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(B) Let $R = 30\,\Omega$ be the actual resistance and $R_v$ be the internal resistance of the voltmeter.In the given circuit,the voltmeter is connecte

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$570$

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(B) Let $R = 30\,\Omega$ be the actual resistance and $R_v$ be the internal resistance of the voltmeter.In the given circuit,the voltmeter is connected in parallel with the resistor $R$.The equivalent resistance $R_{eq}$ of the parallel combination is given by:$R_{eq} = \frac{R \cdot R_v}{R + R_v}$The measured value of resistance is $R_{measured} = R_{eq} = \frac{30 R_v}{30 + R_v}$.According to the problem,the measured value is $5\%$ less than the actual value:$R_{measured} = R 	imes (1 - 0.05) = 30 	imes 0.95 = 28.5\,\Omega$.Equating the two expressions for $R_{measured}$:$\frac{30 R_v}{30 + R_v} = 28.5$$30 R_v = 28.5(30 + R_v)$$30 R_v = 855 + 28.5 R_v$$1.5 R_v = 855$$R_v = \frac{855}{1.5} = 570\,\Omega$.Thus,the internal resistance of the voltmeter is $570\,\Omega$.

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