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NEET PHYSICSRotational motionMedium

Question

Three identical spherical shells, each of mass mm and radius rr are placed as shown in the figure. Consider an axis XXXX', which is touching two shells and passing through the diameter of the third shell. The moment of inertia of the system consisting of these three spherical shells about the XXXX' axis is:

A

115mr2\frac{11}{5}mr^2

B

3mr23mr^2

C

165mr2\frac{16}{5}mr^2

D

4mr24mr^2

Step-by-Step Solution

The total moment of inertia of the system is the sum of the moments of inertia of the three individual spherical shells about the given axis XXXX'. The axis XXXX' passes through the diameter of one shell and is tangent to the other two shells. Moment of inertia of a thin spherical shell about its diameter is Idiameter=23mr2I_{diameter} = \frac{2}{3}mr^2. Moment of inertia of a spherical shell about a tangent can be found using the parallel axis theorem: Itangent=Idiameter+mr2=23mr2+mr2=53mr2I_{tangent} = I_{diameter} + mr^2 = \frac{2}{3}mr^2 + mr^2 = \frac{5}{3}mr^2. Therefore, the total moment of inertia of the system is: Itotal=Idiameter+2×ItangentI_{total} = I_{diameter} + 2 \times I_{tangent} Itotal=23mr2+2×(53mr2)=23mr2+103mr2=123mr2=4mr2I_{total} = \frac{2}{3}mr^2 + 2 \times \left(\frac{5}{3}mr^2\right) = \frac{2}{3}mr^2 + \frac{10}{3}mr^2 = \frac{12}{3}mr^2 = 4mr^2

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from Rotational motion. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSRotational motionidenticalsphericalshellsradiusplaced

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