A solid steel ball of diameter 3.6 mm acquired terminal velocity 2.45 × 10^(-2) m/s while falling under gravity through an oil of density ​

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

(1) 2.18 

(2) 2.38

(3) 1.68

(4) 1.99

Answer 1

Correct option is : (4) 1.99

\(v_T \Rightarrow \frac{2}{9} \frac{(\rho_0-\rho_\ell)r^2 g}{\eta}\)

\(\eta = \frac{2}{9} \left( \frac{7825 - 925}{2.45 \times 10^{-2}}\right) \times (1.8)^2 \times 10^{-6} \times 9.8\)

\(\eta \approx 1.99\)