Question

Jo mujhe ine 4 question ka correct answer dega main use 4000 thanks dunga​

Answer 1

✦ 1st QUESTION ✦

♣GIVEN :

• $\sf{m_1}$ = 3 kg

• $\sf{m_2}$ = 12 kg

• $\sf{r}$ = 12 m

• $\sf{m_3}$ = 0.5 kg

♣TO FIND :

• Position from $\sf{m_1}$ to $\sf{m_2}$

♣SOLUTION :

❍ Let it's distance from $\sf{m_1}$ be x m and from $\sf{m_2}$ be (12 - x) m

Force of attraction on $\sf{m_3}$ by $\sf{m_1}$1($\sf{F_1}$)=$\sf{GMm/r^2}$

= $\sf{ \dfrac{Gm_1m_3}{ {x}^{2} }}$....➊

Force of attraction on $\sf{m_3}$ by $\sf{m_2}$1($\sf{F_2}$)=$\sf{GMm/r^2}$

= $\sf{ \dfrac{Gm_2m_3}{ {12-x}^{2} }}$....➋

❏ NOTE :-

Net force on the body is 0.

⛬ $\sf{F_1 = F_2}$

✯ Putting all values :-

$: \longmapsto\bf{Gm_1m_ 3/x^2=Gm_ 2m_3/(12-x)}$

$: \longmapsto\bf{m_ 1/x^2=m_2 /(12-x)2}$

$: \longmapsto\bf{(12-x)^2=x^2 \times m_ 2/m_ 1}$

$: \longmapsto\bf{144+x^2-24x=x^2 \times 12/4}$

$: \longmapsto\bf{144+x^2-24x=4x^2}$

$: \longmapsto\bf{4x^2-x^2+24x-144=0}$

$: \longmapsto\bf{3x^2+24-144=0}$

$: \longmapsto\bf{3x^2+36x-12x-144=0 }$

$: \longmapsto \bf{3 x(x+12)-12(x+12)=0}$

$: \longmapsto \bf{(x+12)(3x-12)=0}$

$:\leadsto \boxed{\bf{x= \dfrac{12}{3} = 4}}$

♣ANSWER :

• The answer is 4m