Moment of inertia of uniform triangular plate about axis passing through sides `AB,AC,BC` are `I_p,I_B` and `I_H` respectively and about an axis perpendicular to the plane and passing through point `C` is `I_C`. Then :
.

Options

A. `I_C gt I_p gt I_B gt I_H`
B. `I_H gt I_B gt I_C gt I_p`
C. `I_p gt I_H gt I_B gt I_C`
D. none of these

Answer

A

Explanation

(a) Moment of inertia is more when mass is farther from the axis. In case of axis `BC`, mass distribution is closet to it and in case of axis `AB` mass dritribution is farthest. Hence
`I_(BC) lt I_(AC) lt I_(AB) rArr I_p gt I_B gt I_H`
`I_C = I_(Cm) + my^2 = I'_B - ms^2 + my^2`
Hence `I'_B` is moment of inertia of the plate about an axis perpendicular to it and passing through `B`
`rArr I_C = I'_B+ m(y^2 - x^2) = I_p + I_B + m(y^2 - x^2)`
It means `I_C gt I_p + I_B` also `I_C gt I_p`
:. `I_C gt I_p gt I_B gt I_H`.
.