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…

Question

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/s^2. The viscosity of the oil in SI unit is

Accepted Answer

Correct answer: D. (d) To determine the viscosity of the oil, we use the formula for the terminal velocity of a sphere falling through a viscous fluid: v_T= 29 ( _s- _f)r^2g Where: v_T is the terminal velocity, _s is the density of the steel ball, _f is the density of the fluid, r is the radius of the ball, g is the acceleration due to gravity, is the viscosity of the fluid. Given: Diameter of the steel ball =3.6 mm, therefore radius r=1.8 mm=1.8 10^-3 m, Terminal velocity v_T=2.45 10^-2 m/s, Density of oil _f=925 kg/m^3, Density of steel _s=7825 kg/m^3, Acceleration due to gravity g=9.8 m/s^2. Rearranging the formula to solve for : = 29 ( _s- _f) r^2 gv_T Substituting the known values into the equation: = 29 (7825-925 ) (1.8 10^-3)^2 9.82.45 10^-2 Calculating this, we approximate the viscosity to be: ≈1.99 Pa s

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

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