JEE Challenger
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PhyMedAdv 2016 • Paper 1

Ratio of Terminal Velocities of Two Spheres in Viscous Liquids

Consider two solid spheres PP and QQ each having a density of 8 gm cm−38\ \text{gm cm}^{-3} with diameters 1 cm1\ \text{cm} and 0.5 cm0.5\ \text{cm}, respectively. Sphere PP is dropped into a liquid with density 0.8 gm cm−30.8\ \text{gm cm}^{-3} and viscosity η=3 poiseulles\eta = 3\ \text{poiseulles}. Sphere QQ is dropped into another liquid of density 1.6 gm cm−31.6\ \text{gm cm}^{-3} and viscosity η=2 poiseulles\eta = 2\ \text{poiseulles}. Find the ratio of terminal velocities corresponding to PP and QQ.

Official Numerical Answer3

Topics & Concepts

Mechanical Properties of FluidsFluid Dynamics

Step-by-Step Solution

To find the ratio of the terminal velocities of sphere PP and sphere QQ, we use Stokes' Law. The formula for the terminal velocity vTv_T of a solid sphere falling through a viscous fluid under gravity is given by:

vT=29r2(ρ−σ)gηv_T = \frac{2}{9} \frac{r^2 (\rho - \sigma) g}{\eta}

where:

  • rr is the radius of the sphere,
  • ρ\rho is the density of the sphere,
  • σ\sigma is the density of the liquid,
  • η\eta is the coefficient of viscosity of the liquid,
  • gg is the acceleration due to gravity.

Let vPv_P and vQv_Q be the terminal velocities of spheres PP and QQ, respectively. Taking the ratio of their terminal velocities, we get:

vPvQ=(rPrQ)2×(ρ−σPρ−σQ)×(ηQηP)\frac{v_P}{v_Q} = \left( \frac{r_P}{r_Q} \right)^2 \times \left( \frac{\rho - \sigma_P}{\rho - \sigma_Q} \right) \times \left( \frac{\eta_Q}{\eta_P} \right)

Given data for the two spheres and liquids:

  • Sphere PP:

    • Diameter dP=1 cm  ⟹  radius rP=0.5 cmd_P = 1\ \text{cm} \implies \text{radius } r_P = 0.5\ \text{cm}
    • Density of sphere ρ=8 gm cm−3\rho = 8\ \text{gm cm}^{-3}
    • Density of liquid σP=0.8 gm cm−3\sigma_P = 0.8\ \text{gm cm}^{-3}
    • Viscosity of liquid ηP=3 poiseulles\eta_P = 3\ \text{poiseulles}

  • Sphere QQ:

    • Diameter dQ=0.5 cm  ⟹  radius rQ=0.25 cmd_Q = 0.5\ \text{cm} \implies \text{radius } r_Q = 0.25\ \text{cm}
    • Density of sphere ρ=8 gm cm−3\rho = 8\ \text{gm cm}^{-3}
    • Density of liquid σQ=1.6 gm cm−3\sigma_Q = 1.6\ \text{gm cm}^{-3}
    • Viscosity of liquid ηQ=2 poiseulles\eta_Q = 2\ \text{poiseulles}

Now, evaluating each ratio term individually:

  1. Ratio of radii: rPrQ=0.5 cm0.25 cm=2  ⟹  (rPrQ)2=22=4\frac{r_P}{r_Q} = \frac{0.5\ \text{cm}}{0.25\ \text{cm}} = 2 \implies \left(\frac{r_P}{r_Q}\right)^2 = 2^2 = 4

  2. Ratio of effective densities: ρ−σPρ−σQ=8−0.88−1.6=7.26.4=98\frac{\rho - \sigma_P}{\rho - \sigma_Q} = \frac{8 - 0.8}{8 - 1.6} = \frac{7.2}{6.4} = \frac{9}{8}

  3. Ratio of viscosities: ηQηP=23\frac{\eta_Q}{\eta_P} = \frac{2}{3}

Substituting these individual ratios into the combined ratio expression:

vPvQ=4×98×23\frac{v_P}{v_Q} = 4 \times \frac{9}{8} \times \frac{2}{3}

vPvQ=4×9×28×3=7224=3\frac{v_P}{v_Q} = \frac{4 \times 9 \times 2}{8 \times 3} = \frac{72}{24} = 3

Thus, the ratio of the terminal velocities corresponding to PP and QQ is 33.

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