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NEET PHYSICSRotational motionEasy

Question

Two particles of mass 5 kg5 \text{ kg} and 10 kg10 \text{ kg} respectively are attached to the two ends of a rigid rod of length 1 m1 \text{ m} with negligible mass. The centre of mass of the system from the 5 kg5 \text{ kg} particle is nearly at a distance of:

A

50 cm50 \text{ cm}

B

67 cm67 \text{ cm}

C

80 cm80 \text{ cm}

D

33 cm33 \text{ cm}

Step-by-Step Solution

Let the particle of mass m1=5 kgm_1 = 5 \text{ kg} be placed at the origin, so its position coordinate is x1=0x_1 = 0. The particle of mass m2=10 kgm_2 = 10 \text{ kg} is at a distance of 1 m=100 cm1 \text{ m} = 100 \text{ cm}, so its position coordinate is x2=100 cmx_2 = 100 \text{ cm}. The position of the centre of mass from the 5 kg5 \text{ kg} particle is given by: Xcm=m1x1+m2x2m1+m2X_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} Xcm=5(0)+10(100)5+10X_{cm} = \frac{5(0) + 10(100)}{5 + 10} Xcm=100015=2003 cm66.67 cmX_{cm} = \frac{1000}{15} = \frac{200}{3} \text{ cm} \approx 66.67 \text{ cm} Therefore, the centre of mass of the system from the 5 kg5 \text{ kg} particle is nearly at a distance of 67 cm67 \text{ cm}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Rotational motion. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSRotational motionparticlesrespectivelyattachedlengthnegligible

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