If linear density of a rod of length 3m varies as λ= 2 + x , then the position of the centre of gravity of the rod is:

If the linear density of a rod of length \( 3 \, \text{m} \) varies as \( \lambda = 2 + x \), what is the position of the center of gravity of the rod?

A) \( \frac{7}{3} \, \text{m} \)
B) \( \frac{12}{7} \, \text{m} \)
C) \( \frac{10}{7} \, \text{m} \)
D) \( \frac{9}{7} \, \text{m} \)

Answer: B) \( \frac{12}{7} \, \text{m} \)

The center of gravity is calculated using:

\[ M = \int_0^3 (2 + x) \, dx = \frac{21}{2}, \quad \bar{x} = \frac{1}{M} \int_0^3 x(2 + x) \, dx = \frac{12}{7} \, \text{m} \]

MCQ On Motion In Plane | Newton's Laws Of Motion | CBSE, IIT JEE, NEET, CET | Physics PYQ

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