$\mathrm{dQ}=\lambda \mathrm{Rd} \theta=\lambda_{0} \cos \frac{\theta}{2} \mathrm{Rd} \theta$
Potential at the centre of ring due to charge $\mathrm{dQ}$
$\mathrm{dV}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{d} \mathrm{Q}}{\mathrm{R}}=\frac{\lambda_{0} \cos \frac{\theta}{2} \cdot \mathrm{Rd} \theta}{4 \pi \varepsilon_{0} \mathrm{R}}$
$\mathrm{V}=\int \mathrm{d} \mathrm{V} \Rightarrow \mathrm{V}=\frac{\lambda}{4 \pi \varepsilon_{0}} \int_{0}^{2 \pi} \cos \frac{\theta}{2} \mathrm{d} \theta$
$=\frac{\lambda}{4 \pi \varepsilon_{0}}\left[\frac{\sin \frac{\theta}{2}}{\frac{1}{2}}\right]_{0}^{2 \pi}=0 \mathrm{\,V}$
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