Water flows in a streamline manner through a capillary tube of radius a. The pressure difference being P and the rate of flow is Q . If the radius is reduced to `a//2` and the pressure difference is increased to 2P, then find the rate of flow.

Explanation

`"Rate of flow Q"=(P)/((8etal//pir^(4)))`
`rArr Q prop (Pr^(4))rArr (Q_(1))/(Q_(2))=(P_(1))/(P_(2))((r_(1))/(r_(2)))^(4)`
`rArr (Q_(1))/(Q_(2))=(P)/(2P)((a)/(a//2))^(4)rArrQ_(2)=(Q)/(8)`
`"Rate of flow will become "(1)/(8)" times"`