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Terminal Velocity

A sphere of r...

Question

A sphere of radius $10$ cm and density $500 k g / m^{3}$ is under water of density $1000 k g / m^{3}$ . The acceleration of the sphere is $9.80 m / s^{2}$ upward. Viscosity of water is $1.0$ centipoises. If $g = 9.81 m / s^{2}$ , the velocity of sphere is:

9

m/s

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10

m/s

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11

m/s

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12

m/s

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Solution

The correct option is C $11$ m/s

$Weknow that$

$f_{n e t} = B o y u n c y - m g - d r a g f o r c e$

$\frac{4}{3} π (0.1)^{3} \times 500 \times 9.8 = 1000 \times \frac{4}{3} π (0.1)^{3} \times 9.81 - 500 \times \frac{4}{3} π (0.1)^{3} \times 9.81 - 6 π \times 10^{- 3} \times V \times 0.1$

$6 π \times 10^{- 3} v = 500 \times \frac{4}{3} π (0.1)^{2} (0.01)$

$V = \frac{100}{9} = 11 m / s e c$

Ans $: (c)$

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