JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
Three particles of masses \(50\, g\), \(100\, g\) and \(150\, g\) are placed at the vertices of an equilateral triangle of side \(1\, m\) (as shown in the figure). The \((x, y)\) coordinates of the centre of mass will be
- A \(\left( {\frac{{\sqrt 3 }}{7}\,m,\,\frac{7}{{12}}\,m} \right)\)
- B \(\left( {\frac{7}{{12}}m,\,\frac{{\sqrt 3 }}{8}\,m} \right)\)
- C \(\left( {\frac{{\sqrt 3 }}{4}\,m,\,\frac{5}{{12}}\,m} \right)\)
- D \(\left( {\frac{7}{{12}}m,\,\frac{{\sqrt 3 }}{4}\,m} \right)\)
Answer & Solution
Correct Answer
(D) \(\left( {\frac{7}{{12}}m,\,\frac{{\sqrt 3 }}{4}\,m} \right)\)
Step-by-step Solution
Detailed explanation
The co-ordinates of the center of mass \({\overrightarrow r _{cm}} = \frac{{0 + 150 \times \left( {\frac{1}{2}i + \frac{{\sqrt 3 }}{2}\hat j} \right) + 100 \times \hat i}}{{300}}\) \({\overrightarrow r _{cm}} = \frac{7}{{12}}\hat i + \frac{{\sqrt 3 }}{4}\hat j\)…
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