A uniform ring of mass m and radius r is placed directly above a uniform sphere of mass M and of equal radius. The centre of the ring is directly above the centre of the sphere at a distance r√3 as shown in the figure. The gravitational force exerted by the sphere on the ring will be

Question

A uniform ring of mass m and radius r is placed directly above a uniform sphere of mass M and of equal radius. The centre of the ring is directly above the centre of the sphere at a distance r√3 as shown in the figure. The gravitational force exerted by the sphere on the ring will be (A) (GMm/8r2) (B) (GMm/4r2) (C) √3(GMm/8r2) (D) (GMm/8r3√3)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/da9f64dc8f2efa0d2c7a2c67bc31e95c-.jpeg" /> dF = G (Mdm/4r2) F = Sigma dF cos θ = Sigma (GMdm/4r2) cos θ = (GM/4r2) × (√3r/2r) Sigma dm = (√3GMm/8r2)

Q. A uniform ring of mass m and radius $r$ is placed directly above a uniform sphere of mass $M$ and of equal radius. The centre of the ring is directly above the centre of the sphere at a distance $r 3$ as shown in the figure. The gravitational force exerted by the sphere on the ring will be

$\frac{GM m}{8 r ^{2}}$

$\frac{GM m}{4 r ^{2}}$

$3 \frac{GM m}{8 r ^{2}}$

$\frac{GM m}{8 r ^{3} 3}$

$d F = G \frac{M d m}{4 r ^{2}}$
$F = Σ d F cos θ$
$= Σ \frac{GM d m}{4 r ^{2}} cos θ$
$= \frac{GM}{4 r ^{2}} \times \frac{3 r}{2 r} Σ d m$
$= \frac{3 GM m}{8 r ^{2}}$

Q. A uniform ring of mass m and radius r is placed directly above a uniform sphere of mass M and of equal radius. The centre of the ring is directly above the centre of the sphere at a distance r3​ as shown in the figure. The gravitational force exerted by the sphere on the ring will be

8r2GMm​

4r2GMm​

3​8r2GMm​

8r33​GMm​