Question

1. Two objects of mass 1 kg and 2 kg are fixed at a
certain distance between them. If a mass of 4 kg.
placed between them, attains equilibrium at a
distance r from the smaller mass, then its
distance from the larger mass will be
(1) r
(2) 2r
(3) √2r
(4) 4r

pls give an explanation how did you find the answer so that I can do like these question by myself. thank you ​

Answer 1

GiveN :-

• Two objects of mass 1 kg and 2 kg are fixed at a certain distance between them.
• A mass of 4 kg placed between them, attains equilibrium at adistance r from the smaller mass .

To FinD :-

• The distance from the larger mass .

SolutioN :-

As the system attains equilibrium after placing a mass of 4kg between them , the gravitational forces acting of mass of 4kg from 1 kg and 2kg must be equal . Let the distance of mass of 4kg from the mass of 2kg be x . Already we are given that the distance of mass of 1kg ( smaller mass) from 4kg is r .

Also we know the gravitational force acting between mass m1 and m2 , with r as distance between them is the product of the masses of object along with Gravitational constant (G) divided by the square of distance between them . Let the forces be F1 and F2 respectively .

$\sf:\implies \pink{ Force_1 = Force_2}\\\\\sf:\implies G\dfrac{m_1 m_2}{r_1^2} =G\dfrac{m_3 m_4}{r_2^2}\\\\\sf:\implies \dfrac{m_1 m_2}{r_1^2} =\dfrac{m_3 m_4}{r_2^2}\\\\\sf:\implies \dfrac{ (1)(4)}{r^2}=\dfrac{(4)(2)}{x^2}\\\\\sf:\implies \dfrac{4}{r^2}=\dfrac{8}{x^2}\\\\\sf:\implies x^2= \dfrac{8r^2}{4}\\\\\sf:\implies x^2= 2r^2 \\\\\sf:\implies x= \sqrt{2r^2} \\\\\sf:\implies \underset{\blue{\sf Option \ 3 }}{\underbrace{\boxed{\pink{\frak{ Distance_{(From\ 2kg \ mass)}= \sqrt{2}r }}}}}$