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If a ball of steel ( density p = 7.8 g c...

If a ball of steel ( density p = 7.8 g `cm^(-3)`) attains terminal velocity of 10 `cm s^(-1)` when falling in a water ( coefficient of viscosity `eta_("water") = 8.5 xx 10^(-4)` Pa.s) then its terminal velocity in glycerine `(p = 1.2 g cm^(-3) , eta = 13.2 Pa.s)` would be, nearly :

A

`6.25xx10^(-4)cms^(-1)`

B

`6.45xx10^(-4)cms^(-1)`

C

`1.5xx10^(-5)cms^(-1)`

D

`1.6xx10^(-3)cms^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A
`Vpg = 6pietarv + Vp_(1)g`
`Vg(p-p-(l))=6pietarv`
`Vg(p-p_(l))=6pieta^(1)rv^(1)`
`V^(1)eta^(1)=((p-p_(l)^(1)))/((p-p_(l)))xxveta`
`V^(1)=((p-p_(l)^(1)))/((p-p_(l)))xx(veta)/(eta^(1))`
`=((7.8-1.2))/((7.8-1))xx(10xx8.5xx10^(-4))/(13.2)`
`V^(1)=6.25xx19^(-4)cm`/s
Correct choice is (a)
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