JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
The identical spheres each of mass \(2 \mathrm{M}\) are placed at the corners of a right angled triangle with mutually perpendicular sides equal to \(4 \mathrm{~m}\) each. Taking point of intersection of these two sides as origin, the magnitude of position vector of the centre of mass of the system is \(\frac{4 \sqrt{2}}{x}\), where the value of \(x\) is _______.
- A \(2\)
- B \(3\)
- C \(4\)
- D \(5\)
Answer & Solution
Correct Answer
(B) \(3\)
Step-by-step Solution
Detailed explanation
\(\text { Position vector } \overrightarrow{\mathrm{r}}_{\text {COM }}=\frac{\mathrm{m}_1 \overrightarrow{\mathrm{r}}_1+\mathrm{m}_2 \overrightarrow{\mathrm{r}}_2+\mathrm{m}_3 \overrightarrow{\mathrm{r}}_3}{\mathrm{~m}_1+\mathrm{m}_2+\mathrm{m}_3}\)…
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