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Properties of Matter question

2022 · 25 Jun · Shift 1 · Q49
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  5. /2022 · 25 Jun · Shift 1 · Q49

Properties of Matter question

2022 · 25 Jun · Shift 1 · Q49

JEE MainPhysicsProperties of MatterMCQ+4 / −1
The terminal velocity (vt) of the spherical rain drop depends on the radius (r) of the spherical rain drop as :
  1. A
    r1/2
  2. B
    r
  3. C
    r2
  4. D
    r3
View written solutionFree

Correct answer: C

  1. Use Stokes' law for a small spherical drop moving through air

For a sphere of radius rrr moving with terminal velocity vtv_tvt​ in a viscous medium,

  • Weight of drop: W=43πr3ρgW = \frac{4}{3}\pi r^3 \rho gW=34​πr3ρg
  • Buoyant force of air: B=43πr3ρairgB = \frac{4}{3}\pi r^3 \rho_{air} gB=34​πr3ρair​g
  • Viscous drag (Stokes' law): Fv=6πηrvtF_v = 6\pi \eta r v_tFv​=6πηrvt​

At terminal velocity, net force is zero: W−B=FvW - B = F_vW−B=Fv​

So, 43πr3(ρ−ρair)g=6πηrvt\frac{4}{3}\pi r^3 (\rho - \rho_{air}) g = 6\pi \eta r v_t34​πr3(ρ−ρair​)g=6πηrvt​

  1. Solve for terminal velocity

vt=43πr3(ρ−ρair)g6πηrv_t = \frac{\frac{4}{3}\pi r^3 (\rho - \rho_{air}) g}{6\pi \eta r}vt​=6πηr34​πr3(ρ−ρair​)g​

Cancel common factors: vt=29(ρ−ρair)gηr2v_t = \frac{2}{9}\frac{(\rho - \rho_{air})g}{\eta} r^2vt​=92​η(ρ−ρair​)g​r2

Thus, vt∝r2v_t \propto r^2vt​∝r2

  1. Match with options
  • A: r1/2r^{1/2}r1/2 — incorrect
  • B: rrr — incorrect
  • C: r2r^2r2 — correct
  • D: r3r^3r3 — incorrect

Hence, the correct option is C.

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