Question

A wide tank of cross section area a containing a liquid to a height h has an orifice at its base of area a/3 the initial acceleration of top surface of liquid is

Answer 1

Answer:

The answer will be g/8(m/s^2)

Explanation:

According to the problem the area of the particular cross section is a .

Now from the principle of continuity for the upper surface and bottom,

a1×u1=a2×u2  (where u1 and  u2 are the velocity of upper surface and at bottom respectively)

now the bottom has the area of a/3, therefore,

=> a x u1 = a/3 x u2

=> u1 = u2/3

Now we will apply bernoulii's theorem for the upper surface and bottom,

P1+1/2ρu1^2 +ρgh1=P2+1/2ρu2^2 +ρgh2

now let P is the atmospheric pressure,

therefore,

P+1/2ρu1^2 +ρgh1=P+1/2ρu2^2 +ρgh2

now let the height of the liquid from the bottom is l

=>1/2ρu1^2 +ρgx = 1/2ρu2^2[ h2= 0]

=> 1/2ρu1^2 +ρgx =  1/2ρ(3u1)^2

=> 1/2u1^2+gx =1/2(3u1)^ 2

=>gx = 4v1^2

Therefore,

g = 4 x 2u1 x du1/dx

 =>g/8 = u1 x du1/dx

Therefore the acceleration of the top surface is g/8(m/s^2)