Two spheres each of mass M and radius `(R )/(2)` are connected with a massless rod of length 2R as shown in the figure.

The moment fo inertia of the system about an axis passing through the centre of one of the spheres and perpendicular to the rod about yy' axis will be :

Options

A. `21/4MR^(2)`
B. `2/5MR^(2)`
C. `5/2MR^(2)`
D. `5/21MR^(2)`

Answer

A

Explanation

We have to calculate moment oif inertia of the system about yy. axis
Suppose `I_("net")to`Moment of inertia of system aboy yy. axis
`I_(p)to` Moment of inertia sphere P about yy. axis
`I_(Q)to` Moment of inertia sphere Q about yy. axis
`I_(R)to` Moment of rod about yy. axis.
w.k.t
Moment of inertia for a solid sphere of rasdius r and mass m about central axis
`I_(r)=2/5mr^(2)`
[`rto` radius , `mto` mass]
applying this fomula
`I_(P)=2/5m(R/2)^(2)+m(2R)^(2)`
[parallel axis theorem]
`I_(P)=41/10mR^(2)`.....i
Now
for sphere Q
`I_(Q)=2/5m(R/2)^(2)`
`I_(Q)=(mR^(2))/10`.........ii
Now for masses and
`I_(R)=0`........iii
`[:.m=0]`
So using eq i , ii and iii
`I_("net")=I_(P)+I_(Q)+I_(R)`
`:.I_("net")=41/10MR^(2)+(MR^(2))/10`
`I_("net")=21/5MR^(2)`
Option 1 is correct