Q53 / 270 · NDA 2025 (Phase 2) · General Ability
One parsec in astronomical units (A.U.) is about
- A$2 \times 10^3$ A.U.
- B$2 \times 10^4$ A.U.
- Correct answerC$2 \times 10^5$ A.U.
- D$2 \times 10^6$ A.U.
Solution
One parsec is defined as the distance at which an arc of length one astronomical unit (A.U.) subtends an angle of one second of arc.
$\theta = 1'' = \left(\dfrac{1}{3600}\right)^\circ = \dfrac{\pi}{180 \times 3600} \text{ rad}$
$r = \dfrac{l}{\theta} = \dfrac{1 \text{ A.U.}}{\dfrac{\pi}{180 \times 3600}}$
$r = \dfrac{180 \times 3600}{\pi} \text{ A.U.} \approx 206265 \text{ A.U.}$
$r \approx 2 \times 10^5 \text{ A.U.}$
Answer: $2 \times 10^5$ A.U.
$\theta = 1'' = \left(\dfrac{1}{3600}\right)^\circ = \dfrac{\pi}{180 \times 3600} \text{ rad}$
$r = \dfrac{l}{\theta} = \dfrac{1 \text{ A.U.}}{\dfrac{\pi}{180 \times 3600}}$
$r = \dfrac{180 \times 3600}{\pi} \text{ A.U.} \approx 206265 \text{ A.U.}$
$r \approx 2 \times 10^5 \text{ A.U.}$
Answer: $2 \times 10^5$ A.U.
Hence, the correct answer is option: 'C'.