Question

ex .19
an irn ball of radis 0.3 cm falls through a colum of oil of densioty 0.94 g/cm3 it is found to attaim a termianl velocify of 0.5 cm/s reltemjne the vs8cpity of the oil given thaf demstiy of irm is 7.8 g/cm^3 .​

Answer 1

Correct Question

An iron ball of radius 0.3 cm falls through a column of oil of density 0.94 g/cm³. It is found to attain a terminal velocity of 0.5 cm/s. Determine the viscosity of the oil. Given that the density of iron is 7.8 g/cm³.

Answer

Given :-

• Radius (r) = 0.3 cm
• Density of oil (σ) = 0.94 g/cm³
• Density of iron (ρ) = 7.8 g/cm³

To find :-

• Viscosity of oil

Formula used :-

$\boxed{\sf \eta = \dfrac{2}{9} \frac{ {r}^{2} }{v} (\rho - \sigma )g}$

Solution

• r = 0.3 cm³
• ρ = 7.8 g/cm³
• σ = 0.94 g/cm³
• g = 9.8 m/s² = 980 cm/s²

Substituting the value in formula :-

$\sf \eta = \dfrac{2}{9} \frac{ {r}^{2} }{v} (\rho - \sigma )g$

$\sf \eta = \dfrac{2}{9} \frac{ {(0.3)}^{2} }{0.5} (7.8 - 0.94) \times 980$

$\sf \eta = \dfrac{2 \times ( {0.3})^{2} \times (7.8 - 0.94) \times 980}{9 \times 0.5}$

$\sf \eta = \dfrac{2 \times 0.09 \times 6.86 \times 980}{4.5}$

$\sf \eta = \dfrac{1,210.104}{4.5}$

$\boxed{\sf\eta = 286.9}$