In the potentiometer circuit shown in the figure the balance length AJ=60cm when switch S is open. When switch S is closed and the value of R=`5Omega`, the b alance length AJ=50cm. The internal resistance of the cell C' is

Options

A. `1.2Omega`
B. `1.0Omega`
C. `0.8Omega`
D. `0.6Omega`

Answer

2

Explanation

Here `I_(1)=60cmI_(2)=50cm,R=5Omega`
`e=kl_(1)`
where k is the potential gradient ac ross the wire
`V=kl_(2)`
`therefore epsilon_(V)=(I_(1))/(I_(2))=(60)/(50)=(6)/(5)`
Internal resistance of the cell is
`r=((epsilon)/(V)-1)R=(( 6)/(5)-1)=1.0Omega`