question_answer

In a Millikan oil drop experiment an oil drop carrying a charge of 2e is kept balanced between two parallel plates 2 cm apart, when a potential difference of 12000 V is applied to them, the radius of the oil drop is : \[(\rho =2\times {{10}^{3}}kg/{{m}^{2}})\]

A)  \[1.3\times {{10}^{-6}}m\]                         

B)  \[3.4\times {{10}^{-5}}m\]

C) \[~4\times {{10}^{-6}}m\]                           

D)  \[5\times {{10}^{-6}}m\]

Correct Answer: A

Solution :

                In the oildrop experiment, in order that the drop stays stationary. Force due to electric field = weight of drop \[\therefore \]  \[F=qE=mg\]                     ...(i) Also,      \[E=\frac{V}{d}\]                                               ?. (ii) where V is potential difference and d the distance between the plates. From Eqs. (i) and (ii), we get                 \[mg=q.\frac{V}{d}\] Since, mass (m) = volume (V)\[\times \]density (p)                 \[m=\frac{4}{3}\pi {{r}^{3}}\rho \] \[\therefore \]  \[\frac{4}{3}\pi {{r}^{3}}\rho g=q.\frac{V}{d}\] \[\Rightarrow \]\[\rho \frac{4}{3}\pi {{r}^{3}}\times 9.8=2\times 1.6\times {{10}^{-19}}\times \frac{12000}{2\times {{10}^{-2}}}\] \[\therefore \]  \[\rho {{r}^{3}}=4.67\times {{10}^{-15}}\] \[\Rightarrow \]               \[{{r}^{3}}=\frac{4.67\times {{10}^{-15}}}{2\times {{10}^{3}}}=1.3\times {{10}^{-6}}m\]