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JEE Main 2022 (27 Jun, Shift 2) · Kepler's Laws & Planetary Motion

Kepler's Laws & Planetary Motion question from JEE Main 2022 (27 Jun, Shift 2)

Q13JEE Main 2022 · 27 JUN · SHIFT 2Easy

Kepler's Laws & Planetary Motion

The distance of the Sun from earth is 1.5 ×\times 1011 m and its angular diameter is (2000)″ when observed from the earth. The diameter of the Sun will be:

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Answer: (c)

  1. θ=2000′′=2000×4.85×10−6 rad=9.7×10−3 rad\theta = 2000'' = 2000 \times 4.85\times10^{-6}\text{ rad} = 9.7\times10^{-3}\text{ rad} (since 1′′=4.85×10−61'' = 4.85\times10^{-6} rad).
  2. Diameter D=θ×D = \theta \times distance =9.7×10−3×1.5×1011= 9.7\times10^{-3} \times 1.5\times10^{11}
  3. D≈1.45×109 mD \approx 1.45\times10^{9}\text{ m}
  4. So the answer is (c).
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