JEE Mains · Physics · STD 11 - 6. system of particles and rotational motion

M and R be the mass and radius of a disc. A small disc of radius \(R / 3\) is removed from the bigger disc as shown in figure. The moment of inertia of remaining part of bigger disc about an axis AB passing through the centre O and perpendicular to the plane of disc is \(\frac{4}{x} M R^2\). The value of \(x\) is ________.

A 9
B 10
C 11
D 12

Verified Solution

Answer & Solution

Correct Answer

(A) 9

Step-by-step Solution

Detailed explanation

Without cavity \(\mathrm{I}_1=\frac{\mathrm{MR}^2}{2}\) \(\begin{aligned} \text { Mass of removed disc }= & \frac{\mathrm{M}}{\pi \mathrm{R}^2} \times\left(\frac{\mathrm{R}}{3}\right)^2 \pi \\ & =\left(\frac{\mathrm{M}}{9}\right) \end{aligned}\) M.I. of removed disc…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Match List \(I\) with List \(II\)

LIST\(-I\)	LIST\(-II\)
\((A)\) Surface tension	\((I)\) \(Kg m ^{-1} s ^{-1}\)
\((B)\) Pressure	\((II)\) \(Kg ms^{-1 }\)
\((C)\) Viscosity	\((III)\) \(Kg m ^{-1} s ^{-2}\)
\((D)\) Impulse	\((IV)\) \(Kg s ^{-2}\)

Choose the correct answer from the options given below :

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