Question

liquid enters a tube of variable cross section with a velocity 3 m/s through the wider end and leaves from the narrower end whose radius is half of that wider one the velocity with which it leaves the tube is

Answer 1

we know that the amount of water that enters will come out it will not be increased or decreased

so according to the law of continuity

A(area) × V(velocity) will be constant

we can say

water entering =water exiting

area of broader end × velocity of water in broader end = area of smaller end × velocity of water in narrow end

pie

${ve} \times \pi \: {r}^{2} = \pi {r \div 2}^{2} \times {vi}$

where {ve} is the velocity of water entering

and {vi} is the velocity leaving

after putting the values and calculations

we get

{vi} = 12 m/s

Answer 2

we know that the amount of water that enters will come out it will not be increased or decreased

so according to the law of continuity

A(area) × V(velocity) will be constant

we can say

water entering =water exiting

area of broader end × velocity of water in broader end = area of smaller end × velocity of water in narrow end

π

where {ve} is the velocity of water entering

and {vi} is the velocity leaving

after putting the values and calculations

we get

= 12 m/s