An unknown resistance `R_1` is connected in series with a resistance of `10 ohm`. This combination is connected to one gap of meter bridge while a resistance `R_2` is connected in other gap. The balance point is at `50 cm`. Now, when the `10 ohm` resistance is removed balance point shifts to `40 cm`. The value of `R_1` is (in ohm)

Options

A. 20
B. 10
C. 60
D. 40

Answer

A

Explanation

To measure unknown resistance in meter bridge `(R_(1))/(R_(2))=(l_(1))/((100-l_(1)))`
When resistance 10`Omega`  connected in series with `R_(1)` then `(R_(1)+10)/(R_(2))=(50)/(50)rArrR_(1)+10=R_(2)`.....(i)When 10`Omega`  resistance is removed, `(R_(1))/(R_(2))=(40)/(60)rArr(R_(1))/(R_(1)+10)=(40)/(60)` (using en (i))
`rArr`    `60R_(1)=40R_(1)+400`  
`rArr`   `20R_(1)=400rArrR_(1)=20Omega`.