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(C) According to Kepler's third law of planetary motion,the square of the time period of revolution $(T)$ is directly proportional to the cube of the

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$3$

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(C) According to Kepler's third law of planetary motion,the square of the time period of revolution $(T)$ is directly proportional to the cube of the semi-major axis $(R)$ of the orbit: $T^2 \propto R^3$.Given:For Earth: $T_1 = 1 \, 	ext{year}$,$R_1 = 1.5 	imes 10^8 \, 	ext{km}$.For the imaginary planet: $T_2 = 2.83 \, 	ext{years}$,$R_2 = ?$.Using the ratio formula:$\left(\frac{T_2}{T_1}ight)^2 = \left(\frac{R_2}{R_1}ight)^3$Substituting the values:$\left(\frac{2.83}{1}ight)^2 = \left(\frac{R_2}{1.5 	imes 10^8}ight)^3$Since $2.83 \approx \sqrt{8}$,then $(2.83)^2 \approx 8$:$8 = \left(\frac{R_2}{1.5 	imes 10^8}ight)^3$Taking the cube root on both sides:$2 = \frac{R_2}{1.5 	imes 10^8}$$R_2 = 2 	imes 1.5 	imes 10^8 = 3 	imes 10^8 \, 	ext{km}$.Thus,the distance is $3 	imes 10^8 \, 	ext{km}$.

If the distance of the earth from the Sun is $1.5 \times 10^8 \, km$. Then the distance of an imaginary planet from the Sun,if its period of revolution is $2.83$ years,is $............. \times 10^8 \, km$.

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