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Answer 1

$\huge{\text{\underline{Solution:-}}}$

A mass of both of the object are not equal thus centre of mass will be more towards 3m than m.

Let distance of centre of mass from mass m be x & that from 3m be (r - x)

$\implies{\boxed{\tt{mx = 3m (r - x)}}}$

★ On solving we will get:-

• x= 3r / 4
• r - x = r / 4

★ Now centre of mass of the system will be:-

3m (r - x)² + mx²

(3mr² / 16) + (m9 r² / 16)

★ On putting x = 3r/4

$\large\implies{\boxed{\tt{3/4 mr^2}}}$

Hence, the correct answer is option (2) 3/4 mr²

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Answer 2

Answer:

Option => 2.

$\bf \implies \dfrac{3}{4}\;mr^2$

Explanation:

Given :

Two point masses M and 3M are placed at a distance r.

Now,

• As mass of both of the object are not equal :

Center of mass will be toward 3M and than M.

Consider the :

Distance from center of mass by m as - $\boxed{\bf y}$

Distance from center of mass by 3M as - $\boxed{\bf r-y}$

Therefore,

$\sf\\ \\\implies My = 3M (r-y)\\ \\ \implies My= 3Mr - 3My\\ \\ \implies y =\dfrac{3r}{4}\\ \\ \implies r-y= \dfrac{r}{4}</p><p>$

Now,

The Center of mass of the system,

$\sf\\ \\\implies M(r-y)^2 + My^2\\ \\ \implies \dfrac{3Mr^2}{16}+\dfrac{M9r^2}{16}\\ \\ \implies \dfrac{3}{4}\;mr^2$

∴ The Option 2 is the right answer.