A galvanometer of resistance `50Omega` is connected to a b attery of 3V alongwith a resistance of `2950Omega` in series. A full scale deflection of 30 division is obtained in the galvanometer in order to reduce this deflection to 20 division. The resistance in sereis. should be:-

Options

A. `4450 Omega `
B. `5050Omega`
C. `5550 Omega `
D. `6050 Omega `

Answer

A

Explanation

Current in case of 30 division = `(3V)/((2950 + 50 ) Omega) = 3/3000 = 10^(-3) A`
So current for 20 division
` = (10^(-3) )/(30) xx 20 = 2/3 xx 10^(-3) A`
` rArr 2/3 xx 10^(-2) = (3)/( (R + 50) Omega) rArr R = 4450 Omega `