Fluid Mechanics

JEE Main level, numerical answer, medium

58 of 192

Question 58 of 192

Figure below shows a liquid being pushed out of the tube by a piston having area of cross section 2.0cm22.0{cm}^{2}. The area of cross section at the outlet is 10mm210{mm}^{2}. If the piston is pushed at a speed of 4cms−14cm{s}^{-1}, the speed of outgoing fluid is _____ cms−1cm{s}^{-1}

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Answer: 80

The given data is
A1=2×10−4m2A2=10×10−6m2v1=4cms−1{A}_{1}=2\times {10}^{-4}{m}^{2} {A}_{2}=10\times {10}^{-6}{m}^{2} {v}_{1}=4cm{s}^{-1}
Using equation of continuity which is a conservation of mass principle,
A1v1=A2v2{A}_{1}{v}_{1}={A}_{2}{v}_{2}
⇒v2=A1v1A2=2×10−4×410×10−6\Rightarrow {v}_{2}=\dfrac{{A}_{1}{v}_{1}}{{A}_{2}}=\dfrac{2\times {10}^{-4}\times 4}{10\times {10}^{-6}}
=80cms−1=80cm{s}^{-1}

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