Question

The moment of inertia of a thin rod about an axis passing through its mid point and perpendicular to the rod is 2400 g cm². The length of the 400 g rod is nearly :

• A. 8.5 cm
• B. 17.5 cm
• C. 20.7 cm
• D. 72.0 cm

Answer 1

The Correct Option is A

The moment of inertia \( I \) of a thin rod about an axis passing through its midpoint and perpendicular to the rod is given by the formula:

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

where \( M \) is the mass of the rod, and \( L \) is the length of the rod.

Given:

• \( I = 2400 \) g cm\(^2\)
• \( M = 400 \) g

Substituting the given values into the formula:

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

Multiplying both sides by 12 to remove the fraction:

$$ 2400 \times 12 = 400 \times L^2 $$

$$ 28800 = 400 \times L^2 $$

Dividing both sides by 400 to solve for \( L^2 \):

$$ L^2 = \frac{28800}{400} $$

$$ L^2 = 72 $$

Taking the square root of both sides to solve for \( L \):

$$ L = \sqrt{72} $$

$$ L \approx 8.5 \text{ cm} $$

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

Answer 2

The moment of inertia of a rod is given by:

$I = \frac{1}{12}ML^2$.

Here, $I = 2400 \, g cm^2 = 2400 \times 10^{-7} \, kg m^2$, $M = 400 \, g = 0.4 \, kg$.

Rearrange to find $L$:

$L = \sqrt{\frac{12I}{M}} = \sqrt{\frac{12 \times 2400 \times 10^{-7}}{0.4}} = 8.5 \, cm$.