Gravitation

JEE Main level, single correct, medium

47 of 223

Question 47 of 223

Two spherical bodies of mass MM and 5M&5 M \& radii R&2RR \& 2 R respectively are released in free space with initial separation between their centres equal to 12R12 \mathrm{R}. If they attract each other due to gravitational force only, then the distance covered by the smaller body just before collision is

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Answer: (C) 7.5R7.5 \mathrm{R}

Distance between the surface of the spherical bodies =12R−R−2R=9R=12 R-R-2 R=9 R
Force ∝\propto Mass, \quad Acceleration ∝\propto Mass, \quad Distance ∝\propto Acceleration
⇒a1a2=MSM=15⇒S1 S2=15⇒S2=5 S1 S1+S2=9⇒6 S1=9⇒S1=96=1.5,\begin{aligned} & \Rightarrow \frac{\mathrm{a}_1}{\mathrm{a}_2}=\frac{\mathrm{M}}{\mathrm{SM}}=\frac{1}{5} \Rightarrow \frac{\mathrm{S}_1}{\mathrm{~S}_2}=\frac{1}{5} \Rightarrow \mathrm{S}_2=5 \mathrm{~S}_1 \\ & \mathrm{~S}_1+\mathrm{S}_2=9 \Rightarrow 6 \mathrm{~S}_1=9 \Rightarrow \mathrm{S}_1=\frac{9}{6}=1.5, \end{aligned}
Note: Maximum distance will be travelled by smaller bodies due to the greater acceleration caused by the same gravitational force

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