JEE Mains · Physics · STD 11 - Fluid mechanics
- JEE Mains 2023
- Hard
- Numerical
The figure shows a liquid of given density flowing steadily in horizontal tube of varying cross-section. Cross sectional areas at \(A\) is \(1.5\,cm ^2\), and \(B\) is \(25\,mm ^2\), if the speed of liquid at \(B\) is \(60\,cm / s\) then \(\left( P _{ A }- P _{ B }\right)\) is :(Given \(P _{ A }\) and \(P _{ B }\) are liquid pressures at \(A\) and \(B\) points.Density \(\rho=1000\,kg\,m ^{-3}\) \(A\) and \(B\) are on the axis of tube \(............\,Pa\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(175\)
Step-by-step Solution
Detailed explanation
From continuity theorem \(A_1 V_1=A_2 V_2\)
\(1.5 \times V_1=25 \times 10^{-2} \times 60\)
\(V_1=\frac{25 \times 60 \times 10^{-2} \times 10}{1.5}\)
\(V_1=10\,cm / s\)
By Bernoulli's theorem
\(P_1+\frac{1}{2} \times 1000 \times(0.1)^2=P_2+\frac{1}{2} \times 1000 \times(0.6)^2\)
\(P_1+5=P_2+\frac{1}{2} \times 1000 \times 36 \times 10^{-2}\)
\(P_1+5=P_2+180\)
\(P_1-P_2=175\,Pa\)
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