Four identical particles of equal masses 1 kg made to move along the circumference of a circle of radius 1 m under the action of their own mutual gravitational attraction. The speed of each particle will be :

Gravitation PYQ — JEE Main 2021 · 24 Feb · Shift 1 · Q54 with Solution

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Correct answer: D

Configuration of the particles

Since four identical masses move on the circumference of a circle of radius $1\,\text{m}$ under their mutual gravitational attraction, they must be placed symmetrically at the corners of a square inscribed in the circle.

Radius of circle: $R=1$
Mass of each particle: $m=1\,\text{kg}$

For a square inscribed in a circle of radius $1$ :

Side length: $a=\sqrt{2}$
Diagonal length: $d=2$

Force on one particle due to the other three

Take one particle at a vertex of the square.

It experiences gravitational attraction due to:

the two adjacent particles, each at distance $\sqrt{2}$
the opposite particle, at distance $2$

Force due to each adjacent particle

Magnitude of force from one adjacent particle: $F_1=\frac{G(1)(1)}{(\sqrt{2})^2}=\frac{G}{2}$

There are two such forces, symmetric about the diagonal through the center.

The angle between each such force and the diagonal is $45^\circ$ , so their resultant along the diagonal is:

=G\cdot \frac{1}{\sqrt{2}}=\frac{G}{\sqrt{2}}$$ --- 4. **Force due to the opposite particle** Distance to opposite particle is $2$, so: $$F_{\text{opp}}=\frac{G(1)(1)}{2^2}=\frac{G}{4}$$ This force is also along the same diagonal toward the center. --- 5. **Total inward force** Thus total gravitational force toward the center on one particle is: $$F=\frac{G}{\sqrt{2}}+\frac{G}{4}$$ Write with common factor: $$F=G\left(\frac{1}{\sqrt{2}}+\frac14\right) =G\left(\frac{\sqrt{2}}{2}+\frac14\right) =\frac{G}{4}(2\sqrt{2}+1)$$ --- 6. **Use centripetal force condition** Each particle moves in a circle of radius $R=1$, so required centripetal force is: $$\frac{mv^2}{R}=v^2$$ (since $m=1$ and $R=1$) Hence, $$v^2=F=\frac{G}{4}(2\sqrt{2}+1)$$ Therefore, $$v=\sqrt{\frac{G}{4}(2\sqrt{2}+1)} =\frac12\sqrt{G(1+2\sqrt{2})}$$ --- 7. **Compare with options** This matches: $$\boxed{\text{D: } \frac12\sqrt{G(1+2\sqrt{2})}}$$

More from Gravitation

View written solutionFree

Correct answer: D

Configuration of the particles

Radius of circle: $R=1$
Mass of each particle: $m=1\,\text{kg}$

For a square inscribed in a circle of radius $1$ :

Side length: $a=\sqrt{2}$
Diagonal length: $d=2$

Force on one particle due to the other three

Take one particle at a vertex of the square.

It experiences gravitational attraction due to:

the two adjacent particles, each at distance $\sqrt{2}$
the opposite particle, at distance $2$

Force due to each adjacent particle

Magnitude of force from one adjacent particle: $F_1=\frac{G(1)(1)}{(\sqrt{2})^2}=\frac{G}{2}$

There are two such forces, symmetric about the diagonal through the center.

The angle between each such force and the diagonal is $45^\circ$ , so their resultant along the diagonal is:

=G\cdot \frac{1}{\sqrt{2}}=\frac{G}{\sqrt{2}}$$ --- 4. **Force due to the opposite particle** Distance to opposite particle is $2$, so: $$F_{\text{opp}}=\frac{G(1)(1)}{2^2}=\frac{G}{4}$$ This force is also along the same diagonal toward the center. --- 5. **Total inward force** Thus total gravitational force toward the center on one particle is: $$F=\frac{G}{\sqrt{2}}+\frac{G}{4}$$ Write with common factor: $$F=G\left(\frac{1}{\sqrt{2}}+\frac14\right) =G\left(\frac{\sqrt{2}}{2}+\frac14\right) =\frac{G}{4}(2\sqrt{2}+1)$$ --- 6. **Use centripetal force condition** Each particle moves in a circle of radius $R=1$, so required centripetal force is: $$\frac{mv^2}{R}=v^2$$ (since $m=1$ and $R=1$) Hence, $$v^2=F=\frac{G}{4}(2\sqrt{2}+1)$$ Therefore, $$v=\sqrt{\frac{G}{4}(2\sqrt{2}+1)} =\frac12\sqrt{G(1+2\sqrt{2})}$$ --- 7. **Compare with options** This matches: $$\boxed{\text{D: } \frac12\sqrt{G(1+2\sqrt{2})}}$$

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