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Mass $M$ is divided into two parts $x m$ and $(1-x) m$. For a given separation the value of $x$ for which the gravitational attraction between the two pieces becomes maximum is
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The correct answer is:
$\frac{1}{2}$
Force of gravitation $F \propto m_1 m_2$
$\begin{aligned} & F=k m_1 m_2 \\ & F=k(x m)(1-x) \times m\end{aligned}$
For $F$ to be maximum $\frac{d F}{d x}=0$
$\begin{aligned} \frac{d}{d x}[k(x m)(1-x) \times m] & =0 \\ m^2-2 x m^2 & =0 \\ \therefore \quad x & =\frac{1}{2}\end{aligned}$
$\begin{aligned} & F=k m_1 m_2 \\ & F=k(x m)(1-x) \times m\end{aligned}$
For $F$ to be maximum $\frac{d F}{d x}=0$
$\begin{aligned} \frac{d}{d x}[k(x m)(1-x) \times m] & =0 \\ m^2-2 x m^2 & =0 \\ \therefore \quad x & =\frac{1}{2}\end{aligned}$
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