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Question:

Two tubes of radii r1 and r2, and lengths l1 and l2, respectively, are connected in series and a liquid flows through each of them in streamline conditions. P1 and P2 are pressure differences across the two tubes. If P2 is 4P1 and l2 is l1/4, then the radius r2 will be equal to:

r1

r1/2

2r1

4r1

Solution:

Given, P2 = 4P1 l2 = l1/4
By the equation of continuity:
A1v1 = A2v2
A1/A2 = v2/v1
Pressure is inversely proportional to velocity.
So, P1/P2 = v2/v1
So, v2/v1 = 1/4
Thus, A1/A2 = 1/4
πr1²/πr2² = 1/4
r1²/r2² = 1/4
r2² = 4r1²
r2 = 2r1