JEE Main 12 Jan 2019 Morning Question 24
Question: A proton and an $ \alpha - $ particle (with their masses in the ratio 1 : 4 and charges in the ratio of 1 : 2) are accelerated from rest through a potential difference V. If a uniform, magnetic field is set up perpendicular to their’ velocities, the ratio of the radii $ r _{p}:{r _{\alpha }} $ of the circular paths described by them will be [JEE Main Online Paper Held On 12-Jan-2019 Morning]
Options:
A) $ 1:3 $
B) $ 1:\sqrt{2} $
C) $ 1:2 $
D) $ 1:\sqrt{3} $
Show Answer
Answer:
Correct Answer: B
Solution:
We know that $ r=\frac{mv}{Bq}=\frac{\sqrt{2mqV}}{Bq}=\frac{1}{B}\sqrt{\frac{2mV}{q}}\Rightarrow r\propto \sqrt{\frac{m}{q}} $
Given $ \frac{m _{p}}{{m _{\alpha }}}=\frac{1}{4};\frac{q _{p}}{{q _{\alpha }}}=\frac{1}{2} $ $ \frac{r _{p}}{{r _{\alpha }}}=\sqrt{\frac{m _{p}{q _{\alpha }}}{q _{p}{m _{\alpha }}}}=\sqrt{( \frac{1}{4} )( \frac{2}{1} )}=\frac{1}{\sqrt{2}} $
