Question

Two circular discs and $B$ are of equal masses and thickness but made of metals with densities and $d _ { B }$ $\left( d _ { A } > d _ { B } \right)$. If their moments of inertia about an axis passing through centres and normal to the circular faces be $I _ { A }$ and $I _ { B }$ , then

• A. $I _ { A } = I _ { B }$
• B. $I _ { A } > I _ { B }$
• C. $I _ { A } < I _ { B }$
• D. $I _ { A } > = < I _ { B }$

Answer 1

Answer: (C)

$I _ { A } < I _ { B }$

Answer 2

Moment of inertia of circular disc about an axis passing through centre and normal to the circular face

$I = \frac { 1 } { 2 } M R ^ { 2 }$ $= \frac { 1 } { 2 } M \left( \frac { M } { \pi t \rho } \right)$[As $M = V \rho$ $= \pi R ^ { 2 } t \rho$

∴ $R ^ { 2 } = \frac { M } { \pi t \rho }$]

⇒ $I = \frac { M ^ { 2 } } { 2 \pi t \rho }$ or $I \propto \frac { 1 } { \rho }$ If mass and thickness are constant.

So, in the problem $\frac { I _ { A } } { I _ { B } } = \frac { d _ { B } } { d _ { A } }$ $\therefore I _ { A } < I _ { B }$[As $d _ { A } > d _ { B }$]