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Question:

Consider a water jar of radius R that has water filled up to height H and is kept on a stand of height h (see figure). Through a hole of radius r (r≪R) at its bottom, the water leaks out and the stream of water coming down towards the ground has a shape like a funnel as shown in the figure. If the radius of the cross-section of water stream when it hits the ground is x. Then:

x=r(H/(H+h))^(1/4)

x=r(H/(H+h))^(1/2)

x=r(H/(H+h))^2

x=r(H/(H+h))

Solution:

By energy conservation, 1/2ρv₁² = ρgH and 1/2ρv₂² = ρg(H+h). Hence, v₂/v₁ = √((H+h)/H). Also, since by mass conservation, A₁v₁ = A₂v₂, we have πr²v₁ = πx²v₂. From the two relations, we get, x/r = (H/(H+h))^(1/4). x = r(H/(H+h))^(1/4).