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Q.

When an ammeter of negligible internal resistance is inserted in series with circuit it reads 1 A. When the voltmeter of very large resistance is connected across X it reads 1 V. When the point A and B are shorted by a conducting wire, the voltmeter measures 10 V. The internal resistance of the battery (r) is equal to 

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a

zero

b

0.2  Ω

c

0.1  Ω

d

0.5  Ω

answer is C.

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Detailed Solution

I = 1 A

r is the internal resistance of the battery

12=(X+Y+r)(1)..............(i)

Also 1=(X)(1)          or   X=1  Ω

Voltage across X (when A and B are shorted)

10=(12X+r)X   or  10=121+r  or  1+r=65

or    r=15  Ω

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