सही विकल्प है: (a) 0.1m/s2
\(\frac{1}{2}mv^2=8\times 10^{-4}\) जूल
\(v^2=\frac{16\times 10^{-4}}{m}=\frac{16\times 10{-4}}{0.01}=16\times10^{-2}\) m2 / c2
\(\frac{1}{2}mv^2=ma_T \times 2 \times 2\pi r\)
\(\therefore \ \ a_T=\frac{v^2}{8\pi r}= \frac{16 \times 10{-2}}{8 \times 3.14\times 6.4 \times 10^{-2}}\)
\(=0.099=0.1\) m / c2