A gardener is watering plants at the rate 0.1 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=Adv^()=((Av)^(2)d)/(A)`. If the 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.
`therefore" Additional force = 3 times initial force"`
`=3Adv^(2)=3((Av)^(2))/(A)d`
`=(3xx0.1xx0.1xx10^(3))/(10^(-4))=3xx10^(5)N`