Four capacitors of capacitance 8 µF, 16 µF, 32 µF and 64 µF are connected in parallel. The resultantcapacitance will be-

Correct option is D

​$$\text{For capacitors in \textbf{parallel}, the \textbf{equivalent capacitance} is the sum of all individual capacitances:} \\C_{\text{eq}} = C_1 + C_2 + C_3 + C_4 \\[10pt]\textbf{Given:} \\C_1 = 8\,\mu F,\quad C_2 = 16\,\mu F,\quad C_3 = 32\,\mu F,\quad C_4 = 64\,\mu F \\[10pt]C_{\text{eq}} = 8 + 16 + 32 + 64 = 120\,\mu F$$​​