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Q.

Water (density ρ) is flowing through the uniform tube of cross-sectional area A with a constant speed v as shown in the figure. The magnitude of force exerted by the water on the curved corner of the tube is (neglect viscous forces)

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a

3ρAv2

b

2ρAv2

c

2ρAv2

d

ρAv22

answer is A.

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Detailed Solution

ΔP→x=mvsin⁡60∘=32mvΔP→y=mv2+mv=32mv⇒ ΔP→net =ΔPx2+ΔPy2=94+34mvΔP→net=3mv⇒ ΔF→net =3dmdt⋅v=3ρAv2 Since, dm=A(vdt)ρ⇒ dmdt=Aρv

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Water (density ρ) is flowing through the uniform tube of cross-sectional area A with a constant speed v as shown in the figure. The magnitude of force exerted by the water on the curved corner of the tube is (neglect viscous forces)