question_answer

The tube shown is of non-uniform cross-section. The cross-section area at A is half of the cross-section area at B, C and D. A liquid is flowing through in steady state. The liquid exerts on the tube is

Statement I: A net force towards right.

Statement II: A net force towards left.

Statement III: A net force in some oblique direction.

Statement IV: Zero net force.

Statement V: A net clockwise torque.

Statement VI: A net counter-clockwise torque.

Out of these:

A) Only statements I and V are correct    

B)           Only statements II and VI are correct

C) Only statements IV and VI are correct               

D) Only statements III and VI are correct

Correct Answer: A

Solution :

[a] The force has been exerted by liquid on the tube due to change in momentum at the comers, i.e., when liquid is taking turn from A to B and from B to C. As cross-section area at ^ is half of that of B and C, so velocity of liquid flow at B and C is half to that of velocity at A. Let velocity of flow of liquid at .4 be v and cross section area at A be S, the velocity of flow of liquid at B and C would be v/2 [from continuity equation] and cross section area at B and C would be 2S. Due to flow of liquid, it is exerting a force per unit time of \[\rho S{{v}^{2}}\] on the tube, where \[\rho \] is the density of liquid, S is cross section are a and v is velocity of flow of liquid. The force exerted by liquid on tube is shown in the figure which clearly shows that a net force is acting on tube due to flowing liquid towards right and a clockwise torque sets in.