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System of Particles and Rotational Motion · Question 3

JEE Main 2026 (Online) 6th April Morning ShiftMCQMedium📈 Topic in 10% papers+4−1
1

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 . [figure]

  1. ( 34, 1.25 )
  2. ( 34, 1.0 )
  3. (0, 0)
  4. (1.25, 0)

Answer

(1) ( 34, 1.25 )

Solution

[figure]

Each half of the 120° angle is 60°:

Half base =10 60 =5 3 m, height =10 60 =5 m

P is the mid-point of the altitude ⇒ 2.5 m above the base.

Origin at P: 2 kg at (-5 3, -2.5), 3 kg at (5 3, -2.5), 15 kg at (0, 2.5); total mass 20 kg

xcm= 2(-5 3)+3(5 3)+15(0)20= 5 320= 34

ycm= 2(-2.5)+3(-2.5)+15(2.5)20= -5-7.5+37.520= 2520=1.25

∴ Centre of mass = ( 34, 1.25 )

Answer: (1)

JEE Main 2026 (Online) 6th April Morning Shift

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