Question

Two point masses A and B having masses in the ratio 4 : 3 are separated by a distance of lm. When another point mass c of mass M is placed in between A and B the forces A and C is (1 / 3rd) of the force between B and C, Then the distance C from A is = ………. m 

(A) (2/3) 

(B) 1/3 

(C) 1/4 

(D) 2/7

Answer 1

Correct option: (A) (2/3)

Explanation:

Given (m_A / m_B) = 4/3

d(AB) = 1m

Let point C be at distance x from A.

F_AC = [(G ∙ m_A ∙ m) / x²] and

F_BC = [(m_B ∙ m ∙ G) / (1 – x)²]---------- F = [(G ∙ m₁ ∙ m₂) / r²]

Given : F_AC = (1/3)F_BC 

hence

[(G ∙ m_A ∙ m) / x²] = (1/3) ∙ [(G m_B ∙ m) / (1 – x)²]

3 ∙ (m_A / m_B) = [x² / (1 – x²)]

3 × (4/3) = [x² / (1 – x)²]

hence 2 = [x / (1 – x)]

hence 2 – 2x = x

x = (2/3)