Find the gravitational force of attraction between the ring and sphere as shown in the diagram,

Question:

Find the gravitational force of attraction between the ring and sphere as shown in the diagram, where the plane of the ring is perpendicular to the line joining the centres. If $\sqrt{8} \mathrm{R}$ is the distance between the centres of a ring (of mass ' $\mathrm{m}$ ') and a sphere (mass 'M') where both have equal radius 'R'.

  1. (1) $\frac{\sqrt{8}}{9} \cdot \frac{G m M}{R}$

  2. (2) $\frac{\sqrt{8}}{27} \cdot \frac{\mathrm{GmM}}{\mathrm{R}^{2}}$

  3. (3) $\frac{2 \sqrt{2}}{3} \cdot \frac{\mathrm{GMm}}{\mathrm{R}^{2}}$

  4. (4) $\frac{1}{3 \sqrt{8}} \cdot \frac{\text { GMm }}{R^{2}}$


Correct Option: , 2

Solution:

(2)

We know that

$F=M E=M\left(\frac{G M \sqrt{8} R}{\left(R^{2}+(\sqrt{8} R)^{2}\right)^{3 / 2}}\right)$

$\mathrm{F}=\frac{\mathrm{GMm} \sqrt{8 \mathrm{R}}}{\left(9 \mathrm{R}^{2}\right)^{3 / 2}} \Rightarrow \frac{2 \sqrt{2} \mathrm{GmM}}{\left(9 \mathrm{R}^{2}\right)^{3 / 2}}$

$=\frac{2 \sqrt{2} \mathrm{GmM}}{27 \mathrm{R}^{2}}$

$F=\frac{\sqrt{8} \mathrm{GMm}}{27 \mathrm{R}^{2}}$

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now