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(B) The capacitors $C_1$ and $C_2$ are connected in parallel to the battery. Therefore,the potential difference across both capacitors is the same,i.e

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$Q_1 < Q_2$

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(B) The capacitors $C_1$ and $C_2$ are connected in parallel to the battery. Therefore,the potential difference across both capacitors is the same,i.e.,$V_1 = V_2 = V$.The charge on a capacitor is given by the formula $Q = CV$.For capacitor $C_1$,the charge is $Q_1 = C_1 V$.For capacitor $C_2$,the charge is $Q_2 = C_2 V$.Since the space between the plates of $C_2$ is filled with a dielectric material of dielectric constant $K > 1$,its capacitance increases such that $C_2 = K C_1$,which implies $C_2 > C_1$.Since $V$ is constant and $C_2 > C_1$,it follows that $C_2 V > C_1 V$,which means $Q_2 > Q_1$ or $Q_1 < Q_2$.

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