Question

A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the sun and the earth in terms of the new unit if light takes $8$min and $20$sec to cover this distance?
A. $300$
B. $400$
C. $500$
D. $600$

Answer 1

Hint: concept of relation between speed, distances and time is to be used with the given values of the new unit.
Speed $ = $ $\dfrac{{Dis\tan ce}}{{Time}}$
$ \Rightarrow $ Distance $ = $ Speed $ \times $ Time

Complete step by step answer:
As we know that speed is a scalar quantity and is defined as the time rate of distance, that is
     Speed $ = $ $\dfrac{{Dis\tan ce}}{{Time}}$
So, Distance $ = $ Speed $ \times $ Time
Now, as we have to compute the distance between sun and the earth in new units from light, new speed and distance.
Now, New speed of light in vacuum $ = $ unity $ = $ 1. unit
and Time taken by light to cover distance between Sun and Earth
$ = \;8\;\min \;\;\;20\;\sec $
$1\;\min ute\;\; = \;\;60\;\sec $
So, $8\;\min \;\; = \;\,8\; \times \;60\; = \;480\;\sec $
So, Total Time $ = \;8\;\min \;\;20\,\sec $
     $ = \;8\; \times 60\; + \;20$
     $ = \;480\; + \;20$
     $500\;\sec $
So, Distance between Sun and Earth $ = $1 unit $ \times \;500\;\sec $
     $ = \;500$ unit sec

So, the correct answer is “Option C”.

Additional Information:
Some of the practical units for measuring large distances are-
1. light year – It is distance travelled by light in vacuum in one year.
     $1$ light year $ = $ $9.467 \times {10^{15}}\,m$
2. Astronomical units – It is the mean distance of earth from the sun.
     $1$ AU $ = $ $1.496 \times {10^{11}}\,m$
3. Parsec – It is the distance at which an arc of length $1$ AU subtends an angle of $1$ second of arc
     $1$ Parsec $ = $ $3.08 \times {10^{16}}\,m = 3.26\,\,ly$
Also, $1$ ly $ = $ $6.3 \times {10^4}\,AU$
So, $1$ Parsec > $1$ ly > AU.

Note:
 This question can also be solved by unitary method which is as given below:-
Speed of light in new units $ = \;1$
This means,
Distance covered in one second $ = \;1$
Distance covered in $8\;\min \;\;20\;\sec \;\;or\;(500\;\sec )$
$
   = \;1 \times 500 \\
   = 500 \\
 $