Question

The force of attraction between the two bodies
at a certain separation is 10 N. What will be the
force of attraction between them if the separation
is reduced to half ?​

Answer 1

Answer:

The force of attraction (F₂) will be 40 N

Given:

1. Initial Force (F) = 10 N

2. Let the initial Separation be " r "

Explanation:

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From the formula we know,

⇒ F = ( G m₁ m₂ ) / r²

Where,

• F Denotes Gravitational force.

• m₁ & m₂ Denotes Masses.

• r Denotes Distance of separation.

From the above formula we can derive inference,

⇒ F ∝ 1 / r²

Therefore, For first case,

⇒ F₁ ∝ 1 / (r₁)² __[1]

And, for second case,

⇒ F₂ ∝ 1 / (r₂)² __[2]

_____________________

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Dividing both the equations

⇒ F₁ / F₂ = (r₂)² / (r₁)²

Now, from the statement of question,

The separation  is reduced to half ​

Therefore,

⇒ r₂ = r₁ / 2

⇒ r₁ = 2 r₂

Substituting,

⇒ 10 / F₂ = (r₂)² / (2 r₂)²

         ∵ [ F₁ = 10 N ]

⇒ 10 / F₂ =  r₂² / 4 r₂²

⇒  10 / F₂ =  1 / 4

⇒ F₂ = 10 × 4

⇒ F₂ = 40 N

∴ The force of attraction (F₂) will be 40 N.

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Answer 2

$\huge\underline\mathbb\red{Answer:-}$

the force of attraction(f2)will be 40n.