The graph shows the variation of potential difference between the plates of two capacitors A and B with the increase of charge stored by them. From the graph we can conclude:

Capacitance of A is greater than B

The correct answer is Option (1) → Capacitance of A is greater than B

Relation between charge and potential difference:

$Q = CV \quad \Rightarrow \quad V = \frac{Q}{C}$

In $V$–$Q$ graph, slope = $\frac{V}{Q} = \frac{1}{C}$

From the graph, slope of B is greater than slope of A.

$\Rightarrow \frac{1}{C_B} \gt \frac{1}{C_A} \quad \Rightarrow \quad C_A \gt C_B$

Therefore, capacitance of A is greater than capacitance of B.