JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
A uniform rectangular thin sheet \(ABCD\) of mass \(M\) has length \(a\) and breadth \(b\), as shown in the figure. If the shaded portion \(HBGO\) is cut off, the coordinates of the centre of mass of the remaining portion will be
- A \(\left( {\frac{{5a}}{3},\frac{{5b}}{3}} \right)\)
- B \(\left( {\frac{{2a}}{3},\frac{{2b}}{3}} \right)\)
- C \(\left( {\frac{{3a}}{4},\frac{{3b}}{4}} \right)\)
- D \(\left( {\frac{{5a}}{12},\frac{{5b}}{12}} \right)\)
Answer & Solution
Correct Answer
(D) \(\left( {\frac{{5a}}{12},\frac{{5b}}{12}} \right)\)
Step-by-step Solution
Detailed explanation
\(x = \frac{{M\frac{a}{2} - \frac{M}{4} \times \frac{{3a}}{4}}}{{M - \frac{M}{4}}}\) \( = \frac{{\frac{a}{2} - \frac{{3a}}{{16}} \times \frac{{3a}}{4}}}{{\frac{3}{4}}} = \frac{{\frac{{5a}}{{16}}}}{{\frac{3}{4}}} = \frac{{5a}}{{12}}\)…
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