In the figure, the velocity `v_(3)` will be:

Options

A. zero
B. 4 `ms^(-1)`
C. 1 `ms^(-1)`
D. 3 `ms^(-1)`

Answer

C

Explanation

According to steady flow,
`A_(1)V_(1)=A_(2)V_(2)+A_(3)V_(3)`
(or) `A_(3)V_(3)=A_(1)V_(1)-A_(2)V_(2)`
(or) `V_(3)=(1)/(A_3)[A_(1)V_(1)-A_(2)V_(2)]`
`=(1)/(0.4)[0.2xx4-0.2xx2]`
`=1 ms^(-1)`