(solved)Question 2.19 of NCERT Class XI Physics Chapter 2

A parsec is a convenient unit of length on the astronomical scale. It is the distance of an object that will show a parallax of 1" (second of arc) from opposite ends of a baseline equal to the distance from the Earth to the Sun. How much is a parsec in terms of metres ?

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Parveen,

Question 2.19
NCERT Class XI Physics

A parsec is a convenient unit of length on the astronomical scale. It is the distance of an object that will show a parallax of 1" (second of arc) from opposite ends of a baseline equal to the distance from the Earth to the Sun. How much is a parsec in terms of metres ?

Physical Concept

Stars are far away and 1" is a very small angle so we can take $\displaystyle \sin \theta \approx \theta $ and apply the parallax formula $\displaystyle b = D \theta$

Video Explanation

Please watch this youtube video for a quick explanation of the solution:

Solution

$ \begin{aligned} &\text{given parallax angle } \theta = 1^{\rq\rq}\\ &\text{memorize }1^{\rq\rq} \equiv 4.85 \times 10^{-6} \text{ rad}\\\\ &\text{we have also been given dia of}\\ &\text{earth sun orbit } = 3 \times 10^{11} \text{ m(SF = 1)}\\\\ &\text{but basis } b \text{ has to }\\ &\text{be taken as earth-sun distance }\\ &\therefore b = (1/2) \times 3 \times 10^{11} \text{ m } \end{aligned} $

$ \begin{aligned} \text{by formula D} &= \frac{b}{\theta}\\\\ &= \frac{(1/2) \times 3 \times 10^{11}}{4.85 \times 10^{-6} }\\\\ &= 3.092 \times 10^{16}\\ \end{aligned} $

$ \begin{aligned} &\text{but S.F. in } b = 1\\ &\text{and in } \theta \text{ also 1}\\ &\therefore \text{on division SF will be } = 1\\\\ &\text{rounding to S.F. = 1}\\ &3 \times 10^{16}\text{ m}\:\underline{Ans}\\ \end{aligned} $

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