A metal ball B1 (density 3.2 g cm−3) is dropped in water while another metal ball B2 (density 6.0 g cm−3) is dropped in a liquid of density 1.6 g cm−3. If both the balls have the same diameter and attain the same terminal velocity, the ratio of viscosity of water to that of the liquid is:

Question

A metal ball B1​ (density 3.2 g cm−3) is dropped in water while another metal ball B2​ (density 6.0 g cm−3) is dropped in a liquid of density 1.6 g cm−3. If both the balls have the same diameter and attain the same terminal velocity, the ratio of viscosity of water to that of the liquid is:

Filo · Accepted Answer

The terminal velocity of the body of radius r, density

ρ

falling through a medium of density

σ

is given by

v = \frac{2}{9} \frac{r ^{2} ( ρ - σ ) g}{η}

where

η

is the coefficient of viscosity of the medium

∴ v_{B_{1}} = \frac{2}{9} \frac{r _{B_{1}}^{2}}{η _{w a t er}} (ρ_{B_{1}} - σ_{w a t er}) g

....(i)

and

v_{B_{2}} = \frac{2}{9} \frac{r _{B_{2}}^{2}}{η _{l i q u i d}} (ρ_{B_{2}} - σ_{l i q u i d}) g

....(ii)

where the subscripts

B_{1}

and

B_{2}

represent metal ball

B_{1}

and metal ball

B_{2}

respectively.

∵ r_{B_{1}} = r_{B_{2}}

and

∵ v_{B_{1}} = v_{B_{2}}

(Given)

Substituting these values in (i) and (ii), we get

\frac{η _{w a t er}}{η _{l i q u i d}} = \frac{( ρ _{B_{1}} - σ _{w a t er} )}{( ρ _{B_{2}} - σ _{l i q u i d} )}

Substituting the given values, we get

\frac{η _{w a t er}}{η _{l i q u i d}} = \frac{( 3.2 - 1 )}{( 6.0 - 1.6 )}

(

∵

Density of water

= 1 g c m^{- 3}

)

= \frac{2.2}{4.4} = 0.5

Question Text	A metal ball $B_{1}$ (density 3.2 g $c m^{- 3}$ ) is dropped in water while another metal ball $B_{2}$ (density 6.0 g $c m^{- 3}$ ) is dropped in a liquid of density 1.6 g $c m^{- 3}$ . If both the balls have the same diameter and attain the same terminal velocity, the ratio of viscosity of water to that of the liquid is:
Answer Type	Text solution:1
Upvotes	150

Text solutionVerified

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