\(\begin{aligned}
& \therefore \quad \mathrm{d}_{\mathrm{n}}^2=\mathrm{d}^2 \frac{\mathrm{V}}{\mathrm{V}_1} \\
& \therefore \quad \mathrm{d}_{\mathrm{n}}=\mathrm{d} \sqrt{\frac{\mathrm{V}}{\mathrm{V}_1}}
\end{aligned}\)