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Two particles of mass 5 kg and 10 kg respectively are attached to the two ends of a rigid rod of length 1 m with negligible mass. The centre of mass of the system from the 5 kg particle is nearly at a distance of:

1. 50 cm
2. 67 cm
3. 80 cm
4. 33 cm

Correct answer: 67 cm

Solution

The center of mass is calculated using the formula: \( x_{cm} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2} \). Taking the 5 kg particle as the origin (\(x_1 = 0\)) and the 10 kg particle at \(x_2 = 1\) m, \(x_{cm} = \frac{(5 \times 0) + (10 \times 1)}{5 + 10} = \frac{10}{15} = 0.67\) m or 67 cm.

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