Fluid Mechanics

JEE Main level, single correct, medium

162 of 192

Question 162 of 192

A small metallic sphere of diameter 2 mm and density 10.5 g/cm310.5 \mathrm{~g} / \mathrm{cm}^{3} is dropped in glycerine having viscosity 10 Poise and density 1.5 g/cm31.5 \mathrm{~g} / \mathrm{cm}^{3} respectively. The terminal velocity attained by the sphere is ____\_\_\_\_ cm/s\mathrm{cm} / \mathrm{s}.
(π=227\pi=\frac{22}{7} and g=10 m/s2g=10 \mathrm{~m} / \mathrm{s}^{2})

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Answer: (C) 2.0

Using Stokes' Law for terminal velocity:
vt=2r2g(ρs−ρf)9ηv_t = \frac{2r^2g(\rho_s - \rho_f)}{9\eta}.
Given: r = 0.1 cm, g = 1000 cm/s², ρs=10.5\rho_s = 10.5 g/cm³, ρf=1.5\rho_f = 1.5 g/cm³, η = 10 Poise = 10 g/(cm⋅s).
Substituting:
vt=2×(0.1)2×1000×(10.5−1.5)9×10v_t = \frac{2 \times (0.1)^2 \times 1000 \times (10.5 - 1.5)}{9 \times 10}
=2×0.01×1000×990= \frac{2 \times 0.01 \times 1000 \times 9}{90}
=18090=2.0= \frac{180}{90} = 2.0 cm/s.

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