Laws of Motion

JEE Main level, single correct, medium

52 of 210

Question 52 of 210

Water is flowing at a speed of 1.5 m s-1 through a horizontal tube of cross-sectional area 10-2 m2 and you are trying to stop the flow by your palm. Assuming that the water stops immediately after hitting the palm, the minimum force that you must exert should be (density of water = 103 kg m-3)

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Answer: (D) 22.5 N

υ=1.5 m/sυ=\text{1.5 m/s} A=10−2m2\text{A}=1{0}^{-2}{\text{m}}^{2}

Now, F=ddtP\text{F}=\dfrac{\text{d}}{\text{dt}}\text{P}

=ddt(mυ)=υdmdt=\dfrac{\text{d}}{\text{dt}}\left(\text{m}υ\right)=υ\dfrac{\text{dm}}{\text{dt}}

=(Aυ⋅ρ)⋅υ=\left(\text{A}υ\cdot \rho \right)\cdot υ

=Aρυ2=\text{A}\rho {υ}^{2}

=10−2×1000×(1.5)2=1{0}^{-2}\times 1000\times {\left(\text{1.5}\right)}^{2}

=10−2×1000×2.25=1{0}^{-2}\times 1000\times \text{2.25}

=22.5N=\text{22.5}N

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