Question

The force of attraction between two objects of masses 2m amd m is F₁.When the object with mass 2m is replaced with another object of mass m/2 ,the new force of attraction is found to be F₂.The ratio of F₁:F₂ IS:-
(A)3:2 (B)2:3 (C)1:4 (D)=4:1

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Answer 1

Answer :-

Required ratio is 4 : 1 [Option.D]

Explanation :-

For the 1st case :-

We have :-

→ 1st mass = 2m

→ 2nd mass = m

→ Gravitational force = F₁

→ Distance between the masses = r

According to Newton's Law of Gravitation, we have the 1st equation as :-

⇒ F₁ = (G × 2m × m)/r²

⇒ F₁ = 2Gm²/r² ---(1)

For the 2nd case :-

We have :-

→ 1st mass = m/2

→ 2nd mass = m

→ Gravitational force = F₂

→ Distance between the masses = r

Again by the law of Gravitation, we have the 2nd equation as :-

⇒ F₂ = (G × m/2 × m)/r²

⇒ F₂ = (Gm²/2)/r²

⇒ F₂ = Gm²/2r² ----(2)

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On dividing eq.1 by eq.2, we get :-

⇒ F₁/F₂ = (2Gm²/r²) ÷ (Gm²/2r²)

⇒ F₁/F₂ = 2Gm²/r² × 2r²/Gm²

⇒ F₁/F₂ = (2 × 2)/1

⇒ F₁/F₂ = 4/1

⇒ F₁ : F₂ = 4 : 1

Answer 2

Answer:

Given :-

• The force of attraction between two objects of masses 2m and m is F₁.
• When the object of mass 2m is replaced with another object of mass m/2, the new force of attraction is found to be F₂ .

To Find :-

• What is the ratio of F₁ : F₂.

Solution :-

$\clubsuit\: \: \sf\bold{\blue{In\: the\: first\: case\: :-}}$

As we know that :

$\bigstar\: \: \sf\boxed{\bold{\pink{F_g =\: \dfrac{Gm_1m_2}{r^2}}}}\: \: \bigstar$

Given :

• Gravitational Force ($\sf F_g$) = F₁
• First Mass (m₁) = 2m
• Second Mass (m₂) = m
• Distance between the centre of masses (r²) = r

According to the question by using the formula we get,

$\implies \sf F_1 =\: \dfrac{G \times 2m \times m}{r^2}$

$\implies \sf F_1 =\: \dfrac{G \times 2m^2}{r^2}$

$\implies \sf\bold{\purple{F_1 =\: \dfrac{2Gm^2}{r^2}\: ------\: (Equation\: No\: 1)}}$

$\clubsuit\: \: \sf\bold{\blue{In\: the\: second\: case\: :-}}$

Given :

• Gravitational Force ($\sf F_g$) = F₂
• First Mass (m₁) = m/2
• Second Mass (m₂) = m
• Distance between the centre of masses (r²) = r

According to the question by using the formula we get,

$\implies \sf F_2 =\: \dfrac{G \times \dfrac{m}{2} \times m}{r^2}$

$\implies \sf F_2 =\: \dfrac{G \times \dfrac{m^2}{2}}{r^2}$

$\implies \sf F_2 =\: \dfrac{\dfrac{Gm^2}{2}}{r^2}$

$\implies \sf F_2 =\: \dfrac{Gm^2}{2} \times \dfrac{1}{r^2}$

$\implies \sf\bold{\purple{F_2 =\: \dfrac{Gm^2}{2r^2}\: ------\: (Equation\: No\: 2)}}$

From the equation no 1 and the equation no 2 we get,

$\longrightarrow \sf \dfrac{F_1}{F_2} =\: \dfrac{\bigg\{\dfrac{2Gm^2}{r^2}\bigg\}}{\bigg\{\dfrac{Gm^2}{2r^2}\bigg\}}$

$\longrightarrow \sf \dfrac{F_1}{F_2} =\: \bigg\{\dfrac{2\cancel{Gm^2}}{\cancel{r^2}}\bigg\} \times \bigg\{\dfrac{2\cancel{r^2}}{\cancel{Gm^2}}\bigg\}$

$\longrightarrow \sf \dfrac{F_1}{F_2} =\: \dfrac{2}{1} \times \dfrac{2}{1}$

$\longrightarrow \sf \dfrac{F_1}{F_2} =\: \dfrac{2 \times 2}{1 \times 1}$

$\longrightarrow \sf \dfrac{F_1}{F_2} =\: \dfrac{4}{1}$

$\longrightarrow \sf\bold{\red{F_1 : F_2 =\: 4 : 1}}$

${\small{\bold{\underline{\therefore\: The\: ratio\: of\: F_1 : F_2\: is\: 4 : 1\: .}}}}$

Hence, the correct options is option no (D) 4 : 1.