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If a ball of steel (density p=7.8g cm-3) attains a terminal velocity of 10cm s-1 when falling in water (Coefficient of viscosity ?water = 8.5*10-4 Pa.s), then, its terminal velocity in glycerine (p=1.2g cm-3, ?=13.2 Pa.s) would be, nearly
(a) 6.25 * 10-4 cm s-1
(b) 6.45 *10-4 cm s-1
(c) 1.5 * 10-5cm s-1
(d) 1.6 * 10-5cm s-1

Simran Bhatia , 11 Years ago
Grade 11
anser 1 Answers
Kevin Nash

Last Activity: 11 Years ago

(a) Vpg = 6p?rv = Vplg

? Vg (p -pl) = p?rv

Also Vg (p - pl) = 6 p?’rv’

? v’?’ = - (p - pl)/ (p - pl) * v?

? v’ = (p - pl)/ (p - pl) * v?/?’

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

? v’ = 6.25 * 10-4 cm/s

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