JEE Advanced · Physics · 7. COM & Collisions
Look at the drawing given in the figure, which has been drawn with ink of uniform line-thickness.

The mass of ink used to draw each of the two inner circles, and each of the two line segments is \(m\). The mass of the ink used to draw the outer circle is \(6 \mathrm{~m}\). The coordinates of the centres of the different parts are, outer circle 10 , \(0)\), left inner circle \((0,0)\), right inner circle \((a, a)\), vertical line \((0,0)\) and horizontal line \((0,-a)\). The \(y\)-coordinate of the centre of mass of the ink in this drawing is
- A \(\frac{a}{10}\)
- B \(\frac{a}{8}\)
- C \(\frac{a}{12}\)
- D \(\frac{a}{3}\)
Answer & Solution
Correct Answer
(A) \(\frac{a}{10}\)
Step-by-step Solution
Detailed explanation
\(y_{C M}\)
\(
\begin{gathered}
=\frac{m_1 y_1+m_2 y_2+m_3 y_3+m_4 y_4+m_5 y_5}{m_1+m_2+m_3+m_4+m_5} \\
=\frac{(6 m)(0)+(m)(a)+m(a)+m(0)}{6 m+m+m+m+m}=\frac{a}{10}
\end{gathered}
\)
\(
\begin{gathered}
=\frac{m_1 y_1+m_2 y_2+m_3 y_3+m_4 y_4+m_5 y_5}{m_1+m_2+m_3+m_4+m_5} \\
=\frac{(6 m)(0)+(m)(a)+m(a)+m(0)}{6 m+m+m+m+m}=\frac{a}{10}
\end{gathered}
\)
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