Rotational Motion

JEE Main level, numerical answer, medium

12 of 280

Question 12 of 280

Suppose there is a uniform circular disc of mass M kgM \mathrm{~kg} and radius r mr \mathrm{~m} shown in figure. The shaded regions are cut out from the disc. The moment of inertia of the remainder about the axis AA of the disc is given by x256Mr2\frac{x}{256} M r^{2}. The value of xx is ____\_\_\_\_.

Show answer and solution

Answer: 109

The moment of inertia of the complete disc about axis A is
Idisc=12Mr2=128Mr2256I_{disc} = \frac{1}{2}Mr^2 = \frac{128Mr^2}{256}.
Each removed circular region has radius r/4r/4 and mass mi=M/(r2/16r2)=M/16m_i = M/(r^2/16r^2) = M/16.
For each removed disc at distance 3r/43r/4 from axis A, using the parallel axis theorem:
Iremoved=12mi(r/4)2+mi(3r/4)2=Mr2512+9Mr2256=19Mr2512I_{removed} = \frac{1}{2}m_i(r/4)^2 + m_i(3r/4)^2 = \frac{Mr^2}{512} + \frac{9Mr^2}{256} = \frac{19Mr^2}{512}.
For two removed discs:
I2removed=38Mr2512=19Mr2256I_{2removed} = \frac{38Mr^2}{512} = \frac{19Mr^2}{256}.
Therefore:
Iremainder=128Mr2256−19Mr2256=109Mr2256I_{remainder} = \frac{128Mr^2}{256} - \frac{19Mr^2}{256} = \frac{109Mr^2}{256}, so x=109x = 109.

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Rotational dynamics and rolling simulationDrag the sliders and watch the numbers in this question change.Revise Rotational Motion firstFormula sheet, solved examples and the traps students fall into.

More Physics practice

JEE Main Physics practice question on Rotational Motion (Moment of inertia of rigid bodies), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.