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Physics Class 12th Mechanical Properties of Fluids Physics NCERT Exemplar Solutions Class 12th Chapter Ten

A water drop of radius $1 μ m$ falls in a situation where the effect of buoyant force is negligible. Co-efficient of viscosity of air is 1.8 × 10^-⁵ Nsm^-² and its density is negligible as compared to that of water 10⁶ gm^-³. Terminal velocity of the water drop is:

(Take acceleration due to gravity = 10 ms^-²)

Option 1 -

145.4 * 10^-6 ms^-1

Option 2 -

118.0 * 10^-6 ms^-1

Option 3 -

132.6 * 10^-6 ms^-1

Option 4 -

123.4 * 10^-6 ms^-1

3 Views | Posted 3 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

3 months ago

Correct Option - 4

Detailed Solution:

NLM 1

$ρ_{ω} \frac{4}{3} π r^{3} g = 6 π η r V_{T}$

$\frac{ρ_{ω} 4}{3} π r^{2} g =$ $6 π η v_{T}$

$V_{T} = \frac{4}{3} \frac{ρ_{ω} π r^{2} g}{6 π η}$

$= \frac{2}{9} \times 1 0^{3} \times \frac{{(1 0^{- 6})}^{2} \times 10}{1.8 \times 1 0^{- 5}}$

$= 0.1234 \times 1 0^{- 3}$

vT = 123.4 × 10^-6m/sec

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