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 925 kg m-3. Take density of steel as 7825 kg m-3 and g as 9.8 m/s2. The viscosity of the oil in SI unit is
- 2.18
- 1.68
- 2.38
- 1.99
Answer
(4) 1.99
Solution
Terminal velocity (Stokes' law): vt = 2r2( - )g9
= 2r2( - )g9vt
r = 1.8 10-3 m, - = 7825 - 925 = 6900 kg m-3
= 2 (1.8 10-3)2 6900 9.89 2.45 10-2
= 2 3.24 10-6 6900 9.80.2205
= 0.43820.2205
1.99 Pa s
Answer: (4)