Consider two solid spheres P and Q each of density 8 g cm -3 and diameters 1 cm and 0-5 cm respectively- Sphere P is dropped into a liquid of density 0-8 g cm -3 and viscosity -3 poiseulles- Sphere Q is dropped into a liquid of density 1-6 g-cm3 and viscosity -2 poiseulles- The ratio of the terminal velocities of P and Q is-

Terminal velocity #160; #160; #160; #160; v #160;= 2 gr^2 (ρ σ)9η #160; #160; v_pv_Q #160;= r_p^2 #160;(ρ_p σ_1 )η ...

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Standard XII

Physics

Terminal Velocity

Consider two ...

Question

Consider two solid spheres P and Q each of density $8 g$ ${c m}^{- 3}$ and diameters $1$ cm and $0.5$ cm respectively. Sphere P is dropped into a liquid of density $0.8 g$ ${c m}^{- 3}$ and viscosity $η$ =3 poiseulles. Sphere Q is dropped into a liquid of density $1.6 g / c m^{3}$ and viscosity $η$ =2 poiseulles. The ratio of the terminal velocities of P and Q is:

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Solution

Terminal velocity $v = \frac{2 g r^{2} (ρ - σ)}{9 η}$
$⟹$ $\frac{v_{p}}{v_{Q}} = \frac{r_{p}^{2} (ρ_{p} - σ_{1}) η_{2}}{r_{Q}^{2} (ρ_{Q} - σ_{2}) η_{1}}$

Given : $η_{1} = 3$ PI $= 30 P$ $η_{2} = 2$ PI $= 20 P$ $ρ_{p} = ρ_{Q} = 8$ $g c m^{- 3}$ $σ_{1} = 0.8$ $g c m^{- 3}$ $σ_{2} = 1.6$ $g c m^{- 3}$
Radius of sphere P $r_{p} = \frac{1}{2} c m = 0.005 m$

Radius of sphere Q $r_{Q} = \frac{0.5}{2} c m = 0.0025 m$

$∴$ $\frac{v_{p}}{v_{Q}} = \frac{(0.005)^{2} (8 - 0.8) (20)}{(0.0025)^{2} (8 - 1.6) (30)} = 3$

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Terminal velocity v=2gr2(ρ−σ)9η⟹ vpvQ=r2p(ρp−σ1)η2r2Q(ρQ−σ2)η1Given : η1=3 PI =30P η2=2 PI =20P ρp=ρQ=8 gcm−3 σ1=0.8 gcm−3 σ2=1.6 gcm−3 Radius of sphere P rp=12cm=0.005mRadius of sphere Q rQ=0.52cm=0.0025m∴ vpvQ=(0.005)2(8−0.8)(20)(0.0025)2(8−1.6)(30)=3