Momentum and CollisionHard

Question

The tube shown is of non-uniform 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-

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

Options

A.Only statement I and V are correct
B.Only statement II and VI are correct
C.Only statement IV and VI are correct
D.Only statement III and VI are correct

Solution

The force has been exerted by liquid on the tube due to change in momentum at the corners i. e. , when liquid is taking turn from A to B and from B to C.As cross-section area at A 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 A be v and cross section area at A be S, the velocity of flow of liquid at B and C would be [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 ρSv2 on the tube, where r is the density of liquid, S is cross section area 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

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