(Take density of water $=1 \mathrm{~g} / \mathrm{cc}$ )
In water $\tan \frac{\theta}{2}=\frac{\mathrm{F}^{\prime}}{\mathrm{mg}^{\prime}}=\frac{\mathrm{q}^2}{4 \pi \varepsilon_0 \varepsilon_{\mathrm{r}} \mathrm{r}^2 \mathrm{mg}_{\text {eff }}}$
Equate both equations
$\varepsilon_0 g=\varepsilon_0 \varepsilon_{\mathrm{r}} \mathrm{g}\left[1-\frac{1}{1.5}\right]$
$\varepsilon_{\mathrm{T}}=3$
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$[A]$ The $x$ component of displacement of the center of mass of the block $M$ is : $-\frac{m R}{M+m}$.
[$B$] The position of the point mass is : $x=-\sqrt{2} \frac{\mathrm{mR}}{\mathrm{M}+\mathrm{m}}$.
[$C$] The velocity of the point mass $m$ is : $v=\sqrt{\frac{2 g R}{1+\frac{m}{M}}}$.
[$D$] The velocity of the block $M$ is: $V=-\frac{m}{M} \sqrt{2 g R}$.
