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) [0.2 xx 4 -0.2 xx 2]= 1ms^(-1)`