Q129 / 150 · MHT CET 2026 (16 April Shift 2) · Physics
A horizontal pipe carries water in a streamline flow. At point along the pipe, where the cross-sectional area is $A_1$, the velocity of water is $V_1$ and the pressure is $P_1$. What is the pressure of water at another point where the cross-sectional area is $A_2$?
- A$P_1 - \dfrac{\varrho V_2^2}{2A_1^2}(A_2^2 - A_1^2)$
- Correct answerB$P_1 + \dfrac{\varrho V_1^2}{2A_2^2}(A_2^2 - A_1^2)$
- C$P_1\varrho V_1^2(A_1^2 - A_2^2)$
- D$P/\varrho V_1^2 A_2^2$
Solution
Using the equation of continuity:
$A_1 V_1 = A_2 V_2 \Rightarrow V_2 = \dfrac{A_1 V_1}{A_2}$
Using Bernoulli's theorem for a horizontal pipe:
$P_1 + \dfrac{1}{2}\varrho V_1^2 = P_2 + \dfrac{1}{2}\varrho V_2^2$
$P_2 = P_1 + \dfrac{1}{2}\varrho (V_1^2 - V_2^2)$
Substituting the value of $V_2$:
$P_2 = P_1 + \dfrac{1}{2}\varrho \left(V_1^2 - \dfrac{A_1^2 V_1^2}{A_2^2}\right)$
$P_2 = P_1 + \dfrac{\varrho V_1^2}{2A_2^2}(A_2^2 - A_1^2)$
$A_1 V_1 = A_2 V_2 \Rightarrow V_2 = \dfrac{A_1 V_1}{A_2}$
Using Bernoulli's theorem for a horizontal pipe:
$P_1 + \dfrac{1}{2}\varrho V_1^2 = P_2 + \dfrac{1}{2}\varrho V_2^2$
$P_2 = P_1 + \dfrac{1}{2}\varrho (V_1^2 - V_2^2)$
Substituting the value of $V_2$:
$P_2 = P_1 + \dfrac{1}{2}\varrho \left(V_1^2 - \dfrac{A_1^2 V_1^2}{A_2^2}\right)$
$P_2 = P_1 + \dfrac{\varrho V_1^2}{2A_2^2}(A_2^2 - A_1^2)$
Hence, the correct answer is option: 'B'.