Sankalp NEET Full Test-1 Question-29 Solution

Question: 29. The moment of inertia of a thin rod about an axis passing through its mid point and perpendicular to the rod is 2400 \mathrm{~g} \mathrm{~cm}^{2}. The length of the 400 \mathrm{~g} \operatorname{rod} is nearly :

(1) 8.5 cm

(2) 17.5 cm

(3) 20.7 cm

(4) 72.0 cm

Answer: Option (1)

Explanation:

The moment of inertia of a thin uniform rod about an axis passing through its midpoint

and perpendicular to its length is given by

I = \frac{1}{12} M L^{2}

Given moment of inertia I = 2400 \, \mathrm{g\,cm^{2}}.

Mass of the rod M = 400 \, \mathrm{g}.

Substituting the given values,

2400 = \frac{1}{12} \times 400 \times L^{2}

Simplifying,

2400 = \frac{400}{12} L^{2} 2400 = \frac{100}{3} L^{2} L^{2} = \frac{2400 \times 3}{100} L^{2} = 72 L = \sqrt{72} L \approx 8.49 \, \mathrm{cm}

Thus, the length of the rod is nearly 8.5 cm.

Hence, the correct answer is Option (1).

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