Q166 / 225 · JEE Main 2003 · Physics
Two spherical bodies of mass $M$ and $5 M \&$ radii $R \& 2 R$ respectively are released in free space with initial separation between their centres equal to $12 \mathrm{R}$. If they attract each other due to gravitational force only, then the distance covered by the smaller body just before collision is
- A$2.5 \mathrm{R}$
- B$4.5 \mathrm{R}$
- Correct answerC$7.5 \mathrm{R}$
- D$1.5 \mathrm{R}$
Solution
Distance between the surface of the spherical bodies $=12 R-R-2 R=9 R$
Force $\propto$ Mass, $\quad$ Acceleration $\propto$ Mass, $\quad$ Distance $\propto$ Acceleration
$$
\begin{aligned}
& \Rightarrow \frac{\mathrm{a}_1}{\mathrm{a}_2}=\frac{\mathrm{M}}{\mathrm{SM}}=\frac{1}{5} \Rightarrow \frac{\mathrm{S}_1}{\mathrm{~S}_2}=\frac{1}{5} \Rightarrow \mathrm{S}_2=5 \mathrm{~S}_1 \\
& \mathrm{~S}_1+\mathrm{S}_2=9 \Rightarrow 6 \mathrm{~S}_1=9 \Rightarrow \mathrm{S}_1=\frac{9}{6}=1.5,
\end{aligned}
$$
Note: Maximum distance will be travelled by smaller bodies due to the greater acceleration caused by the same gravitational force
Hence, the correct answer is option: 'C'.