Momen of inertia (M.I) of four bodies, having same mass and radius, are reported as;I1=M.I. of thin circular disc about its diameter,I2=M.I of circular disc about an axis perpendicular to disc and going through the centre,I3=M.I of solid cylinder about its axis andI4=M.I of solid sphere about its diameter. Then

Book Online Demo

Check Your IQ

Try Test

Courses

Offline Centres

Momen of inertia (M.I) of four bodies, having same mass and radius, are reported as;

$I_{1} = M . I .$ of thin circular disc about its diameter,

$I_{2} = M . I$ of circular disc about an axis perpendicular to disc and going through the centre,

$I_{3} = M . I$ of solid cylinder about its axis and

$I_{4} = M . I$ of solid sphere about its diameter. Then

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$I_{1} = I_{2} = I_{3} > I_{4}$

$I_{1} = I_{2} = I_{3} < I_{4}$

$I_{1} + I_{3} < I_{2} + I_{4}$

$I_{1} + I_{2} = I_{3} + \frac{5}{2} I_{4}$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$I_{1} = \frac{M R^{2}}{4}$

$I_{2} = \frac{M R^{2}}{2}$

$I_{3} = \frac{M R^{2}}{2}$

$I_{4} = \frac{2}{5} M R^{2}$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE