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(D) To increase the range of a voltmeter,a high resistance $R$ must be connected in series with the voltmeter.Let $V$ be the original range $(10 \ V)$

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$25 \ \Omega$ resistance in series

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(D) To increase the range of a voltmeter,a high resistance $R$ must be connected in series with the voltmeter.Let $V$ be the original range $(10 \ V)$,$V'$ be the new range $(15 \ V)$,$G$ be the internal resistance of the voltmeter $(50 \ \Omega)$,and $I_g$ be the full-scale deflection current.First,calculate the full-scale current: $I_g = V / G = 10 \ V / 50 \ \Omega = 0.2 \ A$.Now,for the new range $V'$,the total resistance becomes $G + R$. Thus,$V' = I_g(G + R)$.Substituting the values: $15 = 0.2(50 + R)$.$15 / 0.2 = 50 + R$.$75 = 50 + R$.$R = 75 - 50 = 25 \ \Omega$.Therefore,a $25 \ \Omega$ resistance must be connected in series.

The voltmeter has a range of $10 \ V$ and its internal resistance is $50 \ \Omega$. To increase the range of the voltmeter to $15 \ V$,the resistance which is to be connected is:

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