A galvanometer having a resistance of `50 Omega`, gives a full scale deflection for a current of 0.05 A. The length (in metres) of a resistance wire of area of cross section `3 xx 10^(-2) cm` 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=(I_(g)G)/(I-I_(g)) = (0.05 xx 50)/(5-0.05) = 2.5/4.95 = 50/99 Omega`
`therefore S=(rho l)/(A)` or `l = (SA)/rho therefore I=50/99 xx (3 xx 10^(-6))/(5 xx 10^(-7)) = 3.0 m`