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A cell can be balanced against 110 cm and 100 cm of potentiometer wire, respectively with and without being short circuited through a resistance of 10Ω. Its internal resistance is
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- 1.0 ohm
- 0.5 ohm
- 2.0 ohm
- zero
Correct Option: A
Here
| E > | ||
| R + r |
hence the lengths 110 cm and 100 cm are interchanged.
Without being short-circuited through R, only the battery E is balanced.
| E = | × l1 = | × 110 ................(i) | ||
| L | L |
| When R is connected across E, Ri = | × l2 | |
| L |
| Or, R | ![]() | ![]() | = | × 100..............(ii) | ||
| R + r | L |
Dividing (i) by (ii), we get
| = | |||
| R | 100 |
or, 100 R + 100 r = 110 R

