Grade 12Mechanics

Water stands at a depth of 15 m behind a reservoir dam. A horizontal pipe 4 cm in diameter passes through the dam 6 m below the surface of water as shown. There is plug which secures the pipe opening. Then the friction between the plug and pipe wall is:

sangram patra

15 Years agoGrade 12

3 Answers

Chetan Mandayam Nayakar

15 Years ago

pressure=dgh, d=1000kg/m^3, h=6m, g=9.8m/s^2, area=pi*r²,r=2cm,friction force=pressure*area

Neha

6 Years ago

Friction=force applied

So, force applied= area×pressure

R= d\2= 0.04/2=0.02m

Area= pie× R*2=3.14×0.02×0.02=0.001256m*2

And pressure=P2- P1=P0+ rho×g×h - P0= rho× g× h= 1000×10×6=60000

Therefore, force applied= friction = area× pressure

=0.001256×60000

=75.36N

Vikas TU

6 Years ago

Dear student

Above ans are incorrect

P = rho * g*h
= 58800N/m^2
Splug = pi*R^2
Magnitude of the frictional force b/n plug and pipe wall
F = S* P = 295.41N
V = sqrt(2*g*h)
=7.67 m/s

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Water stands at a depth of 15 m behind a reservoir dam. A horizontal pipe 4 cm in diameter passes through the dam 6 m below the surface of water as shown. There is plug which secures the pipe opening. Then the friction between the plug and pipe wall is:

sangram patra

3 Answers

Chetan Mandayam Nayakar

Neha

Vikas TU

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