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Q.

The distance of the Sun from earth is $1.5 \times 10^{11} m$ and its angular diameter is $(2000) \sec$ when observed from the earth. The diameter of the Sun will be :

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a

$1.45 \times 10^{9} m$

b

$0.14 \times 10^{9} m$

c

$1.45 \times 10^{10} m$

d

$2.45 \times 10^{10} m$

answer is C.

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$θ = \frac{d}{r} \Rightarrow \frac{2000}{60 \times 60} \times \frac{π}{180} = \frac{d}{1.5 \times 10^{11}} \Rightarrow d = \frac{2000}{60 \times 60} \times \frac{π}{180} \times 1.5 \times 10^{11} \Rightarrow d = \frac{π \times 1.5}{3 \times 6 \times 18} \times 10^{11} = 1.45 \times 10^{9} m$

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