One light-year is defined as the distance travelled by light in one year. The speed of light is $3 \times {10^8}\,m/s$. The same in the meter is?
(A) $3 \times {10^{12}}m$
(B) $9.461 \times {10^{15}}m$
(C) $3 \times {10^{15}}m$
(D) None of these

Hint: A light-year is a unit of distance denoted by the symbol ‘ly’. It is used to measure the distance between two astronomical objects.
Formula used: In this solution, we will use the following formula
\[v = \dfrac{d}{t}\] where $v$ is the velocity of the light, $d$ is the distance it travels in time $t$

Complete step by step answer:
As mentioned in the question, one light-year is defined as the distance travelled by light in one year. We’ve also been given that the speed of light is $3 \times {10^8}\,m/s$. Since the relation between the velocity, distance, and time is given as:
\[v = \dfrac{d}{t}\]
\[ \Rightarrow d = vt\]
Hence the distance travelled by light in one year will be the product of the velocity of light and the time it travels for. We’ve been given light ray travels for one year.
Now we know that in one year the number of seconds will be
\[t = 365\,\,{\text{days}}\,\, \times \,24\,{\text{hours}}\, \times \,60\,\min {\text{s}}\, \times \,60\,\sec \]
\[ \Rightarrow t = 31536000\,\sec \]
Now that we know the time, the light travels for, we can substitute $v = 3 \times {10^8}\,m/s$ in the formula for velocity and get
$d = 3 \times {10^8} \times 31536000$
$ \Rightarrow d = 9.46\, \times \,{10^{15}}\,m$
Hence the distance covered by light in one year is $9.46\, \times \,{10^{15}}\,m$. This is a very large distance as compared to the radius of the Earth which is 6400 km so a light-year is only used to measure large distances.

Hence the correct answer is option B.

Note: The distance travelled by light in a given amount of time is the largest compared to any other travelling object hence the term light-year is used as a unit for measuring large distances. The speed of light mentioned in the question is only in a vacuum while in different mediums, light can have lower speeds.