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 √(8) R is the distance between the centres of a ring (of mass 'm') and a sphere (mass 'M') where both have equal radius 'R'. (a) (2√(2))/(3)·GMmR² (b) √(8)/9/R (c) (√(8))/(27)·GmMR² (d) (1)/(3√(8))·GMmR²

Answer: (c). The sphere acts as a point mass M at its centre, on the axis of the ring. Field of a ring on its axis: E = Gm x(R² + x²)3/2, with x = √(8)R. R² + x² = 9R² ⇒ (9R²)3/2 = 27R³ F = (GMm√(8)R)/(27R³) = (√(8))/(27)·(GMm)/(R²) So the answer is (c).

Find the gravitational force of attraction between the ring and… | JEE Main 2021 (26 Feb, Shift 1)

Q14JEE Main 2021 · 26 FEB · SHIFT 1Difficult

Newton's Law of Gravitation

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 $\sqrt{8}$ R is the distance between the centres of a ring (of mass 'm') and a sphere (mass 'M') where both have equal radius 'R'.

Show answer and solution

Answer: (c)

The sphere acts as a point mass $M$ at its centre, on the axis of the ring.
Field of a ring on its axis: $E = \frac{Gm x}{(R^2 + x^2)^{3/2}}$ , with $x = \sqrt{8}R$ .
$R^2 + x^2 = 9R^2 \Rightarrow (9R^2)^{3/2} = 27R^3$
$F = \frac{GMm\sqrt{8}R}{27R^3} = \frac{\sqrt{8}}{27}\cdot\frac{GMm}{R^2}$
So the answer is (c).

Report a mistake

Newton's Law of Gravitation question from JEE Main 2021 (26 Feb, Shift 1)

More questions on Newton's Law of Gravitation