Manipal Entrance Test 2014 Physics Paper

Given, [C2LR]=[C2L2

R

L

]=[(LC)2(

R

L

)]
and we know that frequency of LC circuits
f=

1

2π

1

√LC

. Here the dimension of LC is equal to [T2].
[

L

R

] gives the time constant of L−R circuit, so that the dimension of

L

R

is equal to [T]
Hence the required dimensions
[(LC)2(

R

L

)]= [T2]2[T−1]=[T3].