If E, p, m and c denote the energy, linear momentum, mass and speed of light, then the equation representing the correct…

\(\mathrm{In}\mathrm{relativistic}\mathrm{case}\\ \mathrm{m}=\frac{{\mathrm{m}}_{\mathrm{o}}}{\sqrt{1-\frac{{\mathrm{v}}^{2}}{{\mathrm{c}}^{2}}}}\\ {\mathrm{p}}^{2}=\frac{{\mathrm{m}}_{\mathrm{o}}^{2}{\mathrm{v}}^{2}}{\left(1-\frac{{\mathrm{v}}^{2}}{{\mathrm{c}}^{2}}\right)}=\frac{{\mathrm{m}}_{\mathrm{o}}^{2}\frac{{\mathrm{v}}^{2}}{{\mathrm{c}}^{2}}{\mathrm{c}}^{2}}{\left(1-\frac{{\mathrm{v}}^{2}}{{\mathrm{c}}^{2}}\right)}\\ {\mathrm{p}}^{2}=\frac{{\mathrm{m}}_{\mathrm{o}}^{2}\left(\frac{{\mathrm{v}}^{2}}{{\mathrm{c}}^{2}}-1+1\right){\mathrm{c}}^{2}}{\left(1-\frac{{\mathrm{v}}^{2}}{{\mathrm{c}}^{2}}\right)}={\mathrm{m}}_{\mathrm{o}}^{2}{\mathrm{c}}^{2}+\frac{{\mathrm{m}}_{\mathrm{o}}^{2}{\mathrm{c}}^{2}}{\left(1-\frac{{\mathrm{v}}^{2}}{{\mathrm{c}}^{2}}\right)}\\ {\mathrm{p}}^{2}{\mathrm{c}}^{2}={\mathrm{m}}_{\mathrm{o}}^{2}{\mathrm{c}}^{2}+\left(\frac{{\mathrm{m}}_{\mathrm{o}}}{\sqrt{\left(1-\frac{{\mathrm{v}}^{2}}{{\mathrm{c}}^{2}}\right)}}\right){\mathrm{c}}^{4}\\ {\mathrm{p}}^{2}{\mathrm{c}}^{2}=-{\left({\mathrm{m}}_{\mathrm{o}}{\mathrm{c}}^{2}\right)}^{2}+{\left({\mathrm{mc}}^{2}\right)}^{2}\\ {\left({\mathrm{mc}}^{2}\right)}^{2}={\mathrm{p}}^{2}{\mathrm{c}}^{2}+{\left({\mathrm{m}}_{\mathrm{o}}{\mathrm{c}}^{2}\right)}^{2}\\ {\mathrm{E}}^{2}={\mathrm{p}}^{2}{\mathrm{c}}^{2}+\left({\mathrm{m}}_{\mathrm{o}}^{2}{\mathrm{c}}^{4}\right)\\ \mathrm{E}=\sqrt{{\mathrm{p}}^{2}{\mathrm{c}}^{2}+\left({\mathrm{m}}_{\mathrm{o}}^{2}{\mathrm{c}}^{4}\right)}\\\)