If cross- sectional area of limb I is `A_(1)` and that of limb II is `A_(2)` then velocity of the liquid in the tube will be, (cross- sectional area of tube is very small)

Options

A. `sqrt(2g(x-y))`
B. `(A_(1))/(A_(2)) sqrt(2g(x-y))`
C. `(A_(2))/(A_(1)) sqrt(2g(x-y))`
D. None of these

Answer

A

Explanation

`p_(2)=p_(3) " and" p_(1) gt p_(2)`

`p_(1)=p_(0)+rho gx implies p_(2)=p_(0)+rho gy`
`therefore Deltap=p_(1)-p_(2)=rho g (x-y)`
Between 1 and 2
Difference in pressure energy = difference in kinetic energy
or `rho g(x-y)=(1)/(2)rho v^(2) " or" v=sqrt(2g(x-y))`