Question

Two point masses A and B having masses in the ratio 4 : 3 are separated by a distance of 1 m. When another point mass C of mass M is placed in between A and B, the force between A and C is $\left( \frac { 1 } { 3 } \right) ^ { \mathrm { rd } }$ of the force between B and C. Then the distance of C from A is

• A. $\frac { 2 } { 3 } \mathrm {~m}$
• B.
• C. $\frac { 1 } { 4 } \mathrm {~m}$
• D. $\frac { 2 } { 7 } \mathrm {~m}$

Answer 1

Answer: (A)

$\frac { 2 } { 3 } \mathrm {~m}$

Answer 2

Let a point mass C is placed at a distance of x m form the point mass A as shown in the figure.

Here, $\frac { M _ { A } } { M _ { B } } = \frac { 4 } { 3 }$

Force between A and C is

$\mathrm { F } _ { \mathrm { AC } } = \frac { \mathrm { GMM } _ { \mathrm { A } } } { \mathrm { x } ^ { 2 } }$ …..(i)

Force between B and C is

$\mathrm { F } _ { \mathrm { BC } } = \frac { \mathrm { GMM } _ { \mathrm { B } } } { ( 1 - \mathrm { x } ) ^ { 2 } }$ …. (ii)

According to given problem

$\mathrm { F } _ { \mathrm { AC } } = \frac { 1 } { 3 } \mathrm {~F} _ { \mathrm { BC } }$

$\therefore \frac { \mathrm { GM } _ { \mathrm { A } } \mathrm { M } } { \mathrm { x } ^ { 2 } } = \frac { 1 } { 3 } \left( \frac { \mathrm { GM } _ { \mathrm { B } } \mathrm { M } } { ( 1 - \mathrm { x } ) ^ { 2 } } \right)$ (Using (i) and (ii))

$\frac { 4 } { 3 } = \frac { M _ { B } } { 3 ( 1 - x ) ^ { 2 } }$ or $4 = \frac { x ^ { 2 } } { ( 1 - x ) ^ { 2 } }$

Or $2 = \frac { x } { 1 - x }$ or 2 -2x = 0

3x = 2 or $\mathrm { x } = \frac { 2 } { 3 } \mathrm {~m}$