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Question: In a potentiometer the balancing with a cell is a length of 220 cm. On shunting the cell with a resi...

In a potentiometer the balancing with a cell is a length of 220 cm. On shunting the cell with a resistance of 33×10−10Vm−13 \times 10^{- 10}Vm^{- 1} balance length becomes 130 cm. What is the internal resistance of this cell?

A

4.5 6.5×10−6m2V−1s−16.5 \times 10^{- 6}m^{2}V^{- 1}s^{- 1}

B

7.82.5×106m2V−1S−12.5 \times 10^{6}m^{2}V^{- 1}S^{- 1}

C

6.32.5×104m2V−1S−12.5 \times 10^{4}m^{2}V^{- 1}S^{- 1}

D

2.086.5×104m2V−1s−16.5 \times 10^{4}m^{2}V^{- 1}s^{- 1}

Answer

2.086.5×104m2V−1s−16.5 \times 10^{4}m^{2}V^{- 1}s^{- 1}

Explanation

Solution

: Here, l1=220cm,l2=130cm,R=3Ωl_{1} = 220cm,l_{2} = 130cm,R = 3\Omega

∴\therefore Internal resistance,

r=(11−l2l2)Rr = \left( \frac{1_{1} - l_{2}}{l_{2}} \right)R =(220−130130)×3=2.08Ω= \left( \frac{220 - 130}{130} \right) \times 3 = 2.08\Omega