If `epsilon_(0), B, V` represent permitibvity of free space, magnitude of magnetic field and volume of space respectively, then the dimension of `epsilon_(0)B^(2)V` is `[M^(a)L^(b)T^(c)]`. Find `a+b+c`.

Answer

1

Explanation

`epsilon_(0)VB^(2)-=(epsilonE^(2))/(v^(2))V [B-=E/v]`
`-=(u_(e))/(2v^(2))V-= ("Energy")/(("speed")^(2))-=` mass, so `a=1, b=0, c=0`