A huge circular arc of length $4.4$ ly subtends an angle $4$ s at the centre of the circle. How long it would take for a body to complete $4$ revolution if its speed is $8$ AU per second? Given: $1$ ly $= 9.46\times10^{15}$ m, $1$ AU $= 1.5\times10^{11}$ m
Answer: (B) $4.5\times10^{10}$ s
$4$ s (seconds of arc) $= \dfrac{4}{3600}\times\dfrac{\pi}{180} = 1.94\times10^{-5}$ rad.
Radius $R = \dfrac{\text{arc}}{\theta} = \dfrac{4.4\times9.46\times10^{15}}{1.94\times10^{-5}} \approx 2.15\times10^{21}$ m.
Distance for $4$ revolutions $= 4\times2\pi R \approx 5.40\times10^{22}$ m; speed $= 8\times1.5\times10^{11} = 1.2\times10^{12}$ m s$^{-1}$.
$$t = \frac{5.40\times10^{22}}{1.2\times10^{12}} \approx 4.5\times10^{10}\ \text{s}$$
Solution by Sreeraj P, M.Sc Physics