Question

Pls solve the third one I will Mark u brainliest Don't say wrong ans

Answer 1

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \boxed{\boxed { \huge \mathcal\red{ solution}}}}$

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$\underline{\bf\red{\mathbb{USING\: FORMULAS}}}$

(1)The relationship between made, volume and density is ,

$\boxed{\mathcal{Mass=Volume\times Density}}$

(2)The gravitational force of attraction between two objects of masses M and m separated by a distance d ,is given by,

$\boxed{\mathcal{F=G\frac{Mm}{d{}^{2}}}}$

Where, G=Gravitational constant

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$\underline{\bf\red{\mathbb{NOW\:LET\:SOLVE\: PROBLEM}}}$

$Given= \begin{cases}</p><p></p><p>\text{two spheres are of same material} \\</p><p></p><p>\text{density of both spheres is} \:\bf\rho \\ \text{but spares have same radius =R} </p><p></p><p>\end{cases}$

$\therefore$Both spheres would have the same mass ,Let's it M.

$\textbf{Now Volume of Each Sphere}\bf\:V=\frac{4}{3}\pi R{}^{3}$

$\therefore$Mass of each Sphere=M

$\Longrightarrow\textbf{M}=V\times \rho\\</p><p>\Longrightarrow M=\frac{4}{3}\pi R{}^{3}\times\rho\\</p><p>\Longrightarrow \boxed{M=\frac{4}{3}\pi R{}^{3}\rho}$

Now here the spheres are touching Each other .

So, the distance between their centre of mass is $\bf\Longrightarrow d=(R+R)\\</p><p>\bf\Longrightarrow \boxed{d=2R}$

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$\underline{\bf\red{\mathbb{GRAVITATIONAL\:FORCE\:OF\: ATTRACTION}}}$

$\Longrightarrow F=G\frac{M\times M}{d{}^{2}}\\</p><p>\bf\Longrightarrow F=G\frac{\frac{4}{3}\pi R{}^{3}\rho\times \frac{4}{3}\pi R{}^{3}\rho}{(2R){}^{2}}\\</p><p>\bf\Longrightarrow F=G\frac{\cancel4\times4\times\pi\times\pi\times\cancel{ R{}^{3}}{}^{\Large{R}}\times R{}^{3}\times\rho\times\rho}{3\times3\times\cancel4\times \cancel{R{}^{2}}}\\</p><p>\bf\Longrightarrow \boxed{\bf\blue{F=\frac{4}{9}\pi{}^{2}G\rho{}^{2}R{}^{4}}}$

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$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$

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