A liquid of density -1000 kg - m 3 - viscosity -0-1 N - S - m 2 is flowing down in vertical pipe of large cross section- A small ball of density -0-100 kg - m 3 and radius 5 cm will be at rest in flowing liquid- The velocity of flowing liquid isA- 50 m - sB- 100 m - sC- 25 m - sD- 12 m - s

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Standard XII

Physics

Pseudo Force

A liquid of d...

Question

A liquid of density $ρ = 1000 k g / m^{3}$ ; viscosity $η = 0.1 N - S / m^{2}$ is flowing down in vertical pipe of large cross section. A small ball of density $ρ_{0} = 100 k g / m^{3}$ and radius 5 cm will be at rest in flowing liquid. The velocity of flowing liquid is

50 m/s

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100 m/s

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25 m/s

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12 m/s

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Solution

The correct option is A 50 m/s
In fluid frame ball is moving with constant velocity (terminal) and resultant of all forces acting on it is zero.

$∴$ terminal velocity $V_{t} = \frac{2 (ρ - ρ_{0}) g r^{2}}{η} = \frac{2}{9} \times \frac{900 \times 10 \times (0.05)^{2}}{0.1}$

$= 50 m / s$

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A liquid of density ρ=1000kg/m3; viscosity η=0.1N−S/m2 is flowing down in vertical pipe of large cross section. A small ball of density ρ0=100kg/m3 and radius 5 cm will be at rest in flowing liquid. The velocity of flowing liquid is

The correct option is A 50 m/sIn fluid frame ball is moving with constant velocity (terminal) and resultant of all forces acting on it is zero. ∴ terminal velocity Vt=2(ρ−ρ0)gr2η=29×900×10×(0.05)20.1 =50 m/s

A liquid of density $ρ = 1000 k g / m^{3}$ ; viscosity $η = 0.1 N - S / m^{2}$ is flowing down in vertical pipe of large cross section. A small ball of density $ρ_{0} = 100 k g / m^{3}$ and radius 5 cm will be at rest in flowing liquid. The velocity of flowing liquid is

The correct option is A 50 m/s
In fluid frame ball is moving with constant velocity (terminal) and resultant of all forces acting on it is zero.

$∴$ terminal velocity $V_{t} = \frac{2 (ρ - ρ_{0}) g r^{2}}{η} = \frac{2}{9} \times \frac{900 \times 10 \times (0.05)^{2}}{0.1}$

$= 50 m / s$