A voltmeter reads the potential difference across the terminals of an old battery as `1.40 V`, while a potentiometer reads its voltage to be `1.55 V`. The voltmeter resistance is `280 Omega`.

Options

A. `60Omega`
B. `45Omega`
C. `35Omega`
D. `30Omega`

Answer

4

Explanation

Let `epsilon` be emf and r be internal resistance of the cell when tno current drawn from the cell, the potential difference b etween its terminals is the emf of the cell.

It is this quantity which the potentionmeter measures.
Therefore, emf of the cell `epsilon=1.55V`
When the `280Omega` voltmeter is connected ac ross the cell as shown in adjacent figure, current flows from the cell.
The potential difference between its terminals is just the voltmeter reading. 140V.
`terefore 1xx280=1.40`
`1=(1.4)/(280)=5xx10^(-3)A`
Also `I=(epsilon)/(280+r)=(1.55)/(28+r)`
`5xx10^(-3)=(1.55)/(28+r)`
`1.4+5xx10^(-3)r=1.55`
`5xx10^(-3)r=0.15`
`r=(0.15)/(5xx10^(-3)=30Omega`
According to m aximum power theoram.