JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2024
- Hard
- Numerical
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 _______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(3\)
Step-by-step Solution
Detailed explanation
\(\text { Position vector } \overrightarrow{\mathrm{r}}_{\text {COM }}=\frac{\mathrm{m}_1 \overrightarrow{\mathrm{r}}_1+\mathrm{m}_2 \overrightarrow{\mathrm{r}}_2+\mathrm{m}_3 \overrightarrow{\mathrm{r}}_3}{\mathrm{~m}_1+\mathrm{m}_2+\mathrm{m}_3}\)
\(\overrightarrow{\mathrm{r}}_{\text {COM }}=\frac{2 \mathrm{M} \times 0+2 \mathrm{M} \times 4 \hat{\mathrm{i}}+2 \mathrm{M} \times 4 \hat{\mathrm{j}}}{6 \mathrm{M}}\)
\(\overrightarrow{\mathrm{r}}=\frac{4}{3} \hat{\mathrm{i}}+\frac{4}{3} \hat{\mathrm{j}}\)
\(|\overrightarrow{\mathrm{r}}|=\frac{4 \sqrt{2}}{3}\)
\(\mathrm{x}=3\)
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