Question

An ideal fluid flows through a pipe of circular cross section with diameters $5\,cm$ and $10\, cm$ as shown. The ratio of velocities of fluid at $A$ and $B$ is

• A. 4 : 1
• B. 1 : 4
• C. 2 : 1
• D. 1 : 2

Answer 1

Answer: (A)

From continuity equation,
$a_{1} u_{1} =a_{2} v_{2} $
$\frac{v_{1}}{v_{2}} =\frac{a_{2}}{a_{1}} $
$a_{1} =\pi\left(\frac{d_{1}}{2}\right)^{2}$
$a_{2} =\pi\left(\frac{d_{2}}{2}\right)^{2}$
where, $d_{1}$ and $d_{2}$ are the diameters of the openings.
$\Rightarrow \frac{v_{1}}{v_{2}}=\frac{\left(\pi d_{2}^{2} / 4\right)}{\left(\pi d_{1}^{2} / 4\right)} $
$=\frac{d_{2}^{2}}{d_{1}^{2}}=\left(\frac{10\, cm }{5 \,cm }\right)^{2} $
$=\frac{10 \times 10}{5 \times 5}=\frac{4}{1} $ or $ 4: 1$