Moment of inertia of an equilateral triangular lamina ABC, about the axis passing through its centre O and perpendicular to its plane is Io as shown in the figure. A cavity DEF is cut out from the lamina, where D, E, F are the mid points of the sides. Moment of inertia of the remaining part of lamina about the same axis is :

Question

Moment of inertia of an equilateral triangular lamina ABC, about the axis passing through its centre O and perpendicular to its plane is Io as shown in the figure. A cavity DEF is cut out from the lamina, where D, E, F are the mid points of the sides. Moment of inertia of the remaining part of lamina about the same axis is : (A) (7/8) Io (B) (15/16) Io (C) (3Io/4)  (D) (3 1Io/32)

Tardigrade Admin · Accepted Answer

I0 = kml2 BC = l IDEF = k (m/4) ((l/2))2 = (k/16) ml2 IDEF = (I0/16) Iremain = I0 = (I0/16) = (15I0/16)

$\frac{7}{8} I_{o}$

$\frac{15}{16} I_{o}$

$\frac{3 I _{o}}{4}$

$\frac{31 I _{o}}{32}$

$I_{0} = km l^{2} BC = l$
$I_{D EF} = k \frac{m}{4} (\frac{l}{2})^{2}$
$= \frac{k}{16} m l^{2}$
$I_{D EF} = \frac{I _{0}}{16}$
$I_{re main} = I_{0} = \frac{I _{0}}{16}$
$= \frac{15 I _{0}}{16}$

87​Io​

1615​Io​

43Io​​

3231Io​​

I0​=kml2BC=l IDEF​=k4m​(2l​)2 =16k​ml2 IDEF​=16I0​​ Iremain​=I0​=16I0​​ =1615I0​​

$\frac{7}{8} I_{o}$

$\frac{15}{16} I_{o}$

$\frac{3 I _{o}}{4}$

$\frac{31 I _{o}}{32}$

$I_{0} = km l^{2} BC = l$
$I_{D EF} = k \frac{m}{4} (\frac{l}{2})^{2}$
$= \frac{k}{16} m l^{2}$
$I_{D EF} = \frac{I _{0}}{16}$
$I_{re main} = I_{0} = \frac{I _{0}}{16}$
$= \frac{15 I _{0}}{16}$