Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle $\theta$ with each other. When suspended in water the angle remains the same. If density of the material of the sphere is $1.5 \mathrm{~g} / \mathrm{cc}$, the dielectric constant of water will be
(Take density of water $=1 \mathrm{~g} / \mathrm{cc}$ )
JEE MAIN 2024, Diffcult
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In air $\tan \frac{\theta}{2}=\frac{F}{m g}=\frac{q^2}{4 \pi \varepsilon_0 \mathrm{r}^2 m g}$
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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