Physics Q22 – NEET-UG 2024

Home Physics Rotational MotionQuestion #440

The moment of inertia of a thin rod about an axis passing through its mid point and perpendicular to the rod is

2400 {g cm}^{2}

. The length of the

400 g

rod is nearly :

Clue 1 of 3: Rotational Dynamics

This problem involves the concept of moment of inertia, which describes an object's resistance to changes in its rotational motion.

🧠 Full Solution Path

● Video Walkthrough

Step 1: Recall the Moment of Inertia Formula✦ Active

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

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

Step 2: Substitute Given Values○ Expand

Given the moment of inertia $I = 2400 {g cm}^{2}$ and the mass of the rod $M = 400 g$ , substitute these values into the formula:

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

💡 Teacher's Secret Hint

Ensure units are consistent. Here, all units are in CGS (g, cm), so no conversion is needed.

Step 3: Solve for Length (

L

)○ Expand

Now, solve the equation for $L^{2}$ and then for $L$ :

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

2400 = \frac{100}{3} L^{2}

L^{2} = \frac{2400 \times 3}{100}

L^{2} = 24 \times 3

L^{2} = 72

L = \sqrt{72} \approx 8.485 cm

Comparing this value with the given options, $8.5 cm$ is the closest.

💡 Teacher's Secret Hint

Remember that $\sqrt{72}$ is between $\sqrt{64} = 8$ and $\sqrt{81} = 9$ . A quick estimation can help narrow down the options.

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Physics Question 22 – NEET-UG 2024

Clue 1 of 3: Rotational Dynamics

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