In the formula, `X = 3 YZ^(2)`, X has dimensions of capacitance and Z has dimensions of magnetic induction. The dimensions of Y are

Options

A. `[M^(-3)L^(-2)T^(-2)A^(4)]`
B. `[ML^(-2)T^(2)A^(2)]`
C. `[M^(-3)L^(-2)A^(4)T^(4)]`
D. `[M^(-3)L^(-2)T^(8)A^(4)]`

Answer

D

Explanation

As q= CV
`:. C= (Q)/(V)= (q^(2))/(W)` [as `V= (W)/(q)`]
`:. [X]= [C]= [(A^(2)T^(2))/(ML^(2)T^(-2))]= [M^(-1)L^(-2)T^(4)A^(2)]`
As `F= BIIsintheta :. B= (F)/(II sintheta)`
`:. [Z]= [B]= [(MLT^(-2))/(AL)]= [ML^(0)T^(-2)A^(-1)]`
Given : `X= 3YZ^(2)` or `Y= (X)/(3Z^(2))`
`:. [Y]= ([X])/([Z]^(2))= ([M^(-1)L^(-2)T^(4)A^(2)])/([ML^(0)T^(-2)A^(-1)]^(2))= [M^(-3)L^(-2)T^(8)A^(4)]`