Question

(who want 100 points )can here someone is really genius who can Answer my question please with correct explanation another I will report .

Answer 1

mass of water flowing volume density per second==(volume X density)/time 
=(area x lenght x density)/time 
=area x density x velocity 
=a z x v 
Therefore,Rate of increase of k.e=mv2/2xtime 
=(A x z x V^3)/2 
Mass m, flowing out per second, can be increased to m1 by increasing to V to V1, then power 
increases from P to P1. 
therefore,p1/p=(A x z x V1^3)/(A x z x V^3) 
=(v1/v)^3 
so, 
m1/m=(A x z x v1)/(a x z x v)=v1/v 
But given m1 = 2m, v1 = 2v 
therefore,p1/p=2^3=8 
so,p1=8p.. 

ans 3rd is right............

Answer 2

mass of water flowing volume density per second==(volume X density)/time .

=(area x lenght x density)/time .

=area x density x velocity .

=a z x v 

Therefore,Rate of increase of k.e=mv2/2xtime .

=(A x z x V^3)/2 

Mass m, flowing out per second, can be increased to m1 by increasing to V to V1, then power .

increases from P to P1. 

therefore,p1/p=(A x z x V1^3)/(A x z x V^3) 

=(v1/v)^3 

so, 

m1/m=(A x z x v1)/(a x z x v)=v1/v 

But given m1 = 2m, v1 = 2v 

therefore,p1/p=2^3=8 

so,p1=8p.

ANS= 3rd is right.