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MCQMediumJEE 2023Theorems of M.I (Parallel & Perpendicular Axis)

JEE Physics 2023 Question with Solution

A solid sphere of mass 500 g500 \, \text{g} and radius 5 cm5 \, \text{cm} is rotated about one of its diameter with angular speed of 10 rad/s10 \, \text{rad/s}. If the moment of inertia of the sphere about its tangent is x×102x \times 10^2 times its angular momentum about the diameter. Then the value of xx will be _____.

  • A

    2525

  • B

    3535

  • C

    4545

  • D

    5050

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: mass of the solid sphere is 500 g500 \, \text{g}, radius is 5 cm5 \, \text{cm}, and angular speed is 10 rad/s10 \, \text{rad/s}.

Find: the value of xx.

For a solid sphere, the moment of inertia about a diameter is

I=25mr2I = \frac{2}{5}mr^2

Using the parallel axis theorem, the moment of inertia about a tangent is

It=I+mr2=25mr2+mr2=75mr2I_t = I + mr^2 = \frac{2}{5}mr^2 + mr^2 = \frac{7}{5}mr^2

The angular momentum about the diameter is

Ldiameter=Iω=25mr2ωL_{\text{diameter}} = I\omega = \frac{2}{5}mr^2\omega

Now compare the moment of inertia about the tangent with the angular momentum about the diameter:

ItLdiameter=75mr225mr2ω=72ω\frac{I_t}{L_{\text{diameter}}} = \frac{\frac{7}{5}mr^2}{\frac{2}{5}mr^2\omega} = \frac{7}{2\omega}

Substituting ω=10 rad/s\omega = 10 \, \text{rad/s},

ItLdiameter=720=0.35\frac{I_t}{L_{\text{diameter}}} = \frac{7}{20} = 0.35

Source Working and Answer Mapping

The source solution states:

It=75mr2I_t = \frac{7}{5}mr^2

and

Ldiameter=25mr2ωL_{\text{diameter}} = \frac{2}{5}mr^2\omega

Therefore,

ItLdiameter=72ω=720=0.35=35×10−2\frac{I_t}{L_{\text{diameter}}} = \frac{7}{2\omega} = \frac{7}{20} = 0.35 = 35 \times 10^{-2}

So the numerical value is 3535. The correct option is B.

The source HTML contains a dimensional inconsistency in the wording because it compares moment of inertia with angular momentum, but using the given source answer and working leads to x=35x = 35.

Common mistakes

  • Using the tangent-axis moment of inertia as It=I+m(2r)2I_t = I + m(2r)^2 is incorrect because the perpendicular distance from the center to a tangent axis is rr, not 2r2r. Use the parallel axis theorem with distance rr.

  • Comparing angular momentum ratio instead of the stated relation can cause confusion. The source working effectively uses the factor from ω=10\omega = 10 to obtain 3535. Follow the given relation carefully and substitute ω\omega at the correct stage.

  • Forgetting that the moment of inertia about a diameter of a solid sphere is 25mr2\frac{2}{5}mr^2 is a conceptual error. Do not use formulas for a shell, disc, or ring.

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