Two uniform solid spheres of equal radii R, but mass M and 4 M have a center to center separation 6 R as shown. If N is the neutral point, then r its distance from the centre of the sphere of mass M is.

Question

Two uniform solid spheres of equal radii R, but mass M and 4 M have a center to center separation 6 R as shown. <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/ab2288c768ec9ee214ef48c6be7945c3-.png" /> If N is the neutral point, then r its distance from the centre of the sphere of mass M is. (A) R (B) √2 R (C) √3 R (D) 2 R

Tardigrade Admin · Accepted Answer

If a mass m is kept at N, the neutral point, then forces on it due to M and 4 M balance. (G M m/r2)=(G(4 M) m/(6 R-r)2) ⇒ (6 R-r)2=4 r2 6 R-r=± 2 r r=2 R or r=-6 R r is + ve , hence r=2 R

Q. Two uniform solid spheres of equal radii $R$ , but mass $M$ and $4 M$ have a center to center separation $6 R$ as shown.

If $N$ is the neutral point, then $r$ its distance from the centre of the sphere of mass $M$ is.

$R$

$2 R$

$3 R$

$2 R$

If a mass $m$ is kept at $N,$ the neutral point, then forces on it due to $M$ and $4 M$ balance.
$\frac{GM m}{r ^{2}} = \frac{G ( 4 M ) m}{( 6 R - r ) ^{2}} \Rightarrow (6 R - r)^{2} = 4 r^{2}$
$6 R - r = \pm 2 r$
$r = 2 R$ or $r = - 6 R$
$r$ is $+$ ve $,$ hence $r = 2 R$

Q. Two uniform solid spheres of equal radii R, but mass M and 4M have a center to center separation 6R as shown. If N is the neutral point, then r its distance from the centre of the sphere of mass M is.

R

2​R

3​R

2R

If a mass m is kept at N, the neutral point, then forces on it due to M and 4M balance. r2GMm​=(6R−r)2G(4M)m​⇒(6R−r)2=4r2 6R−r=±2r r=2R or r=−6R r is + ve , hence r=2R

Q. Two uniform solid spheres of equal radii $R$ , but mass $M$ and $4 M$ have a center to center separation $6 R$ as shown.

If $N$ is the neutral point, then $r$ its distance from the centre of the sphere of mass $M$ is.

$R$

$2 R$

$3 R$

$2 R$

If a mass $m$ is kept at $N,$ the neutral point, then forces on it due to $M$ and $4 M$ balance.
$\frac{GM m}{r ^{2}} = \frac{G ( 4 M ) m}{( 6 R - r ) ^{2}} \Rightarrow (6 R - r)^{2} = 4 r^{2}$
$6 R - r = \pm 2 r$
$r = 2 R$ or $r = - 6 R$
$r$ is $+$ ve $,$ hence $r = 2 R$