Water flows through a hose pipe whose internal diameter is \(4 \mathrm{~cm}\) at a speed of \(1…
Water flows through a hose pipe whose internal diameter is \(4 \mathrm{~cm}\) at a speed of \(1 \mathrm{~ms}^{-1}\). If water has to emerge at a speed of \(4 \mathrm{~ms}^{-1}\), then the diameter of the nozzle should be
\(1 \mathrm{~cm}\)
\(2 \mathrm{~cm}\)
\(4 \mathrm{~cm}\)
\(0.5 \mathrm{~cm}\)
Solution
Internal diameter for hose pipe,
\(d_1=4 \mathrm{~cm}, r_1=\frac{d}{2}=\frac{4}{2}=2 \mathrm{~cm}=2 \times 10^{-2} \mathrm{~m}\)
Speed of water through hose pipe, \(v_1=1 \mathrm{~ms}^{-1}\)
Speed of water through nozzle, \(v_2=4 \mathrm{~ms}^{-1}\)
Diameter, \(d_2=\) ?
According to principle of continuity,
\(\begin{array}{llll}
& A_1 v_1=A_2 v_2 \\
\Rightarrow & \pi r_1^2 v_1=\pi r_2^2 v_2 \\
\Rightarrow & r_2^2=\frac{r_1^2 v_1}{v_2} \Rightarrow r_2^2=\frac{\left(2 \times 10^{-2}\right)^2 \times 1}{4} \\
\Rightarrow & r_2^2=10^{-4} \\
\Rightarrow & r_2=10^{-2} \mathrm{~m} \Rightarrow r_2=1 \mathrm{~cm}
\end{array}\)
\(\therefore\) Diameter of nozzle, \(d_2=2 r_2=2 \times 1=2 \mathrm{~cm}\)