Q 11-06-045JEE MainJEE Main 2026 (6 Apr, Shift 1)Medium
The position of center of mass of three masses $2$ kg, $3$ kg and $15$ kg placed with respect to mid point ($p$) of normal bisector, as shown in the figure is ______.
Answer: (A) $\left(\dfrac{\sqrt{3}}{4}, 1.25\right)$
The two sides of $10$ m meet at $120^\circ$, so each makes $60^\circ$ with the vertical bisector: the base half-width is $10\sin 60^\circ = 5\sqrt{3}$ m and the height is $10\cos 60^\circ = 5$ m.
$P$ is the midpoint of the bisector, $2.5$ m above the base. Taking $P$ as origin: $15$ kg at $(0, 2.5)$, $2$ kg at $(-5\sqrt{3}, -2.5)$, $3$ kg at $(5\sqrt{3}, -2.5)$.
$$x_{cm} = \frac{-10\sqrt{3} + 15\sqrt{3}}{20} = \frac{\sqrt{3}}{4}, \qquad y_{cm} = \frac{37.5 - 5 - 7.5}{20} = 1.25$$
Solution by Sreeraj P, M.Sc Physics