The range of voltmeter is 10\ V and its resistance is 50\ . To...

Current for full scale deflection of the voltmeter is given by $I_g = \dfrac{V_1}{R_v}$.

Substituting the given values, $I_g = \dfrac{10}{50} = 0.2\ \text{A}$.

To increase the range of the voltmeter to $V_2 = 15\ \text{V}$, a resistance $R$ must be connected in series with it.

The new range is given by $V_2 = I_g(R_v + R)$.

Substituting the values, $15 = 0.2(50 + R)$.

$50 + R = \dfrac{15}{0.2} = 75$

$R = 75 - 50 = 25\ \Omega$.

Thus, a $25\ \Omega$ resistor must be connected in series.

Answer: $25\ \Omega$ resistor in series.