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1. Electric Charges and Fields
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An electron falls from rest through a vertical distance $h$ in a uniform and vertically upward directed electric field $E.$ The direction of electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance $h.$ The time of fall of the electron, in comparison to the time of fall of the proton is
A
smaller
B
$5$ times greater
C
equal
D
$10$ times greater
(NEET-2018)
Solution
$h = \frac{1}{2}\frac{{eE}}{m}{t^2}$
$\therefore \,\,t = \sqrt {\frac{{2hm}}{{eE}}} $
$\therefore \,\,t \propto \sqrt m $ as $'e'$ is same for electron and proton.
$\because $ Electron has smaller mass so it will take smaller time.
Standard 12
Physics
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