A galvanometer, having a resistance of `50 Omega` gives a full scale deflection for a current of `0.05 A`. The length in meter of a resistance wire of area of cross-section `2.97 xx 10^(-2) cm^(2)` that can be used to convert the galvanometer into an ammeter which can read a maximum of `5 A` current is (Specific resistance of the wire `5 xx 10^(-7) omega m`)

Options

A. 9
B. 6
C. 3
D. 1.5

Answer

C

Explanation

`S = (0.05/(5-0.05)) xx 50 = 50/99`
`rArr (rho l)/A = 50/99 rArr l = 50/99 xx (2.97 xx 10^(-2) xx 10^(-4))/(5 xx 10^(-7)) = 3m`