A huge circular arc of length 4.4 ly subtends an angle 4s 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 = $$9.46 \times 10^{15}$$ m
1 AU = $$1.5 \times 10^{11}$$ m

We first recall the relation between the arc-length $$s$$, the radius $$r$$ of the circle and the angle $$\theta$$ subtended by the arc at the centre: the formula is $$s = r\,\theta.$$

Here the given arc-length is $$4.4\ \text{ly}$$, and the angle is stated as $$4$$ arc-seconds. We must put every quantity in compatible units.

One light-year is given as $$1\ \text{ly}=9.46 \times 10^{15}\ \text{m},$$ so

$$s = 4.4\ \text{ly}=4.4 \times 9.46 \times 10^{15}\ \text{m} = 41.624 \times 10^{15}\ \text{m} = 4.1624 \times 10^{16}\ \text{m}.$$

Next, we convert the very small angle of $$4$$ arc-seconds to radians. We know

$$1^\circ = \frac{\pi}{180}\ \text{rad},$$

$$1'\ (\text{one arc-minute}) = \frac{1^\circ}{60},$$

$$1''\ (\text{one arc-second}) = \frac{1'}{60}= \frac{1^\circ}{3600}.$$

Therefore

$$4'' = 4 \times \frac{1^\circ}{3600} = \frac{4}{3600}\ ^\circ = \frac{1}{900}\ ^\circ,$$

and in radians

$$\theta = \frac{1}{900}\ ^\circ \times \frac{\pi}{180} = \frac{\pi}{900 \times 180} = \frac{\pi}{162000}\ \text{rad}.$$

Evaluating

$$\theta = \frac{3.1415926}{162000}\ \text{rad} \approx 1.939 \times 10^{-5}\ \text{rad}.$$

Now we can find the radius $$r$$ from $$r = \dfrac{s}{\theta}$$:

$$r = \frac{4.1624 \times 10^{16}}{1.939 \times 10^{-5}}\ \text{m} = 4.1624 \times 10^{16} \times \frac{1}{1.939 \times 10^{-5}}\ \text{m}.$$

The reciprocal $$\dfrac{1}{1.939 \times 10^{-5}} \approx 5.158 \times 10^{4},$$ so

$$r \approx 4.1624 \times 10^{16} \times 5.158 \times 10^{4}\ \text{m} \approx 2.15 \times 10^{21}\ \text{m}.$$

With the radius known, the full circumference of the circle is given by $$C = 2\pi r$$:

$$C = 2\pi (2.15 \times 10^{21})\ \text{m} = 6.283 \times 2.15 \times 10^{21}\ \text{m} \approx 1.35 \times 10^{22}\ \text{m}.$$

The body is required to complete four revolutions, so the total path-length traversed is

$$D = 4C = 4 \times 1.35 \times 10^{22}\ \text{m} \approx 5.40 \times 10^{22}\ \text{m}.$$

The speed of the body is stated as $$8\ \text{AU s}^{-1}.$$ With $$1\ \text{AU} = 1.5 \times 10^{11}\ \text{m},$$ the speed in SI units is

$$v = 8 \times 1.5 \times 10^{11}\ \text{m s}^{-1} = 1.2 \times 10^{12}\ \text{m s}^{-1}.$$

The time required is obtained from the basic relation $$t = \dfrac{D}{v}.$$ Substituting the values just found,

$$t = \frac{5.40 \times 10^{22}}{1.2 \times 10^{12}}\ \text{s} = 4.50 \times 10^{10}\ \text{s}.$$

Expressed in scientific notation this is $$4.5 \times 10^{10}\ \text{s},$$ matching Option B.

Hence, the correct answer is Option B.