ExamBro
ExamBro
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 _______.

  1. A \(2\)
  2. B \(3\)
  3. C \(4\)
  4. D \(5\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(3\)

Step-by-step Solution

Detailed explanation

\(\text { Position vector } \overrightarrow{\mathrm{r}}_{\text {COM }}=\frac{\mathrm{m}_1 \overrightarrow{\mathrm{r}}_1+\mathrm{m}_2 \overrightarrow{\mathrm{r}}_2+\mathrm{m}_3 \overrightarrow{\mathrm{r}}_3}{\mathrm{~m}_1+\mathrm{m}_2+\mathrm{m}_3}\)…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app