1.

If a ball of steel (density p 7.8 g cm-3) attains a terminal velocity of 10 cm s' when falling in a water(Coefficient of Viscosity nwater 8.5 x 104 Pa.s) then its terminal velocity in glycerine (p 1.2g cm[AIEEE 2011, 11 May; 4/120, -1]13.2 Pa.s.) would be, nearly :(1) 6.25 x 10 cms(2) 6.45 x 10 cms(3)1.5 x10 cms1 (4)1.6 x105 cms

Answer»

Vpg = 6πηrv = Vpℓg

Vg (p -pℓ) = πηrv

Also Vg (p - pℓ) = 6 πη’rv’

∴ v’η’ = - (p - pℓ)/ (p - pℓ) * vη

v’ = (p - pℓ)/ (p - pℓ) * vη/η’

= (7.8 – 1.2)/(7.8 – 1) * 10 * 8.5 * 10-4/13.2

∴ v’ = 6.25 * 10-4cm/s



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