If a ball of steel (density ρ = 7.8 g cm -3) attains a terminal velocity of 10 cm s-1 when falling in a water (Coefficient of Viscosity ηwater = 8.5 × 10-4 Pa.s) then its terminal velocity in glycerine (ρ = 1.2 g cm-3, η = 13.2 Pa.s.) would be, nearly :

Question

If a ball of steel (density ρ = 7.8 g cm -3) attains a terminal velocity of 10 cm s-1 when falling in a water (Coefficient of Viscosity ηwater = 8.5 × 10-4 Pa.s) then its terminal velocity in glycerine (ρ = 1.2 g cm-3, η = 13.2 Pa.s.) would be, nearly : (A) 6.25 × 10-4 cm s-1 (B) 6.45 × 10-4 cm s-1 (C) 1.5 × 10-5 cm s-1 (D) 1.6 × 10-4 cm s-1

Tardigrade Admin · Accepted Answer

Vρ g = 6πη rv + vρℓg Vg(ρ - ρℓ) = 6πη rv Vg(ρ-ρℓ') = 6πη'rv' V' η' = ((ρ -ρ ℓ ')/(ρ -ρℓ ))× vη V' = ((ρ -ρ ℓ ')/(ρ -ρℓ ))× (vη/η') = ((7.8 -1.2)/(7.8 - 1))× (10 × 8.5 ×10-4/13.2) v' = 6.25 Ã 10-4 cm/s.

$6.25 \times 1 0^{- 4} c m s^{- 1}$

$6.45 \times 1 0^{- 4} c m s^{- 1}$

$1.5 \times 1 0^{- 5} c m s^{- 1}$

$1.6 \times 1 0^{- 4} c m s^{- 1}$

Q. If a ball of steel (density ρ=7.8gcm−3) attains a terminal velocity of 10cms−1 when falling in a water (Coefficient of Viscosity ηwater​ =8.5×10−4Pa.s) then its terminal velocity in glycerine (ρ=1.2gcm−3,η=13.2Pa.s.) would be, nearly :

6.25×10−4cms−1

6.45×10−4cms−1

1.5×10−5cms−1

1.6×10−4cms−1

Vρg=6πηrv+vρℓ​g Vg(ρ−ρℓ​)=6πηrv Vg(ρ−ρℓ′​)=6πη′rv′ V′η′=(ρ−ρℓ​)(ρ−ρℓ′​)​×vη V′=(ρ−ρℓ​)(ρ−ρℓ′​)​×η′vη​ =(7.8−1)(7.8−1.2)​×13.210×8.5×10−4​ v′=6.25×10−4cm/s.

$6.25 \times 1 0^{- 4} c m s^{- 1}$

$6.45 \times 1 0^{- 4} c m s^{- 1}$

$1.5 \times 1 0^{- 5} c m s^{- 1}$

$1.6 \times 1 0^{- 4} c m s^{- 1}$