A circular disc A of radius r is made from an iron plate of thickness t and another circular disc B of radius 4 r and thickness t / 4. The relation between moments of inertia IA and IB about the same axis is:

Question

A circular disc A of radius r is made from an iron plate of thickness t and another circular disc B of radius 4 r and thickness t / 4. The relation between moments of inertia IA and IB about the same axis is: (A) IA > IB (B) IA = IB (C) IA < IB (D) depends on the actual values of t and r

Tardigrade Admin · Accepted Answer

mA= volume × density =π x2 t ρ IA=(1/2) mA r2=(1/2) π r4 ρ t Similarly IB=(1/2) π(4 r)4 ρ((t/4))=64 IA Clearly IB>IA

Q. A circular disc $A$ of radius $r$ is made from an iron plate of thickness $t$ and another circular disc $B$ of radius $4 r$ and thickness $t /4$ . The relation between moments of inertia $I_{A}$ and $I_{B}$ about the same axis is:

$I_{A} > I_{B}$

$I_{A} = I_{B}$

$I_{A} < I_{B}$

depends on the actual values of $t$ and $r$

$m_{A} =$ volume $\times$ density $= π x^{2} tρ$
$I_{A} = \frac{1}{2} m_{A} r^{2} = \frac{1}{2} π r^{4} ρt$
Similarly $I_{B} = \frac{1}{2} π (4 r)^{4} ρ (\frac{t}{4}) = 64 I_{A}$
Clearly $I_{B} > I_{A}$

Q. A circular disc A of radius r is made from an iron plate of thickness t and another circular disc B of radius 4r and thickness t/4. The relation between moments of inertia IA​ and IB​ about the same axis is:

IA​>IB​

IA​=IB​

IA​<IB​

depends on the actual values of t and r