Find the terminal velocity of a steel ball of diameter $0.2 c m$ when it falls through a tube filled with glycerin. $(g = 9.8 m / s^{2}$ ,density of steel $= 8000 k g / m 63$ ,density of glycerin $= 1.33 g / c m^{3}, η = 8.33$ poise $= 0.833 N s / m^{2}$ )

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Solution

Given : $σ = 8000 k g / m^{3}$ $ρ = 1330 k g / m^{3}$
Radius of ball $r = \frac{d}{2} = \frac{0.2 c m}{2} = 0.1 c m = 10^{- 3} m$
Terminal velocity $v = \frac{2 g r^{2} (σ - ρ)}{9 η}$
Or $v = \frac{2 \times 9.8 \times (10^{- 3})^{2} (8000 - 1330)}{9 \times 0.833} = 1.74 \times 10^{- 2} m / s = 1.74 c m / s$

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Find the terminal velocity of a steel ball of diameter 0.2cm when it falls through a tube filled with glycerin.(g=9.8m/s2,density of steel =8000kg/m63,density of glycerin =1.33g/cm3,η=8.33 poise =0.833Ns/m2)

Given : σ=8000 kg/m3 ρ=1330 kg/m3 Radius of ball r=d2=0.2 cm2=0.1 cm=10−3 mTerminal velocity v=2gr2(σ−ρ)9ηOr v=2×9.8×(10−3)2(8000−1330)9×0.833=1.74×10−2 m/s=1.74 cm/s

Find the terminal velocity of a steel ball of diameter $0.2 c m$ when it falls through a tube filled with glycerin. $(g = 9.8 m / s^{2}$ ,density of steel $= 8000 k g / m 63$ ,density of glycerin $= 1.33 g / c m^{3}, η = 8.33$ poise $= 0.833 N s / m^{2}$ )