Two moving coil metres `M_(1) and M_(2)` have the following particular `R_(1) =10Omega, N_(1) =30, A_(1) = 3.6 xx 10^(-3) m^(2), B_(1) =0.25 T,`
`R_(2) = 14Omega , N_(2) =42, A_(2) = 1.8 xx 10^(-3) m^(2), B_(2) =0.50 T`
The spring constants are identical for the two metres. What is the ratio of current sensitivity and voltage sensitivity of `M_(2)" to" M_(1)?`

Options

A. 1.4,1
B. 1,1.4
C. `1:1`
D. `1:4`

Answer

C

Explanation

Current sensitivity of a moving coil galvanometer is defined as
`C.S = phi/I = (NAB)/(k)`
and voltage sensitivity , `V.S = (NAB)/(kR)`
(i) Ratio of current sensitivity
`(C.S_2)/(C.S_1) = (N_2B_2A_2k)/(N_1B_1A_1k) = (42 xx 1.8 xx 0.5 xx 10^(-3) xx k)/(30 xx 0.25 xx 03.6 xx 10^(-3) xx k) = 1.4 `
(ii) Ratio of voltage sensitivity
`(V.S_2)/(V.S_1) = (C.S_2 xx R_1)/(C.S_1 xx R_2) = 7/5 xx 10/14 = 1`