In the circuit, the battery is ideal. A voltmeter is connected in turn across ${R_1}$ ​and ${R_2}$, giving readings of ${V_1}$ and ${V_2}$​ respectively:
$\left( A \right){V_1} = 80V$
$\left( B \right){V_1} = 60V$
$\left( C \right){V_2} = 30V$
$\left( D \right){V_2} = 40V$

Hint: Apply Kirchhoff’s law to find the potential difference between the points. Consider the sign convention rules while applying Kirchhoff law. Now you will end up with an equation, which includes potential. Simplify in such a way to obtain the potential difference between the two points.

Formula Used:
$I = \dfrac{E}{{{R_1} + {R_2}}}$
Where $E$ is the emf, $I$ is the current,${R_1},{R_2}$ is the resistors.

Complete step by step solution:
Kirchhoff’s first law (junction law or current law): At any junction, the current leaving the junction is equal to the current entering the system. Kirchhoff’s first law tells us that a steady current flows in the circuit then there will be no accumulation of charge at any point.
Kirchhoff’s second law (loop law or voltage law): The algebraic sum of changes in potential around any closed loop is zero.
Sign conventions:
The change in potential passing through resistance in the direction of current is $ - IR$ while in the opposite direction $ + IR$.
The change in potential passing through an emf source from positive to negative is $ - E$ while in the opposite direction $ + E$
The current flowing through the circuit
$I = \dfrac{E}{{{R_1} + {R_2}}}$
$ \Rightarrow I = \dfrac{{120}}{{600 + 300}}$
$ \Rightarrow I = \dfrac{2}{{15}}A$
Now reading of voltmeter across ${R_1}$ is given by
${V_1} = \dfrac{2}{{15}} \times 600 = 80V$
Reading of voltmeter across ${R_2}$ is given by
${V_2} = \dfrac{2}{{15}} \times 300 = 40V$

Hence the correct option is option $\left( D \right) \Rightarrow {V_2} = 40V$.

Note: According to Kirchhoff’s first, the charge is conserved, since no charge can accumulate at a junction. Second law Kirchhoff’s law represents the conservation of energy. The electric energy source of emf is lost in passing through the resistance. Kirchhoff’s second law can be applied to a mesh containing a capacitor as well as an emf.