Q134 / 160 · TS EAMCET 2020 (14 Sep Shift 2) · Physics
A mass $M$ is split into two parts $m_0$ and $M-m_0$. These two masses are then separated by a distance $D$. If the gravitational force between the parts is maximum, then the ratio $\frac{m_0}{M}$ is
- A0.2
- B0.4
- Correct answerC0.5
- D0.6
Solution
According to given situation,
Again, $m_0$ and $M-m_0$ are kept at a distance $D$ gravitational force,
$F=\frac{G m_0\left(M-m_0\right)}{D^2}$ $F$ will be maximum, when $\frac{d F}{d m_0}=0$ $\begin{array}{ll} \Rightarrow & \frac{d}{d m_0} \frac{G m_0\left(M-m_0\right)}{D^2}=0 \\ \Rightarrow & \frac{d}{d m_0}\left(\frac{G m_0 M-G m_0^2}{D^2}\right)=0 \\ \Rightarrow & \frac{G M-2 G m_0}{D^2}=0 \\ \Rightarrow & G M-2 G m_0=0 \Rightarrow M-2 m_0=0 \\ \Rightarrow & 2 m_0=M \Rightarrow \frac{m_0}{M}=\frac{1}{2}=0.5 \end{array}$
Again, $m_0$ and $M-m_0$ are kept at a distance $D$ gravitational force,
$F=\frac{G m_0\left(M-m_0\right)}{D^2}$ $F$ will be maximum, when $\frac{d F}{d m_0}=0$ $\begin{array}{ll} \Rightarrow & \frac{d}{d m_0} \frac{G m_0\left(M-m_0\right)}{D^2}=0 \\ \Rightarrow & \frac{d}{d m_0}\left(\frac{G m_0 M-G m_0^2}{D^2}\right)=0 \\ \Rightarrow & \frac{G M-2 G m_0}{D^2}=0 \\ \Rightarrow & G M-2 G m_0=0 \Rightarrow M-2 m_0=0 \\ \Rightarrow & 2 m_0=M \Rightarrow \frac{m_0}{M}=\frac{1}{2}=0.5 \end{array}$
Hence, the correct answer is option: 'C'.