A galvanometer of resistance `50 Omega`, gives a full scale deflection for current of 50 mA. It is to be converted intyo an ammeter to read a maximum current 5 A. What should be the length of the resistance wire of specific resisrance` 5xx 10^(-7) Omega` and cross sectional area` 2.97 xx 10^(-2)cm^(2)`, reqiured for this conversion?

Options

A. `7.5 m`
B. `6 m`
C. `3 m`
D. `1.5 m`

Answer

C

Explanation

To increase the range of the ammeter, the resistance wire should be joined in parallel. The required shunt is calculated by using the relation,`I_g/I=S/(S+G) therefore(5xx10^(-2))/5=S/(S+50)therefore1/100=S/(S+50) therefore100 S=S+50 therefore99S=50 therefore S=(50)/(99)Omega and R=S=(pL)/Atherefore L=(SA)/p=50/(99)xx(2.97 xx 10^(-6)m^(2))/(5 xx 10^(-7))therefore L=3 m`.