υ=1.5 m/s A=10−2m2
Now, F=dtdP
=dtd(mυ)=υdtdm
=(Aυ⋅ρ)⋅υ
=Aρυ2
=10−2×1000×(1.5)2
=10−2×1000×2.25
=22.5N
JEE Main 2014 — Physics Mechanics
Water is flowing at a speed of 1.5 m s−1 through a horizontal tube of cross-sectional area 10−2 m2and 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)
Held on 9 Apr 2014 · Verified 6 Jul 2026.
33.7 N
45 N
15 N
22.5 N
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