Question:
A liquid is flowing through a tube of diameter 9 mm with a speed of \(10\,cm\,s^{-1}\). If this tube is connected to a narrow tube in which the liquid flows with a speed of \(90\,cm\,s^{-1}\), then the diameter of the narrow tube is:
A liquid is flowing through a tube of diameter 9 mm with a speed of \(10\,cm\,s^{-1}\). If this tube is connected to a narrow tube in which the liquid flows with a speed of \(90\,cm\,s^{-1}\), then the diameter of the narrow tube is:
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In continuity equation problems, always remember \(A \propto d^2\), not \(d\).
Updated On: Jun 19, 2026
- 9 mm
- 1 mm
- 3 mm
- 10 mm
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The Correct Option is C
Solution and Explanation
Step 1: Apply equation of continuity.
For incompressible fluid flow: \[ A_1 v_1 = A_2 v_2 \]
Step 2: Express area in terms of diameter.
Since \(A \propto d^2\), \[ d_1^2 v_1 = d_2^2 v_2 \]
Step 3: Substitute given values.
\[ (9)^2 \times 10 = d_2^2 \times 90 \]
Step 4: Simplify equation.
\[ 81 \times 10 = 90 d_2^2 \Rightarrow 810 = 90 d_2^2 \]
Step 5: Solve for \(d_2\).
\[ d_2^2 = 9 \Rightarrow d_2 = 3\,mm \]
Step 6: Final conclusion.
Thus, the diameter of the narrow tube is \(3\,mm\).
Final Answer: \[ \boxed{3\,mm} \]
For incompressible fluid flow: \[ A_1 v_1 = A_2 v_2 \]
Step 2: Express area in terms of diameter.
Since \(A \propto d^2\), \[ d_1^2 v_1 = d_2^2 v_2 \]
Step 3: Substitute given values.
\[ (9)^2 \times 10 = d_2^2 \times 90 \]
Step 4: Simplify equation.
\[ 81 \times 10 = 90 d_2^2 \Rightarrow 810 = 90 d_2^2 \]
Step 5: Solve for \(d_2\).
\[ d_2^2 = 9 \Rightarrow d_2 = 3\,mm \]
Step 6: Final conclusion.
Thus, the diameter of the narrow tube is \(3\,mm\).
Final Answer: \[ \boxed{3\,mm} \]
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