${i_{rms}} = \frac{{{V_{rms}}}}{{\sqrt {{R^2} + {{\left( {\frac{1}{{\omega C}}} \right)}^2}} }}$ ......$(i)$
Also $\frac{{{i_{rms}}}}{2} = \frac{{{V_{rms}}}}{{\sqrt {{R^2} + {{\left( {\frac{1}{{\frac{\omega }{3}C}}} \right)}^2}} }} = \frac{{{V_{rms}}}}{{\sqrt {{R^2} + \frac{9}{{{\omega ^2}{C^2}}}} }}$ ......$(ii)$
From equation $(i)$ and $(ii)$ we get
$3{R^2} = \frac{5}{{{\omega ^2}{C^2}}} \Rightarrow \frac{{\frac{1}{{\omega C}}}}{R} = \sqrt {\frac{3}{5}} $==> $\frac{{{X_C}}}{R} = \sqrt {\frac{3}{5}} $
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Reason : At Curie point a ferromagnetic substance start behaving as a paramagnetic substance.
$F=F_{0}\left(1-\left(\frac{t-T}{T}\right)^{2}\right)$
Where $F_{0}$ and $T$ are constants. The force acts only for the time internal $2 T$. The velocity $v$ of the particle after time $2 {T}$ is -