Question
Download Solution PDFLet the moment of inertia of a hollow cylinder of length 30 cm (inner radius 10 cm and outer radius 20 cm) about its axis be I. The radius of a thin cylinder of the same mass such that its moment of inertia about its axis is also I, is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
Moment of inertia of hollow cylinder about its axis is given as:
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
Where,
R1 = Inner radius and R2 = Outer radius
Moment of inertia of thin hollow cylinder of radius ‘R’ about its axis is given as:
I2 = MR2
Calculation:
From question, we know that,
⇒ I1 = I2
Both cylinders have same mass (M)
⇒ R2 = 250 = 15.8
∴ R ≈ 16 cm
Last updated on May 23, 2025
-> JEE Main 2025 results for Paper-2 (B.Arch./ B.Planning) were made public on May 23, 2025.
-> Keep a printout of JEE Main Application Form 2025 handy for future use to check the result and document verification for admission.
-> JEE Main is a national-level engineering entrance examination conducted for 10+2 students seeking courses B.Tech, B.E, and B. Arch/B. Planning courses.
-> JEE Mains marks are used to get into IITs, NITs, CFTIs, and other engineering institutions.
-> All the candidates can check the JEE Main Previous Year Question Papers, to score well in the JEE Main Exam 2025.