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

Solution

## The correct option is D $$9 : 4$$According to equation of continuity $$A_1v_1= A_2v_2$$$$\displaystyle \dfrac{v_1}{v_2}=\dfrac{A_1}{A_2}=\dfrac{\pi D_2^2/4}{\pi D_1^2 /4}= \left ( \dfrac{D_2}{D_1} \right )^2$$Here, $$D_1 = 2.5 cm, D_2= 3.75 cm$$$$\therefore \displaystyle \dfrac{v_1}{v_2}= \left ( \dfrac{3.75}{2.5} \right )^2 = \left ( \dfrac {3}{2} \right ) ^2 = \dfrac {9}{4}$$Physics

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