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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Work, Power and Energy
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Water is pumped through a hose-pipe at the rate of 75 lit/min in and issue from a nozzle of cross sectional area of 0.00625 sq-m. The force of reaction of nozzle is

\[(a)\;35\;N \quad (b)\;2.5\;N \quad (c)\;3.5\;N \quad (d)\;0.25\;N\]

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1 Answer

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mass of water pumped per second $ \large\frac{75}{60}$
$\qquad=1.25 kg/s$
$a v=75\; lit/mt$
$\qquad= \large\frac{75 \times 10^{-3}}{60} $$kg/s$
$v=\large\frac{75 \times 10 ^{-3}}{60 \times 6.25 \times 10^{-3}}$
$\qquad=0.2 m/s$
Force of reaction
$F=ma=\large\frac{mv}{t}$
$\quad=\frac{1.25 \times 0.2}{1}=0.25N$
Hence d is the correct answer.
answered Jul 22, 2013 by meena.p
edited Jun 5, 2014 by lmohan717
 

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