Two capacitors $$C_1$$ and $$C_2$$ are charged to 120 V and 200 V respectively. It is found that by connecting them together the potential on each one can be made zero. Then:

We have two capacitors. The first has capacitance $$C_1$$ and is charged to a potential of $$120\;{\rm V}$$, so the magnitude of charge on it is

$$Q_1 = C_1 \times 120.$$

The second has capacitance $$C_2$$ and is charged to $$200\;{\rm V},$$ therefore the magnitude of charge on it is

$$Q_2 = C_2 \times 200.$$

To be able to bring the potential on each capacitor down to zero only by inter-connecting them, the obvious idea is to join plates carrying charges of opposite sign, so that the two initial charges can neutralise each other. Hence we connect

  • the positive plate of the first capacitor to the negative plate of the second, and

  • the negative plate of the first capacitor to the positive plate of the second.

After this cross-connection both capacitors share the same two conducting nodes; let us call the node that now carries the charges $$+Q_1$$ (from the first capacitor) and $$-Q_2$$ (from the second capacitor) node $$P$$. The other node, which we call $$Q,$$ automatically carries the algebraically opposite charge. So the net charge residing on node $$P$$ is

$$Q_{\text{net}} = +Q_1 \;-\; Q_2 = C_1 \times 120 \;-\; C_2 \times 200.$$

Because both capacitors now lie in parallel between the same two nodes, their equivalent capacitance is simply the sum

$$C_{\text{eq}} = C_1 + C_2.$$

The final common potential difference between the two nodes is therefore obtained from the elementary formula

$$V_f = \dfrac{Q_{\text{net}}}{C_{\text{eq}}}.$$

Substituting the expressions just written gives

$$V_f \;=\; \dfrac{\,C_1 \times 120 \;-\; C_2 \times 200\,}{C_1 + C_2}.$$

The statement of the problem tells us that this potential can be made zero. Hence we impose the condition

$$V_f = 0.$$

Setting the numerator equal to zero gives

$$C_1 \times 120 \;-\; C_2 \times 200 = 0.$$

Now we perform the algebra step by step:

$$120\,C_1 = 200\,C_2,$$

$$\dfrac{C_1}{C_2} = \dfrac{200}{120},$$

$$\dfrac{C_1}{C_2} = \dfrac{5}{3}.$$

Cross-multiplying finally gives

$$3\,C_1 = 5\,C_2.$$

Thus the required relation between the two capacitances is exactly the one written in Option D.

Hence, the correct answer is Option 4.