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Cells A and B and a galvanometer G are connected to a slide wire OS by two sliding contacts C and D as shown in figure. The slide wire is 100 cm long and has a resistance of 12 Ω. With OD=75 cm, the galvanometer gives no deflection when OC is 50 cm. If D is moved to touch the end of wire S, the value of OC for which the galvanometer shows no deflection is 62.5 cm. The e.m.f. of cell B is 1.0 V. Calculate the internal resistance & e.m.f. of cell A.

A
3 Ω and 2 V
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B
2 Ω and 2 V
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C
3 Ω and 3 V
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D
4 Ω and 3 V
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Solution

The correct option is A 3 Ω and 2 V
Given:
EB=1 V
Let EA and r be the e.m.f. and internal resistance of cell A, respectively.

When OD=75 cm and OC=50 cm

Resistance of wire OC and OD will be

ROC=(12100)×50=6 Ω

ROD=(12100)×75=9 Ω

Since, EB=1 V is balanced across 50 cm, so potential gradient of wire will be

K=150 V/cm

Therefore, voltage drop across the wire OD of length 75 cm will be

VOD=(1/50)×75=1.5 V

So,

(EA9+r)×9=1.5 ........(1)

Similarly,
When OD=100 cm and OC=62.5 cm

Resistance of wire OD will be

ROD=12 Ω

Since, EB=1 V is balanced across 62.5 cm, so potential gradient of wire will be

K=162.5 V/cm

Therefore, voltage drop across the wire OD of length 75 cm will be

VOD=(1/62.5)×100=1.6 V

So,
(EA12+r)×12=1.6 ........(2)

On solving eqs. (1) & (2), we get

r=3 Ω and EA=2 V

Hence, option (a) is correct.
Key concept:
Principle: The potentiometer is based upon the principle that when a constant current is passed through a wire of uniform area of cross-section, the potential drop across any portion of the wire is directly proportional to the length of that portion.

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