U - Tube moves with constant speed v parallel to the surface of a stationary liquid. The cross section area of the lower part of the tube, lowered into the liquid, is equal to S1 and of the top part located over liquid is S2. Friction and the formation of waves should be neglected. Density fluid ρ. Neglect difference in heights at both opening of the tube.

U - Tube moves with constant speed v parallel to the surface of a stationary liquid. The cross section area of the lower part of the tube, lowered into the liquid, is equal to S₁ and of the top part located over liquid is S₂. Friction and the formation of waves should be neglected. Density fluid ρ. Neglect difference in heights at both opening of the tube.

Linked Question 1

If we turn the U tube upside down so that the narrower part (area S₂) is inside the liquid and broader part (area S₁) is outside the liquid, what will be the effect on the velocity of liquid coming out as seen by a ground observer and force required to be applied to the tube?

velocity will be lesser, force will be more

velocity will be lesser, force will be lesser

velocity will be more, force will be lesser

velocity will be more, force will be more

$F = {ρv}^{2} S_{2} [1 + \frac{S_{2}}{S_{1}}] = {ρv}^{2} [S_{2} + \frac{S_{2}^{2}}{S_{1}}]$

$v \leq v (1 + \frac{S_{1}}{S_{2}})$ .

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Linked Question 2

What force is applied to the tube?

${ρv}^{2} S_{1} (1 + \frac{S_{2}}{S_{1}})$

${ρv}^{2} S_{1} (1 + \frac{S_{1}}{S_{2}})$

${ρv}^{2} S_{2} (1 - \frac{S_{2}}{S_{1}})$

ρv²S₁

$F = \frac{dm}{dt} ({\vec{v}}_{f} - {\vec{v}}_{i})$

$= {pvS}_{1} [\frac{vS}{S_{1}} + v] = {pv}_{2} S_{1} (1 + \frac{S_{1}}{S_{2}})$ .

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Linked Question 3

What is the velocity of the liquid coming out of the top part as seen by an observer on ground?

$v (1 - \frac{S_{1}}{S_{2}})$

None of these

$v (1 + \frac{S_{1}}{S_{2}})$

Zero

As seen by U tube, water moves back with velocity v
vS₁ = v'_rel S₂

$\Rightarrow v_{rel}^{'} = \frac{{vS}_{1}}{S_{2}}$

$y_{gr} = v_{rel} + v = v (1 + \frac{S_{1}}{S_{2}})$ .

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