JEE Mains · Physics · STD 11 - 9.1 fluid mechanics
Two tubes of radii \(r_1\) and \(r_2\), and lengths \(l_1\) and \(l_2\), respectively, are connected in series and a liquid flows through each of them in streamline conditions. \(P_1\) and \(P_2\) are pressure differences across the two tubes. If \(P_2\) is \(4P_1\) and \(l_2\) is \( \frac{l_1}{4}\) , then the radius \(r_2\) will be equal to
- A \(r_1\)
- B \(2r_1\)
- C \(4r_1\)
- D \(\frac{r_1}{2}\)
Answer & Solution
Correct Answer
(D) \(\frac{r_1}{2}\)
Step-by-step Solution
Detailed explanation
The volume of liquid flowing through both the tubes i.e., rate of flow of liquid same. \(Therefore,V = {V_1} = {V_2}\) \(i.e.,\frac{{\pi {p_1}r_1^4}}{{8\eta {l_1}}} = \frac{{\pi {p_2}r_2^4}}{{8\eta {l_2}}}\)…
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