You\'re pumping water into a 2-cm radius hose at ground level with a speed of 5

Question

You're pumping water into a 2-cm radius hose at ground level with a speed of 5 m/s. The other end of the pipe is 4-cm radius and at an elevation of 50 meters. The water will exit the hose with a speed which is... O One-fourth as fast as it entered O None of the above... answer is more complicated O Half as fast as it entered The answer is uncomputable with the information given O Twice as fast as it entered O Four times as fast as it entered If water is to emerge from the top, the pressure of the water at the bottom must, at minimum, be: O Twice atmospheric pressure O The answer is uncomputable with the information given O None of the above... answer is more complicated Half atmospheric pressure O Atmospheric pressure plus pgh (where ? is the density of water, and h is 50 meters)

Explanation / Answer

Q1.

as flow rate is constant,

speed*cross sectional area =constant

==>speed*radius ^2=constant…(1)

so higher the radius, lower the speed and vice versa.

so at exit , as radius is higher, speed will be smaller.

now, using equation 1,

speed at entry*entry radius^2=speed at exit*exit radius^2

==>speed at entry*2^2=speed at exit*4^2

=>speed at exit=(1/4)*speed at entry

hence the first option is correct.

Q2.

let pressure at bottom=P1

pressure at top=P2=1 atm=101325 N/m^2

speed at bottom=v1=5 m/s

speed at top=v2=(1/4)*5=1.25 m/s

height at bottom=h1=0

height at top=h2=50 m

using bernoulli principle:

P1+0.5*rho*v1^2+rho*g*h1=P2+0.5*rho*v2^2+rho*g*h2

where rho=water density

hence P1=101325+0.5*1000*1.25^2+1000*9.8*50-0.5*1000*5^2-1000*9.8*0

=5.7961*10^5 N/m^2

which is 5.72 times atmospheric pressure.

hence at minimum, it will be atmospheric pressure+rho*g*h

last option is correct.

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