JEE Main 27-Aug-2021 Shift 1 Solved Paper

Given, mass of 1 st ion, m1=4‌amu
Mass of 2 nd ion, m2=16‌amu
Charge on 1 st ion, q1=+2e
Charge on 2 nd ion, q2=+3e
Kinetic energy of both ion is same,
KE1=KE2
The radius of path traced by a charged particle in magnetic field is given as
r=

√2‌mKE

qB

⇒r∝

√m

q

⇒

r1

r2

=√

m1

m2

×(

q2

q1

)
⇒

r1

r2

=√

4

16

×(

+3e

2e

)
=

1

2

×

3

2

=

3

4

⇒r1=

3

4

r2
⇒r1<r2 Now, the deflection of ion in region with constant magnetic field as shown in figure is denoted by θ.
From diagram,
sin‌θ=

d

r

θ∝

1

r

[∵ d = constant]
As, r1<r2 from above expression θ′1>θ2.
Thus, deflection of lighter ion will be more than the deflection of heavier ion.