Question

In the absence of electric field the oil drop falls freely under gravity through a distance of 2.0 mm in 35.7 s. The radius of the drop, if the viscosity of air is $1.8\times {10}^{-5}$ $Ns{m}^{-2}$ and density of oil is 880 Kg/ $m^3$, is (neglect the buoyancy)

• A. $8.25\times {10}^{-8}m$
• B. $7.25\times {10}^{-7}m$
• C. $6.25\times {10}^{-8}m$
• D. $6.25\times {10}^{-7}m$

Answer 1

Answer: (B)

$7.25\times {10}^{-7}m$

As oil drop falls $2\times {10}^{-3}m$ in 35.7 seconds.
using $S=vt\Rightarrow v=\frac{2\times {10}^{-3}}{35.7}=5.6022\times {10}^{-5}m/s$
Let the radius of drop be m.
$\eta =1.8\times {10}^{-5}$
So,
The drop is in equilibrium with the velocity as given.
So,
Weight $=$ viscous force
$\Rightarrow (\frac{4}{3}\pi r^3.d)g=6\pi \eta rv$
$\Rightarrow \frac{4}{3}\pi r^2\times d\times g=6\pi \eta v$
$\Rightarrow r^2=\frac{\eta v}{2dg}\Rightarrow r^2=\frac{9\times 1.8\times {10}^{-10}\times 5.6022}{2\times 880\times 9.8}$
$\Rightarrow r^2=5.2618\times {10}^{-13}$
$\Rightarrow r=7.2538\times {10}^{-7}m$

So, the answer is option (B).