A liquid of density 750 kgm-3 flows smoothly through a horizontal pipe that tapers in cross-sectional area from A1=1.2×10−2m2 to A2=A12. The pressure difference between the wide and narrow sections of the pipe is 4500 Pa. The rate of flow of liquid is __________________ ×10−3m3s−1.

A2=A12P1−P2=4500PaP1+12ρV12+ρgh=P2+12ρV22+ρghP1−P2=12ρV22−V12 .....(1)And A1V1=A2V2⇒ V2=2V1 .....(2)4500=12×750×3V12V1=2m/sVolume flow rate =A1V1=24×10−3m3s−1

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

A liquid of density 750 kgm^-3 flows smoothly through a horizontal pipe that tapers in cross-sectional area from $A_{1} = 1 .2 \times 10^{- 2} m^{2}$ to $A_{2} = \frac{A_{1}}{2}$ . The pressure difference between the wide and narrow sections of the pipe is 4500 Pa. The rate of flow of liquid is __________________ $\times 10^{- 3} m^{3} s^{- 1}$ .

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

answer is 24.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} A_{2} = \frac{A_{1}}{2} \\ P_{1} - P_{2} = 4500 Pa \\ P_{1} + \frac{1}{2} {ρV}_{1}^{2} + ρgh = P_{2} + \frac{1}{2} {ρV}_{2}^{2} + ρgh \\ P_{1} - P_{2} = \frac{1}{2} ρ (V_{2}^{2} - V_{1}^{2}) . . . . . (1) \end{array}$