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
Answers
Explanation:
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\: :-}}♣Inthefirstcase:−
As we know that :
\bigstar\: \: \sf\boxed{\bold{\pink{F_g =\: \dfrac{Gm_1m_2}{r^2}}}}\: \: \bigstar★
F
g
=
r
2
Gm
1
m
2
★
Gravitational Force (\sf F_gFg ) = 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,
⟹ \: F < /p > < p > _ 1= \frac{g \: \times \: 2m \: \times \: m}{ {r}^{2} } < /p > < p >⟹F</p><p>
1
=
r
2
g×2m×m
</p><p>
\implies \sf F_1 =\: \dfrac{G \times 2m^2}{r^2}⟹F
1
=
r
2
G×2m
2
\implies \sf\bold{\purple{F_1 =\: \dfrac{2Gm^2}{r^2}\: ------\: (Equation\: No\: 1)}}⟹F
1
=
r
2
2Gm
2
−−−−−−(EquationNo1)
< /p > < p > \clubsuit\: \: \sf\bold{\blue{In\: the\: second\: case\: :-}} < /p > < p ></p><p>♣Inthesecondcase:−</p><p>
Given:-
Gravitational Force (\sf F_gFg ) = 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,
⟹F < /p > < p > _2= \: \frac{g \times \frac{ m }{2} \: \times m }{ {r}^{2} }⟹F</p><p>
2
=
r
2
g×
2
m
×m
\implies \sf F_2 =\: \dfrac{G \times \dfrac{m^2}{2}}{r^2}⟹F
2
=
r
2
G×
2
m
2
⟹F_2= \: \frac{ \frac{ {gm}^{2} }{2} }{ {r}^{2} }⟹F
2
=
r
2
2
gm
2
\implies \sf F_2 =\: \dfrac{Gm^2}{2} \times \dfrac{1}{r^2}⟹F
2
=
2
Gm
2
×
r
2
1
\implies \sf\bold{\purple{F_2 =\: \dfrac{Gm^2}{2r^2}\: ------\: (Equation\: No\: 2)}}⟹F
2
=
2r
2
Gm
2
−−−−−−(EquationNo2)
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\}}⟶
F
2
F
1
=
{
2r
2
Gm
2
}
{
r
2
2Gm
2
}
< /p > < p > < /p > < p > < /p > < p > \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\}</p><p></p><p></p><p>⟶
F
2
F
1
={
r
2
2
Gm
2
}×{
Gm
2
2
r
2
}
< /p > < p > < /p > < p > < /p > < p > \longrightarrow \sf \dfrac{F_1}{F_2} =\: \dfrac{2}{1} \times \dfrac{2}{1}</p><p></p><p></p><p>⟶
F
2
F
1
=
1
2
×
1
2
< /p > < p > < /p > < p > < /p > < p > \longrightarrow \sf \dfrac{F_1}{F_2} =\: \dfrac{2 \times 2}{1 \times 1}</p><p></p><p></p><p>⟶
F
2
F
1
=
1×1
2×2
< /p > < p > < /p > < p > < /p > < p > \longrightarrow \sf \dfrac{F_1}{F_2} =\: \dfrac{4}{1}</p><p></p><p></p><p>⟶
F
2
F
1
=
1
4
\longrightarrow \sf\bold{\red{F_1 : F_2 =\: 4 : 1}}⟶F
1
:F
2
=4:1
{\small{\bold{\underline{\therefore\: The\: ratio\: of\: F_1 : F_2\: is\: 4 : 1\: .}}}}
∴TheratioofF
1
:F
2
is4:1.
Hence, the correct options is option
pls mark the Brainliest and follow the instructions
Answer:
I'm little bit fine ☹︎
Today Maa durga went home .... it's kinda sad