A liquid is flowing in a cylindrical pipe of internal diameter \(4\text{ cm}\) with a velocity of \(5\text{ m/s}\). If this tube is joined with another tube of internal diameter \(2\text{ cm}\) then the velocity of flow of liquid in the smaller tube will be (in \(\text{ms}^{-1}\))
Solution:
Using the equation of continuity, \(A_1 v_1 = A_2 v_2\), which ⇒ \(d_1^2 v_1 = d_2^2 v_2\). Substituting \(d_1 = 4\text{ cm}\), \(v_1 = 5\text{ m/s}\), and \(d_2 = 2\text{ cm}\) gives \(16 \times 5 = 4 \times v_2\), so \(v_2 = 20\text{ m/s}\).
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