From a circular of mass 'M' and radius 'R' an arc corresponding to a sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is 'K' times '

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
From a circular of mass 'M' and radius 'R' an arc corresponding to a $90 °$ sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is 'K' times ' $M R^{2}$ '. Then the value of 'K' is [2021]

1 $\frac{1}{8}$

2 $\frac{3}{4}$

3 $\frac{7}{8}$

4 $\frac{1}{4}$

Ans.
(2)

Here: M is the mass and R is the radius of the ring.

Moment of inertia of a ring, $I = M R^{2}$

If $90 °$ sector is removed, the moment of inertia of the remaining part of the ring is

$I = M R^{2} - \frac{M R^{2}}{4} \Rightarrow K M R^{2} = \frac{3}{4} M R^{2}$

Here $K = \frac{3}{4}$

Want Us to Email you About Special Offers & Updates?