question_answer

A oil drop having a mass \[4.8\times {{10}^{-10}}g\]and charge \[2.4\times {{10}^{-18}}C\]stands still between two charged horizontal plates separated by a distance of 1 cm. If now the polarity of the plates is changed, instantaneous acceleration of the drop is (g = 10 ms-2)

A) 5 ms-2                                   

B) 10 ms-2

C) 15 ms-2                                 

D) 20 ms-2

Correct Answer: B

Solution :

                                  \[a=\frac{F}{m}\] \[=\frac{qE}{m}\] \[=\frac{mg}{m}\] \[g=10\,m/{{s}^{2}}\]