A solid steel ball of diameter 3.6 mm acquired terminal velocity while falling under gravity through an oil of density 925 . Take density of steel as 7825 and g as 9.8 . The viscosity of the oil in SI unit is [2025]

Question

A solid steel ball of diameter 3.6 mm acquired terminal velocity $2.45 \times 10^{- 2} m / s$ while falling under gravity through an oil of density 925 $kg m^{- 3}$ . Take density of steel as 7825 $kg m^{- 3}$ and g as 9.8 $m / s^{2}$ . The viscosity of the oil in SI unit is [2025]

Smart Achievers · Accepted Answer

(4)

$v_{T} \Rightarrow \frac{2}{9} \frac{(ρ_{0} - ρ_{l}) r^{2} g}{η}$

$η = \frac{2}{9} (\frac{7825 - 925}{2.45 \times 10^{- 2}}) \times {(1.8)}^{2} \times 10^{- 6} \times 9.8$

$η = 1.99$

Solution

Q.
A solid steel ball of diameter 3.6 mm acquired terminal velocity $2.45 \times 10^{- 2} m / s$ while falling under gravity through an oil of density 925 $kg m^{- 3}$ . Take density of steel as 7825 $kg m^{- 3}$ and g as 9.8 $m / s^{2}$ . The viscosity of the oil in SI unit is [2025]

1 2.18

2 2.38

3 1.68

4 1.99

Ans.
(4)

$v_{T} \Rightarrow \frac{2}{9} \frac{(ρ_{0} - ρ_{l}) r^{2} g}{η}$

$η = \frac{2}{9} (\frac{7825 - 925}{2.45 \times 10^{- 2}}) \times {(1.8)}^{2} \times 10^{- 6} \times 9.8$

$η = 1.99$

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