Q102 / 180 · NEET 2026 (Re-NEET) · Physics
Water flows in a streamline motion through a horizontal pipe of circular cross-section as shown in the figure. The pressure difference of water between $P$ and $Q$ is $15$ Nm$^{-2}$. The area of cross-section at $P$ and $Q$ are $40$ cm$^2$ and $20$ cm$^2$, respectively. The rate of flow of water through the pipe, in cm$^3$s$^{-1}$, is :
[Take density of water $=1000$ kg m$^{-3}$]
[Take density of water $=1000$ kg m$^{-3}$]
- Correct answerA$400$
- B$100$
- C$200$
- D$300$
Solution
From the equation of continuity, the rate of flow is constant:
$A_P v_P = A_Q v_Q$
Given $A_P = 40 \text{ cm}^2$ and $A_Q = 20 \text{ cm}^2$, we have:
$40 v_P = 20 v_Q \Rightarrow v_Q = 2 v_P$
Applying Bernoulli's equation for a horizontal pipe:
$P_P + \dfrac{1}{2} \rho v_P^2 = P_Q + \dfrac{1}{2} \rho v_Q^2$
$P_P - P_Q = \dfrac{1}{2} \rho (v_Q^2 - v_P^2)$
Substitute the given values ($P_P - P_Q = 15 \text{ N m}^{-2}$, $\rho = 1000 \text{ kg m}^{-3}$) and $v_Q = 2 v_P$:
$15 = \dfrac{1}{2} \times 1000 \times ((2 v_P)^2 - v_P^2)$
$15 = 500 \times (4 v_P^2 - v_P^2)$
$15 = 500 \times 3 v_P^2$
$15 = 1500 v_P^2$
$v_P^2 = \dfrac{1}{100} \Rightarrow v_P = 0.1 \text{ m s}^{-1} = 10 \text{ cm s}^{-1}$
The rate of flow of water is:
$Q = A_P v_P = 40 \text{ cm}^2 \times 10 \text{ cm s}^{-1} = 400 \text{ cm}^3 \text{ s}^{-1}$
Answer: $400$
$A_P v_P = A_Q v_Q$
Given $A_P = 40 \text{ cm}^2$ and $A_Q = 20 \text{ cm}^2$, we have:
$40 v_P = 20 v_Q \Rightarrow v_Q = 2 v_P$
Applying Bernoulli's equation for a horizontal pipe:
$P_P + \dfrac{1}{2} \rho v_P^2 = P_Q + \dfrac{1}{2} \rho v_Q^2$
$P_P - P_Q = \dfrac{1}{2} \rho (v_Q^2 - v_P^2)$
Substitute the given values ($P_P - P_Q = 15 \text{ N m}^{-2}$, $\rho = 1000 \text{ kg m}^{-3}$) and $v_Q = 2 v_P$:
$15 = \dfrac{1}{2} \times 1000 \times ((2 v_P)^2 - v_P^2)$
$15 = 500 \times (4 v_P^2 - v_P^2)$
$15 = 500 \times 3 v_P^2$
$15 = 1500 v_P^2$
$v_P^2 = \dfrac{1}{100} \Rightarrow v_P = 0.1 \text{ m s}^{-1} = 10 \text{ cm s}^{-1}$
The rate of flow of water is:
$Q = A_P v_P = 40 \text{ cm}^2 \times 10 \text{ cm s}^{-1} = 400 \text{ cm}^3 \text{ s}^{-1}$
Answer: $400$
Hence, the correct answer is option: 'A'.