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 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 $L$:

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

Answer 2

Step 1: Recall the Formula for the Moment of Inertia 

The moment of inertia of a thin rod about its midpoint is:

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

• I = Moment of inertia
• M = Mass of the rod
• L = Length of the rod

Step 2: Rearrange to Solve for L

Rearranging the equation:

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

Step 3: Substitute the Given Values

Given:

• I = 2400 g·cm²
• M = 400 g

Substituting the values:

$$ L^2 = \frac{12 \times 2400}{400} = 72 \text{ cm}^2 $$

Step 4: Find L

Taking the square root:

$$ L = \sqrt{72} = 8.5 \text{ cm} $$

The length of the rod is 8.5 cm.