A gardener is watering plants at the rate `0.litre//sec` using a pipe of cross-sectional area `1cm^(2)` What additional force he has to exert if he desires to increase the rate of watering two times ? .

Explanation

`F =Ad v^(2) = ((Av)^(2)d)/(A)` If rate of watering of plant `(Av)` is doubled. It means that the amount of water poured//sec is doubled which is possible only if velocity is doubled Hence, force is to be made 4 times
`:.` Additional force `=3` times initial force
`= 3Adv^(2) = 3 ((Av^(2)))/(A)d`
` = (3 xx 0.1 xx 0.1 xx 10^(3))/(10^(-4)) = 3xx10^(5)N` .