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NEET generalGeneralMedium

Question

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

10 times greater

B

5 times greater

C

Smaller

D

Equal

Step-by-Step Solution

Using h=12at2h = \frac{1}{2}at^2 where a=eEma = \frac{eE}{m}, we get h=12eEmt2h = \frac{1}{2} \frac{eE}{m} t^2, so t=2hmeEt = \sqrt{\frac{2hm}{eE}}. Since t∝mt \propto \sqrt{m} and the mass of an electron is smaller than the mass of a proton, the electron takes less time to fall.

Exam Context & Concepts Covered

This question aligns with the NEET general syllabus, specifically targeting concepts from General. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

generalelectronthroughverticaldistanceuniform

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