A) \[1.6\times {{10}^{-19}}C\]
B) \[2.2\times {{10}^{-19}}C\]
C) \[4.8\times {{10}^{-19}}C\]
D) \[8.0\times {{10}^{-19}}C\]
Correct Answer: D
Solution :
When the electric field is on |
Force due to electric field= weight |
\[qE=mg\] |
\[qE=\frac{4}{3}\pi {{R}^{3}}\rho g\] |
\[qE=\frac{4\pi {{R}^{3}}\rho g}{3E}...(i)\] |
When the electric field is switched off |
Weight=viscous drag force |
\[\operatorname{mg}=6\pi \eta R{{v}_{t}}\] |
\[\frac{4}{3}\pi {{R}^{3}}\rho g=6\pi \eta R{{v}_{t}}\]\[\therefore \,\,\,\,R=\sqrt{\frac{9\nu \eta {{v}_{t}}}{2\rho g}}...\left( \operatorname{ii} \right)\] |
From (i) and (ii), \[q=\frac{4}{3}\pi {{\left[ \frac{9\nu \eta {{v}_{t}}}{2\rho g} \right]}^{\frac{3}{2}}}\times \frac{\rho g}{E}\]\[=\frac{4}{3}\times \pi {{\left[ \frac{9\times 1.8\times {{10}^{-5}}\times 2\times {{10}^{-3}}}{2\times 900\times 9.8} \right]}^{\frac{3}{2}}}\]\[\times \frac{900\times 9.8\times 7}{81\pi \times {{10}^{5}}}\] |
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