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# The tube shown is of a 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 stateThe liquid exerts on the tube:Statement I: A net force towards right Statement II: A net force towards leftStatement III: A net force in some oblique direction Statement IV: Zero net forceStatement V: A net clockwise torqueStatement 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

Last updated date: 04th Mar 2024
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We know that 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 $\rho \mathrm{Sv}^{2}$ 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 the 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.