The position of center of mass of three masses $2 \mathrm{~kg}, 3 \mathrm{~kg}$ and 15 kg placed with respect to mid point (p) of normal bisector, as shown in the figure is ______.

(A) $\left(\frac{\sqrt{3}}{4}, 1.25\right)$

(B) $\left(\frac{\sqrt{3}}{4}, 1.0\right)$

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Conservation of Momentum

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)

Take p as origin with the dashed bisector vertical. Each 10 m side makes

6 0^{\circ}

with the bisector (half the

12 0^{\circ}

apex angle), so the apex-to-base vertical distance is

10 cos 6 0^{\circ} = 5

; as p is its midpoint, the 15 kg apex sits at

(0, 2.5)

and the two base masses at height

- 2.5

with horizontal offsets

\pm 10 sin 6 0^{\circ} = \pm 53

. Thus the 2 kg is at

(- 53, - 2.5)

and the 3 kg at

(53, - 2.5)

. Then

X_{c m} = \frac{2 ( - 5 3 ) + 3 ( 5 3 ) + 15 ( 0 )}{20} = \frac{5 3}{20} = \frac{3}{4}

and

Y_{c m} = \frac{15 ( 2.5 ) + 2 ( - 2.5 ) + 3 ( - 2.5 )}{20} = \frac{25}{20} = 1.25

. The centre of mass is at

(\frac{3}{4}, 1.25)

JEE Main · 2026 · 06 Apr 2026 · Shift 1 · Single correct option