CGPET 2015 Solved Paper

Velocity acquired in 2 s. by the charged particle of mass 4,
v4=

q(E)

4m

×2
where, E is magnitude of electric field
Similarly,
v1=

3q(E)

m

×2
Corresponding kinetic energies are
k4=

1

2

×(4m)(v4)4
=

1

2

×(4m)(

qE

2m

)2
2m(

q2E2

4m2

)=

q2E2

2m

k1=

1

2

×m×v1
=

1

2

m×

36q2E2

m2

=

18q2E2

m

⇒

k4

k1

=

q2E2

2m

×

m

18q2E2

=

1

36

or,k4:k1=1:36