Question

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:
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 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

Answer 1

The correct option is A Only statement I and V are correct

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 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 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 the tube due to flowing liquid towards right and a clockwise torque sets in.