Question

Two copper circular discs are same thickness. The diameter of A is twice that of B. The moment of inertia of A as compared to that of B is​

Answer 1

• HOPE IT WILL HELP YOU

THANK YOU......

Answer 2

Given :

Two copper circular discs of same thickness.

Diameter of A is twice that of B.

Therefore , $r_a=2r_b$

To Find :

The moment of inertia of A as compared to that of B .

Solution :

We know ,  moment of inertia of a circular disc is given by :

$I=\dfrac{Mr^2}{2}$

Now , moment of inertia of of disc B is :

$I_b=\dfrac{Mr_b^2}{2}$

Also , moment of inertia of of disc A is :

$I_a=\dfrac{Mr_a^2}{2}\\\\I_a=2Mr_b^2$

Therefore , $I_a=4I_b$ .

So , the moment of inertia of A as compared to that of B is $I_a=4I_b$ .

Learn More :

Moment of inertia

https://brainly.in/question/6384344