हिंदी

A drop of radius 1.5 x 10-5 and density 2 x 103 kg/m3 falls through air. The viscosity of air is 1.8 x 10-5 N s/m2. Neglecting buoyancy due to air, the terminal speed of the drop is ____________. -

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प्रश्न

A drop of radius 1.5 x 10-5 and density 2 x 103 kg/m3 falls through air. The viscosity of air is 1.8 x 10-5 N s/m2. Neglecting buoyancy due to air, the terminal speed of the drop is ____________.

विकल्प

  • 2.8 cm/s

  • 3.8 cm/s

  • 4.8 cm/s

  • 5.4 cm/s

MCQ
रिक्त स्थान भरें

उत्तर

A drop of radius 1.5 x 10-5 and density 2 x 103 kg/m3 falls through air. The viscosity of air is 1.8 x 10-5 N s/m2. Neglecting buoyancy due to air, the terminal speed of the drop is 5.4 cm/s.

Explanation:

Neglecting buoyancy due to air,

`"v" = (2"r"^2rho"g")/(9eta)`

`= (2 xx(1.5 xx 10^-5)^2 xx 2 xx10^3 xx 9.8)/(9 xx 1.8 xx 10^-5)`

`= 5.4 xx 10^-2  "m/s"`

`"v" approx 5.4  "cm/s"`

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