A symmetrical lamina of mass M consists of a square shape with a semicircular section over each of the edge of the square as shown. The moment of inertia of the lamina about an axis through the centre of mass and perpendicular to the plane is 1.6Ma2. The moment of inertia about the tangent AB in the plane of the lamina is:

Question

A symmetrical lamina of mass M consists of a square shape with a semicircular section over each of the edge of the square as shown. The moment of inertia of the lamina about an axis through the centre of mass and perpendicular to the plane is 1.6Ma2. The moment of inertia about the tangent AB in the plane of the lamina is: <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-gfs3bufg1fjm4vny.jpg" /> (A) 4.8Ma2 (B) 2.6Ma2 (C) 1.8Ma2 (D) 7.2Ma2

Tardigrade Admin · Accepted Answer

Moment of inertia about the line AB is given by, IA B=Ix+M(2 a., where I X = moment of inertia of lamina with respect to X - axis. Therefore, IAB=((1 . 6 M a2/2))+4Ma2 ⇒ IAB=4.8Ma2

$4.8 M a^{2}$

$2.6 M a^{2}$

$1.8 M a^{2}$

$7.2 M a^{2}$

Moment of inertia about the line $A B$ is given by,
$I_{A B} = I_{x} + M (2 a$ , where $I_{X} =$ moment of inertia of lamina with respect to $X -$ axis.
Therefore,
$I_{A B} = (\frac{1.6 M a ^{2}}{2}) + 4 M a^{2}$
$\Rightarrow I_{A B} = 4.8 M a^{2}$

4.8Ma2

2.6Ma2

1.8Ma2

7.2Ma2

Moment of inertia about the line AB is given by, IAB​=Ix​+M(2a, where IX​= moment of inertia of lamina with respect to X− axis. Therefore, IAB​=(21.6Ma2​)+4Ma2 ⇒IAB​=4.8Ma2

$4.8 M a^{2}$

$2.6 M a^{2}$

$1.8 M a^{2}$

$7.2 M a^{2}$

Moment of inertia about the line $A B$ is given by,
$I_{A B} = I_{x} + M (2 a$ , where $I_{X} =$ moment of inertia of lamina with respect to $X -$ axis.
Therefore,
$I_{A B} = (\frac{1.6 M a ^{2}}{2}) + 4 M a^{2}$
$\Rightarrow I_{A B} = 4.8 M a^{2}$