Two circular discs A and B of equal masses and thicknesses but are made of metals with densities dA and dB(d A d B ). If their moments of inertia about an axis passing through the centre and normal to the circular faces be IA and IB, then

Question

Two circular discs A and B of equal masses and thicknesses but are made of metals with densities dA and dB(d A >d B ). If their moments of inertia about an axis passing through the centre and normal to the circular faces be IA and IB, then (A) IA = IB (B) IA > IB (C) IA < IB (D) IA ≥ IB

Tardigrade Admin · Accepted Answer

mass = volume × density ⇒ M=(π R2 t) d ldots (1) where t is thickness I=(M R2/2) ldots (2) From (1) and (2), we get: I=(M2/2 π d t) ⇒ I propto (1/d) ⇒ IA < IB

Q. Two circular discs $A$ and $B$ of equal masses and thicknesses but are made of metals with densities $d_{A}$ and $d_{B} (d_{A} > d_{B})$ . If their moments of inertia about an axis passing through the centre and normal to the circular faces be $I_{A}$ and $I_{B}$ , then

$I_{A} = I_{B}$

$I_{A} > I_{B}$

$I_{A} < I_{B}$

$I_{A} \geq I_{B}$

mass = volume $\times$ density
$\Rightarrow M = (π R^{2} t) d \dots$ (1)
where $t$ is thickness
$I = \frac{M R ^{2}}{2} \dots$ (2)
From (1) and (2), we get:
$I = \frac{M ^{2}}{2 π d t} \Rightarrow I \propto \frac{1}{d} \Rightarrow I_{A} < I_{B}$

Q. Two circular discs A and B of equal masses and thicknesses but are made of metals with densities dA​ and dB​(dA​>dB​). If their moments of inertia about an axis passing through the centre and normal to the circular faces be IA​ and IB​, then

IA​=IB​

IA​>IB​

IA​<IB​

IA​≥IB​