JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion
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 \(\left(\dfrac{\sqrt{3}}{4}, 1.25\right)\)
- B \(\left(\dfrac{\sqrt{3}}{4}, 1.0\right)\)
- C \((0, 0)\)
- D \((1.25, 0)\)
Answer & Solution
Correct Answer
(A) \(\left(\dfrac{\sqrt{3}}{4}, 1.25\right)\)
Step-by-step Solution
Detailed explanation
Let the midpoint \(p\) of the normal bisector be the origin \((0,0)\). The triangle is isosceles with two sides of \(10\text{ m}\) and an included angle of \(120^\circ\). The altitude from the \(15\text{ kg}\) mass bisects the angle into two \(60^\circ\) angles. The length of…
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