Question

You’re at a picnic that has a circular jug of lemonade, measuring 39 cm in diameterand 61 cm tall (which would hold about 10 gallons). It is placed on the very edge ofa 1 meter high picnic table. This cooler has a spigot down at the bottom to fill yourcups, but it has popped off, leaving a 1.15 cm diameter hole open to the air. If thecooler is not sealed (so that the top is open to the air) and if its filled up 4/5 of theway with lemonade,

(a) what is the speed of the lemonade moving through the hole when the spigot firstpops off?Note: the density of the lemonade isρL= 1.1 g/cm2or 1100 kg/m3.

(b) How far along the ground does the lemonade splash from the edge of the table?Note: consider the lemonade to come out of the holehorizontally.

Answer 1

a.

By Bernoulli's equation

Ptop + dgh + 1/2dv^2 = Pbottom + dg(0) + 1/2dv^2

Ptop = Pbottom = Patm

(because open to air)

Where d = density of lemonade

2gh + (V1) ^2 = (V2) ^2

By equation of continuity

A1V1 = A2V2

π(39)^2×V1 = π(1.15)^2×V2

V1 = 0.00087V2

V1 is negligible as compared to V2 so we can neglect it.

So 2gh = (V2)^2

h = (4/5)×0.61 = 0.488 m

2×9.8×0.488 = (V2)^2

V2 = 3.1 m/s

b.

Height of table(h) = 1 m

By equation of motion in vertical direction

S = ut + 1/2at^2

h = 0 + 1/2gt^2

1 = 1/2×9.8×t^2

t = 0.45 sec

So equation of motion in horizontal direction

S = ut ( as acceleration is zero in horizontal direction)

S = V2×t = 3.1×0.45

= 1.4 m

This is the horizontal distance from the table.

If you have any doubt regarding this feel free to ask.