Question

A 0.13 inch diameter hole is punched in the bottom of a 2.1 inch diameter can. The can is open at the top and contains 20 oz of liquid. What is the initial velocity of the liquid coming out of the hole? (include units with answer)

Answer 1

diameter of the hole=0.13 inches = 0.0033 m

diameter of the can = 2.1 inch = 0.05334 m

volume = 20oz = 5.91*10^-4 m^3

so,

let the height of the liquid in the can be h.so,

volume = pi*r^2*h

or pi*(0.05334*0.5)^2*h = 5.91*10^-4

or h=0.2644 m

now, let the velocity with which the heigh of the liquid in the can is decreasing be v and the speed of liquid flowing out be u.so,

applying equation of continuity,

u*A1=v*A2

or u*0.0033^2 = v*0.05334^2

also, conserving energy,(bernouli's equation)

0.5*v^2 + 9.81*0.2644 = 0.5*u^2

solving the above two equation, we get

v=0.00872 m/s

u(required answer)=2.277 m/s