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Question

The velocity of a liquid (v) in steady flow at a location through cylindrical pipe is given by v = v0 (1 − r2 R2), where r is the radial distance of that location from the axis of the pipe and R is the inner radius of the pipe, if R = 10 cm, volume flow rate through the pipe is π/2 × 10−2 m2s−1 and the coefficient of viscosity of the liquid is 0.75 N s m−2

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Solution

The correct options are

**A** Magnitude of the viscous force per unit area at r = 4 cm is 6 N/m2

**B** Maximum velocity of flow of liquid is 1 m/s

**C** Viscous force per unit area increases with radial distance

Magnitude of viscous force, F = ηAdvdr

⇒ Viscous force per unit area τ = FA = −ηdvdr

v = v0 (1 − r2R2)dvdr = −2v0rR2 −−− (i)⇒ τ = η.2v0rR2

From above equation, we can conclude that viscous force per unit area increases with radial distance.

Consider an annular element at distance r from axis having width dr

dA = 2πrdrdQ = V.dA = v0 (1 − r2R2)2πrdrQ = ∫dQ = 2πv0 (r22−r44R2]R0 = π2R2v0⇒v0 = 2QπR2∴ τ = η.4QπR4.r, (from equation (i)) where R = 0.1 m At r = 0.04 m, τ = 0.75×4×π2×10−2×0.04π×10−4 = 6 N/m2

Also, maximum velocity will be at r=0 which is equal to v0=2QπR2=1 m/s

Magnitude of viscous force, F = ηAdvdr

⇒ Viscous force per unit area τ = FA = −ηdvdr

v = v0 (1 − r2R2)dvdr = −2v0rR2 −−− (i)⇒ τ = η.2v0rR2

From above equation, we can conclude that viscous force per unit area increases with radial distance.

Consider an annular element at distance r from axis having width dr

dA = 2πrdrdQ = V.dA = v0 (1 − r2R2)2πrdrQ = ∫dQ = 2πv0 (r22−r44R2]R0 = π2R2v0⇒v0 = 2QπR2∴ τ = η.4QπR4.r, (from equation (i)) where R = 0.1 m At r = 0.04 m, τ = 0.75×4×π2×10−2×0.04π×10−4 = 6 N/m2

Also, maximum velocity will be at r=0 which is equal to v0=2QπR2=1 m/s

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