Question

An ideal fluid flows through a pipe of circular cross-section made of two sections with diameters 2.5 cm and 3.75 cm. The ratio of the velocities in the two pipes is(a) 9 : 4(b) 3 : 2(c) 3: 2(d) 2: 3

Answer 1

Answer: (a). The ratio of the velocities in the two pipes is $\dfrac{v_{1}}{v_{2}} = \dfrac{9}{4}$.

Explanation:

Given that,

Diameter of first section $d_{1} = 2.5 cm$

Diameter of second section $d_{2} = 3.75 cm$

According to the equation of continuity

$A_{1}v_{1} = A_{2}v_{2}$

Where, A = area of the pipe

v = velocity

Now, the ratio of the velocities

$\dfrac{v_{1}}{v_{2}} = \dfrac{A_{2}}{A_{1}}$

$\dfrac{v_{1}}{v_{2}} = (\dfrac{d_{2}}{d_{1}})^{2}$

$\dfrac{v_{1}}{v_{2}} = \dfrac{9}{4}$

Hence, the ratio of the velocities in the two pipes is $\dfrac{v_{1}}{v_{2}} = \dfrac{9}{4}$.