Water flows through a tapering horizontal tube whose radii of cross-section of the ends `r_(1) =20`cm and `r_(2) =10`cm. The velocity of water at the points for the radius of cross-section `r_(1)` is `v_(1)=2 ms^(-1)` . The force imparted by the emergiang water at the other end of teh tube is nearly `2xx10^(n)`N. What is the value of n ?

Answer

3

Explanation

From equation of continuity, we have

`A_(1)v_(1) =A_(2)v_(2)` or `v_(2) =(A_(1)v_(1))/(A_(2)) = (pi r_(1)^(2))/(pi r_(2)^(2))xxv_(1) =(pi_(1)^(2)v_(1))/(r_(2)^(2))`
Force exerted by emerging liquid on the tube is
`F=v_(2)(dm)/(dt) =v_(2)(A_(2) rho v_(2)) =A_(2) rho v_(2)^(2)`
`=(pi r_(2)^(2)) rho((r_(1)^(2))/(r_(2)^(2))v_(1))^(2) = (pi rho v_(1)^(2) r_(1)^(4))/(r_(2)^(2))`
`=2xx10^(3) N=2xx10^(n) N`
Hence, `n=3`