Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

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

$5 \text{ m}$

$\frac{10}{3} \text{ m}$

$\frac{20}{3} \text{ m}$

$10 \text{ m}$

Step-by-Step Solution

Let the position of the $10 \text{ kg}$ mass be at the origin, so $x_1 = 0$ . The position of the $20 \text{ kg}$ mass will be at the other end of the rod, so $x_2 = 10 \text{ m}$ . The position of the center of mass $x_{cm}$ from the origin (which is the $10 \text{ kg}$ mass) is given by the formula: $x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$ Substituting the values: $x_{cm} = \frac{10(0) + 20(10)}{10 + 20}$ $x_{cm} = \frac{200}{30} = \frac{20}{3} \text{ m}$

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