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Q.

A voltmeter with a resistance of 50 x103 ohm is used to measure voltage in a circuit. To increase its range to 3 times, the additional resistance to be put in series is

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a

105ohm

b

150k ohm

c

900 k ohm

d

9x106 ohm

answer is A.

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Detailed Solution

R=V′ig−G=nVV/R0−R0where Ro is the resistance of voltmeter∴ R=nR0−R0=(n−1)R0=(3−1)×50×103=100×103=105Ω
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