JEE Main 2026 Apr 6 Shift 1, Physics Q8: Centre of Mass

JEE Main2026Apr 6, Shift 1Physics

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 ____.

A
$(\frac{3}{4}, 1.25)$
B
$(\frac{3}{4}, 1.0)$
C
$(0, 0)$
D
$(1.25, 0)$

Solution

Place the origin at $p$ (midpoint of the base), $x$ -axis along the base. The apex angle is $12 0^{\circ}$ , so the base angles are $3 0^{\circ}$ each.

From the apex (at $15$ kg), each $10$ cm side projects horizontally $10 cos 3 0^{\circ} = 53$ and vertically $10 sin 3 0^{\circ} = 5$ . So, the apex is at $(0, 5/2)$ cm and the base vertices at $(\pm 53, - 5/2)$ cm.

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

$Y_{c m} = \frac{2 ( - 5/2 ) + 3 ( - 5/2 ) + 15 ( 5/2 )}{20} = \frac{- 5 - 7.5 + 37.5}{20} = \frac{25}{20} = 1.25$ .

$∴$ centre of mass at $(\frac{3}{4}, 1.25)$ .

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JEE Main 2026 Apr 6 Shift 1, Physics Q8: Centre of Mass

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