Ncert Exercises & Solutions

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:
1. \(9:4\)
2. \(3:2\)
3. \(\sqrt{3}:\sqrt{2}\)
4. \(\sqrt{2}:\sqrt{3}\)

Hint: Apply equation of continuity.

Step: Find the ratio of velocities at the two cross-sections.

Consider the diagram where an ideal fluid is flowing through a pipe.

As given,

d_{1}

= Diameter at 1st point is 2.5 cm

d_{2}

= Diameter at 2nd point is 3.75 cm
Applying the equation of continuity for cross-sections

A_{1} and A_{2}

\(\Rightarrow A_1v_1 = A_2v_2\)
\(\Rightarrow \frac{v_1}{v_2} = \frac{A_2}{A_1}= \frac{\pi r_2^2}{\pi r^2_1} = \frac{r_2^2}{r^2_1} = \frac{d^2_2}{d^2_1}\\ \Rightarrow \frac{(3.75)^2}{(2.5)^2} = \frac{3^2}{2^2} = \frac{9}{4}\)
Hence, option (1) is the correct answer.

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