If ball of steel (density ρ=7.8 cm−3) attains a terminal velocity of 10 cms−1 when falling in a tank of water (coefficient of viscosity η water=8.5×10−4 Pa.s) then its terminal velocity in glycerin (ρ=1.2 gcm−3 η=13.2 Pa.s) would be nearly:

Question

If ball of steel (density ρ=7.8 cm−3) attains a terminal velocity of 10 cms−1 when falling in a tank of water (coefficient of viscosity η water​=8.5×10−4 Pa.s) then its terminal velocity in glycerin (ρ=1.2 gcm−3 η=13.2 Pa.s) would be nearly:

Filo · Accepted Answer

Terminal velocity of ball when falling in liquid is given by

(ρ_{b} - ρ_{l}) V g = 6 π ηr v

...eq(1)

1. case

when ball falling in tank of water of density

ρ_{w a t er} = 1 g c m^{- 3}

,

ρ_{ba ll} = 7.8 g c m^{- 3}

η_{w a t er} = 8.5 \times 1 0^{- 4} P a . s

and terminal velocity

v = 10 c m s^{- 1}

Put all the above value in eq(1)

\Rightarrow (7.8 - 1) V g = 6 π \times 8.5 \times 1 0^{- 4} r \times 10

...eq(2)

2. case

when ball falling in tank of glycerin

ρ_{ba ll} = 7.8 g c m^{- 3}

η_{g l ycer in} = 13.2 P a . s

and terminal velocity

v_{1}

put all the value in eq(1)

\Rightarrow (7.8 - 1.2) V g = 6 π \times 13.2 r \times v_{1}

...eq(3)

divide eq(3) b eq(1)

\frac{7.8 - 1.2}{7.8 - 1} = \frac{13.2 \times v _{1}}{8.5 \times 1 0 ^{- 4} \times 10}

\Rightarrow v_{1} = \frac{6.6 \times 8.51 0 ^{- 4} \times 10}{6.8 \times 13.2} = 6.25 \times 1 0^{- 4} c m s^{- 1}

terminal velocity of ball in glycerin is

6.25 \times 1 0^{- 4} c m s^{- 1}

Hence C option is correct.

Question Text	If ball of steel (density $ρ = 7.8 c m^{- 3}$ ) attains a terminal velocity of $10 c m s^{- 1}$ when falling in a tank of water (coefficient of viscosity $η_{w a t er} = 8.5 \times 10^{- 4} P a . s$ ) then its terminal velocity in glycerin ( $ρ = 1.2 g c m^{- 3} η = 13.2 P a . s$ ) would be nearly:
Answer Type	Text solution:1
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If ball of steel (density ρ=7.8 cm−3) attains a terminal velocity of 10 cms−1 when falling in a tank of water (coefficient of viscosity η water​=8.5×10−4 Pa.s) then its terminal velocity in glycerin (ρ=1.2 gcm−3 η=13.2 Pa.s) would be nearly:

Text solutionVerified

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