The waveform shown in solid line is obtained by clipping a full-wave rectified sinusoid (shown dashed). The ratio of the rms value of the full-wave rectified waveform to the rms value of the clipped waveform is ________. (Round off to 2 decimal places,)

The given problem asks for the ratio of the root mean square (rms) value of a full-wave rectified sinusoidal waveform to that of the clipped waveform shown in the diagram below:

From the image, we can identify that the period of the full-wave rectified waveform is $\pi$ radians.
The equation of the unclipped full-wave rectified waveform over one period $(0\le \omega t\le \pi )$ is:
$v\left(\omega t\right)=V_{m}sin\left(\omega t\right)$

The rms value of a standard full-wave rectified sinusoidal wave is:
$V_{\text{rms, full}}=\frac{V_m}{\sqrt{2}}$

From the image, the clipped waveform is clipped at $0.707V_m$ (which is $\frac{V_m}{\sqrt{2}}$). The clipping occurs between the angles $\frac{\pi }{4}$ and $\frac{3\pi }{4}$.
Thus, the mathematical description of the clipped waveform $v_c\left(\omega t\right)$ over one period is:
$v_c\left(\omega t\right)=V_{m}sin\left(\omega t\right)$ for $0\le \omega t\le \frac{\pi }{4}$ and $\frac{3\pi }{4}\le \omega t\le \pi$
$v_c\left(\omega t\right)=\frac{V_m}{\sqrt{2}}$ for $\frac{\pi }{4}\le \omega t\le \frac{3\pi }{4}$

Due to the symmetry of the waveform about $\frac{\pi }{2}$, we can compute the mean-square value of the clipped waveform by integrating from $0$ to $\frac{\pi }{2}$:
$V_{\text{rms, clipped}}^2=\frac{2}{\pi }{\int }_0^{\frac{\pi }{2}}{v}_c^2\left(\theta \right)d\theta$
$V_{\text{rms, clipped}}^2=\frac{2}{\pi }\left[{\int }_0^{\frac{\pi }{4}}{V}_m^2{sin}^2\theta d\theta +{\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}{\left(\frac{V_m}{\sqrt{2}}\right)}^{2}d\theta \right]$

Let us evaluate each integral separately:
1. First integral:
${\int }_0^{\frac{\pi }{4}}{sin}^2\theta d\theta ={\int }_0^{\frac{\pi }{4}}\frac{1-cos\left(2\theta \right)}{2}d\theta ={\left[\frac{\theta }{2}-\frac{sin\left(2\theta \right)}{4}\right]}_0^{\frac{\pi }{4}}=\frac{\pi }{8}-\frac{1}{4}$

2. Second integral:
${\int }_{\frac{\pi }{4}}^{\frac{\pi }{2}}\frac{1}{2}d\theta =\frac{1}{2}\left(\frac{\pi }{2}-\frac{\pi }{4}\right)=\frac{\pi }{8}$

Now, substituting these back into the expression for $V_{\text{rms, clipped}}^2$:
$V_{\text{rms, clipped}}^2=\frac{2V_m^2}{\pi }\left[\left(\frac{\pi }{8}-\frac{1}{4}\right)+\frac{\pi }{8}\right]$
$V_{\text{rms, clipped}}^2=\frac{2V_m^2}{\pi }\left[\frac{\pi }{4}-\frac{1}{4}\right]=\frac{V_m^2}{2\mathrm{mi}}(\pi -1)=\frac{V_m^2}{2}\left(1-\frac{1}{\pi }\right)$

Taking the square root to find $V_{\text{rms, clipped}}$:
$V_{\text{rms, clipped}}=\frac{V_m}{\sqrt{2}}\sqrt{1-\frac{1}{\pi }}$

The ratio of the rms value of the full-wave rectified waveform to the rms value of the clipped waveform is:
$\text{Ratio}=\frac{V_{\text{rms, full}}}{V_{\text{rms, clipped}}}=\frac{\frac{V_m}{\sqrt{2}}}{\frac{V_m}{\sqrt{2}}\sqrt{1-\frac{1}{\pi }}}=\frac{1}{\sqrt{1-\frac{1}{\pi }}}$

Calculating the numerical value:
$1-\frac{1}{\pi }\approx 1-0.3183=0.6817$
$\sqrt{0.6817}\approx 0.8256$
$\text{Ratio}\approx \frac{1}{0.8256}\approx 1.211$

Rounding off to two decimal places, we get:
1.21