Please solve q 132?

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Answer 1 · 2018-04-04T18:05:09

The answer is #"option (2)"#

Apply the following equation (derived from Bernouilli's Equation)

#p_x-p_y=rho/2(v_y^2-v_x^2)#

#Deltap=rho/2(v_y^2-v_x^2)#

The density of water is #rho=1000kgm^-3#

The difference in pressure is

#Deltap=rhogh=1000*10*0.0051=51Pa#

The speed at #x# is #v_x=0.02ms^-1#

Therefore,

#v_y^2-v_x^2=51*2/1000=0.102#

So,

#v_y^2=0.102+0.02^2=0.1024#

#v_y=sqrt(0.1024)=0.32ms^-1#

The speed at #y# is #=32cms^-1#

The answer is #"option (2)"#