(C) The radius of all bodies is $2R$ and mass is $M$.$I_{1} = \frac{2}{5} M (2R)^{2} = \frac{8}{5} MR^{2}$$I_{2} = \frac{1}{2} M (2R)^{2} = 2 MR^{2}$$

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(C) The radius of all bodies is $2R$ and mass is $M$.$I_{1} = \frac{2}{5} M (2R)^{2} = \frac{8}{5} MR^{2}$$I_{2} = \frac{1}{2} M (2R)^{2} = 2 MR^{2}$$I_{3} = \frac{M (2R)^{2}}{4} = MR^{2}$$I_{4} = \frac{M (2R)^{2}}{2} = 2 MR^{2}$Given equation: $2(I_{2} + I_{3}) + I_{4} = x I_{1}$Substitute the values: $2(2 MR^{2} + MR^{2}) + 2 MR^{2} = x (\frac{8}{5} MR^{2})$$2(3 MR^{2}) + 2 MR^{2} = x (\frac{8}{5} MR^{2})$$6 MR^{2} + 2 MR^{2} = x (\frac{8}{5} MR^{2})$$8 MR^{2} = x (\frac{8}{5} MR^{2})$$x = 5$

Class 11 Physics System of Particles and Rotational Motion Moment of Inertia and Radius of gyration

Medium

Moment of Inertia $(M.I.)$ of four bodies having same mass $M$ and radius $2R$ are as follows:
$I_{1} =$ $M.I.$ of solid sphere about its diameter
$I_{2} =$ $M.I.$ of solid cylinder about its axis
$I_{3} =$ $M.I.$ of solid circular disc about its diameter
$I_{4} =$ $M.I.$ of thin circular ring about its diameter
If $2(I_{2} + I_{3}) + I_{4} = x I_{1}$,then the value of $x$ will be...

JEE MAIN 2022 All JEE MAIN

A
$57$
B
$55$
C
$5$
D
$9$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Physics Questions

Chapter

System of Particles and Rotational Motion

Subtopic

Moment of Inertia and Radius of gyration

Previous Year

JEE MAIN 2022 Paper All JEE MAIN years →

When ice at the poles of the Earth melts,its moment of inertia will ........ .

Medium

View Solution

Four spheres of diameter $2a$ and mass $M$ are placed with their centers on the four corners of a square of side $b$. The moment of inertia of the system about an axis along one of the sides of the square is

Difficult

View Solution

Two rings of same mass $m$ and radius $R$ are placed with their planes perpendicular to each other and centres at a common point. The radius of gyration of the system about an axis passing through the centre and perpendicular to the plane of one ring is ............

Medium

View Solution

Four point masses (each of mass $m$) are arranged in the $X-Y$ plane at coordinates $(a, 0)$,$(0, a)$,$(-a, 0)$,and $(0, 2a)$. What is the moment of inertia of this system about the $Y$-axis?

Difficult

View Solution

The ratio of the radius of gyration of a thin uniform disc about an axis passing through its centre and normal to its plane to the radius of gyration of the disc about its diameter is

MediumNEET 2022

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

Similar Questions

When ice at the poles of the Earth melts,its moment of inertia will ........ .

Four spheres of diameter $2a$ and mass $M$ are placed with their centers on the four corners of a square of side $b$. The moment of inertia of the system about an axis along one of the sides of the square is

Two rings of same mass $m$ and radius $R$ are placed with their planes perpendicular to each other and centres at a common point. The radius of gyration of the system about an axis passing through the centre and perpendicular to the plane of one ring is ............

Four point masses (each of mass $m$) are arranged in the $X-Y$ plane at coordinates $(a, 0)$,$(0, a)$,$(-a, 0)$,and $(0, 2a)$. What is the moment of inertia of this system about the $Y$-axis?

The ratio of the radius of gyration of a thin uniform disc about an axis passing through its centre and normal to its plane to the radius of gyration of the disc about its diameter is

Vedclass Test Series

Exam Paper Generator

Online Exam Module