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) \[m=\frac{4}{3}\pi {{r}^{3}}\rho \] \[\therefore \] \[\frac{4}{3}\pi {{r}^{3}}\rho g=q.\frac{V}{d}\] \[\Rightarrow \]\[\rho \frac{4}{3}\pi {{r}^{3}}\times 9.8=2\times 1.6\times {{10}^{-19}}\times \frac{12000}{2\times {{10}^{-2}}}\] \[\therefore \] \[\rho {{r}^{3}}=4.67\times {{10}^{-15}}\] \[\Rightarrow \] \[{{r}^{3}}=\frac{4.67\times {{10}^{-15}}}{2\times {{10}^{3}}}=1.3\times {{10}^{-6}}m\]You need to login to perform this action.
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