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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 η water = 8.5*10 -4 Pa.s), then, its terminal velocity in glycerine (p=1.2g cm -3 , η=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 -5 cm s -1 (d) 1.6 * 10 -5 cm s -1

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 ηwater = 8.5*10-4 Pa.s), then, its  terminal velocity in glycerine (p=1.2g cm-3, η=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

Grade:upto college level

1 Answers

Navjyot Kalra
askIITians Faculty 654 Points
9 years ago
(a) Vpg = 6πηrv = Vpℓg
\RightarrowVg (p -pℓ) = πηrv
Also Vg (p - pℓ) = 6 πη’rv’
∴ v’η’ = - (p - pℓ)/ (p - pℓ) * vη
\Rightarrowv’ = (p - pℓ)/ (p - pℓ) * vη/η’
= (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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