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NEET PHYSICSMedium

Two objects of mass 10 kg10\text{ kg} and 20 kg20\text{ kg} respectively are connected to the two ends of a rigid rod of length 10 m10\text{ m} with negligible mass. The distance of the center of mass of the system from the 10 kg10\text{ kg} mass is

A

103 m\frac{10}{3}\text{ m}

B

203 m\frac{20}{3}\text{ m}

C

10 m10\text{ m}

D

5 m5\text{ m}

Step-by-Step Solution

Let the 10 kg10\text{ kg} mass be at x1=0x_1 = 0 and the 20 kg20\text{ kg} mass be at x2=10 mx_2 = 10\text{ m}. The center of mass Xcm=m1x1+m2x2m1+m2=10(0)+20(10)10+20=20030=203 mX_{cm} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2} = \frac{10(0) + 20(10)}{10 + 20} = \frac{200}{30} = \frac{20}{3}\text{ m}.

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