A galvanometer has a current sensitivity of 1 mA per division. A variable shunt is connected across the galvanometer and the combination is put in series with a resistance of `500Omega` and cell of internal resistance `1Omega`. It gives a deflection of 5 division for shunt of 5 ohm and 20 division for shunt of 25 ohm. The emf of cell is

Options

A. `47.1 ` V
B. `57.1 ` V
C. ` 67.1 ` V
D. `77.1 ` V

Answer

A

Explanation

`"Here", I=(E)/(R+r+(GS)/(G+S))and I_(R)=(IS)/(G+S)`
`I_(R)=(E)/((R+r)+(GS)/((G+S)))xx(S)/((G+S))`
`therefore I_(R)=(ES)/(((R+r)(G+S)+GS))`
For S=5ohm, `I_(R)=5xx10^(-3)A` and for S=25ohm,
`I_(R)=20xx10^(-3)A`
`"Hence", 5xx10^(-3)=(Exx5)/(501(G+5)+5G)......(I)`
`and 20xx10^(-3)=(Exx25)/(501(G+5)+25G)......(II)`
Dividing and solving
`G=88.2Omega`
From, (i) we get
`E=10^(-3) [501 (88.2+5)+5xx88.2]`
=47.1 volt