Question

The angular diameter of the sun is $1920''$ . If the distance of the sun from the earth is $1.5 \times {10^{11}}\,m$ , then the linear diameter of the sun is:
(A) $2.6 \times {10^9}\,m$
(B) $0.7 \times {10^9}\,m$
(C) $5.2 \times {10^9}\,m$
(D) $1.4 \times {10^9}\,m$

Answer 1

Hint Convert the unit of the angular diameter from the seconds to the radians. Use the formula of the linear diameter given below, and substitute the angular diameter in the radian and the distance between the earth and the sun to find the linear diameter of the sun.
Useful formula:
The formula of the linear diameter is given by
$d = l\theta $
Where $d$ is the linear diameter of the sun, $l$ is the distance between the sun and the earth and $\theta $ is the angular diameter of the sun.

Complete step by step answer
It is given that the
Angular diameter of the sun, $\theta = 1920''$
The distance between the sun and the earth, $l = 1.5 \times {10^{11}}\,m$
First let us convert the angular diameter from the seconds into the minutes . $1\,\min = 60\,\sec $
$1'' = \dfrac{1}{{60'}}$
Substitute the above relation for the given angular diameter, we get
$1920'' = \dfrac{{1920}}{{60}}$
By performing division in the right hand side of the equation, we get
$\theta = 32'$
Next, let us convert the minutes into the degrees. $1\,\min = \dfrac{1}{{{{60}^ \circ }}}$ .
$32' = \dfrac{{32}}{{{{60}^ \circ }}} = {0.53^ \circ }$
Let us convert the degree into the radian, we get
${1^ \circ } = \dfrac{\pi }{{180}}\,rad$
${0.53^ \circ } = \dfrac{{{{0.53}^ \circ } \times \pi }}{{180}} = 0.009\,rad$
Hence the angular diameter is obtained as $0.009\,rad$ .
Let us use the formula of the linear diameter of the sun, we get
$d = l\theta $
Substituting the known parameters in the above formula, we get
$d = 1.5 \times {10^{11}} \times 0.009$
By performing various basic arithmetic operations, we get
$d = 1.4 \times {10^9}\,m$
Hence the linear diameter of the sun is obtained as $1.4 \times {10^9}\,m$ .

Thus the option (D) is correct.

Note Remember the step to convert the angular diameter from the unit seconds to the radian. Note that $1\,\sec = \dfrac{1}{{60}}\,\min $ . $1\,\min = {\dfrac{1}{{60\,}}^ \circ }$ . ${1^ \circ } = \dfrac{\pi }{{{{180}^ \circ }}}\,rad$ . The angular diameter depends only upon the observer’s angle. It is known that the sun is very much larger than the moon, but the angular diameter of both sun and moon are the same.