JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Semiconductors

Read the amplitudes off the plot: the envelope runs between $8$ V and $10$ V, a mean of $9$ V with a $\pm 1$ V swing, so the modulating part is $9+\sin ({\omega }_{s}t)$. The slow signal wave has period $100\mu$s, giving ${\omega }_s=2\pi \times 10^4$, while the fast carrier has period $8\mu$s, giving ${\omega }_c=2.5\pi \times 10^5$. The modulated voltage is therefore $(9+\sin (2\pi \times 10^{4}t))\sin (2.5\pi \times 10^{5}t)$ V.