Question

The sun’s angular diameter is measured to be 1920’’. The distance of the sun from the earth is 1.496 × ${10^{11}}$m. What is the diameter of the sun?
$
  {\text{A}}{\text{. 1 }} \times {\text{ 1}}{{\text{0}}^9}{\text{m}} \\
  {\text{B}}{\text{. 1}}{\text{.39 }} \times {\text{ 1}}{{\text{0}}^9}{\text{m}} \\
  {\text{C}}{\text{. 13}}{\text{.9 }} \times {\text{ 1}}{{\text{0}}^9}{\text{m}} \\
  {\text{D}}{\text{. 139 }} \times {\text{ 1}}{{\text{0}}^9}{\text{m}} \\
$

Answer 1

Hint: In order to find the diameter of the sun, we apply the formula of angular diameter using the given data. Angular diameter of the sun is given in ‘seconds’, we should convert it into radians and then use it in the formula.

Formula Used,
${\text{Tan }}\theta {\text{ = }}\dfrac{{\text{d}}}{{\text{l}}}$, where θ is the angular diameter, l is the distance between earth and sun and d is the diameter of the sun.

Complete Step-by-Step solution:
Given Data,
Angular diameter θ = 1920” (seconds)
Distance from earth to sun = 1.496 × ${10^{11}}$m
The angular diameter of the sun is given in the units – ‘seconds’ and represented as ‘’.
We know $\dfrac{{{\text{angle in seconds}}}}{{60}} {\text{ = angle in minutes}}$ and $\dfrac{{{\text{angle in minutes}}}}{{60}} {\text{ = angle in degrees}}$.
Given θ = 1920” ⟹ $\dfrac{{1920}}{{60}} = 32'$minutes ⟹$\dfrac{{32}}{{60}} = 0.533^\circ $ degrees.
We know the SI unit of measurement of angular diameter is “radians”. We know 180° = π radians.
$ \Rightarrow \theta {\text{ = 0}}{\text{.533}}^\circ {\text{ }} \times {\text{ }}\dfrac{\pi }{{180}}{\text{ }} \simeq {\text{ 9}}{\text{.3 }} \times {\text{ 1}}{{\text{0}}^{ - 3}}{\text{ radians}}$
From earth as the point of view, the distance between them, the angular distance and the diameter of the sun form an arc as shown in the figure:

θ is the angular diameter, l is the distance between earth and sun and d is the diameter of the sun. Therefore from the figure: ${\text{Tan }}\theta {\text{ = }}\dfrac{{\text{d}}}{{\text{l}}}$
Since the distance from earth to the sun is very large, then angular distance θ is naturally a very small value. For very small values of an angle θ, we can say
${\text{Tan}}\theta {\text{ }} \simeq {\text{ }}\theta $
$
   \Rightarrow \theta {\text{ = }}\dfrac{{\text{d}}}{{\text{l}}} \\
   \Rightarrow {\text{d = l}}\theta \\
   \Rightarrow {\text{d = 1}}{\text{.496 }} \times {\text{ 1}}{{\text{0}}^{11}}{\text{ }} \times {\text{ 9}}{\text{.3 }} \times {\text{ 1}}{{\text{0}}^{ - 3}} \\
   \Rightarrow {\text{d = 13}}{\text{.91 }} \times {\text{ 1}}{{\text{0}}^8} \\
   \Rightarrow {\text{d = 1}}{\text{.39 }} \times {\text{ 1}}{{\text{0}}^9}{\text{m}} \\
$
Hence diameter of the sun =${\text{1}}{\text{.39 }} \times {\text{ 1}}{{\text{0}}^9}{\text{m}}$.
Option B is the correct answer.

Note – In order to answer this type of question the key is to know the concept of angular diameter and the relevant formula. Converting the angle in given units to its SI units is a very important step to obtain the correct answer.
Always, while substituting any values of variables in a formula it is important that all the values are converted to their SI units.
For very small values of an angle ‘x’, the value of x and tan x are approximately equal for any value of x.
$\dfrac{\pi }{{180}}$Radians = 1 degree = 60 minutes = 60 × 60 seconds.