Question bank/NEET UG/Physics

NEET UGPhysicsModerateRe-NEET 2026

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 cm² and 20 cm², respectively. The rate of flow of water through the pipe, in cm³s⁻¹, is:

[Take density of water = 1000 kg m⁻³]

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Show answer and solution
Sol. APVP=AQVQA_P V_P = A_Q V_Q
40VP=20VQ40V_P = 20V_Q, VQ=2VPV_Q = 2V_P
Using Bernoulli's equation:
PP−PQ=12ρ(VQ2−VP2)P_P - P_Q = \frac{1}{2}\rho(V_Q^2 - V_P^2)
15=12×103(4VP2−VP2)15 = \frac{1}{2} \times 10^3 (4V_P^2 - V_P^2)
VP=151500=0.1 m/sV_P = \sqrt{\frac{15}{1500}} = 0.1 \text{ m/s}
VQ=0.2 m/sV_Q = 0.2 \text{ m/s}
Rate = APVP=40×10−4×0.1 m3/sA_P V_P = 40 \times 10^{-4} \times 0.1 \text{ m}^3/\text{s}
=4×10−4 m3/s= 4 \times 10^{-4} \text{ m}^3/\text{s}
=4×10−4×106 cm3/s= 4 \times 10^{-4} \times 10^6 \text{ cm}^3/\text{s}
=400 cm3/s= 400 \text{ cm}^3/\text{s}

Source: geekyrakshit/indian-exam-questions / reneet2026-code50

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