Two water pipes of diameter \(2 \mathrm{~cm}\) and \(4 \mathrm{~cm}\) are separately connected to a main…
Two water pipes of diameter \(2 \mathrm{~cm}\) and \(4 \mathrm{~cm}\) are separately connected to a main supply line. The velocity of flow of water in the pipe of \(2 \mathrm{~cm}\) diameter is
4 times that in the other pipe
\(\frac{1}{4}\) times that in the other pipe
2 times that in the other pipe
\(\frac{1}{2}\) times that in the other pipe
Solution
Diameter of first pipe, \(d_1=2 \mathrm{~cm}\)
\(\therefore\) Radius, \(r_1=1 \mathrm{~cm}=10^{-2} \mathrm{~cm}\)
Diameter of second pipe, \(d_2=4 \mathrm{~cm}\)
\(\therefore \quad r_2=2 \mathrm{~cm}=2 \times 10^{-2} \mathrm{~m}\)
If \(v_1\) and \(v_2\) are the velocities of water in first and second pipe respectively, then according to principle of continuity,
\(\begin{aligned}
& & A_1 v_1=A_2 v_2 \Rightarrow \pi r_1^2 v_1=\pi r_2^2 v_2 \\
\Rightarrow & v_2 & =\left(\frac{r_1}{r_2}\right)^2 v_1=\left(\frac{10^{-2}}{2 \times 10^{-2}}\right)^2 v_1 \\
\Rightarrow & & v_2=\frac{v_1}{4} \Rightarrow v_1=4 v_2
\end{aligned}\)