NEET-XII-Physics
10: Rotational Mechanics
- #11A circular disc A of radius r is made from an iron plate of thickness t and another circular disc B of radius 4r is made from an iron plate of thickness t/4. The relation between the moments of inertia IA and IB is
(a) IA > IB
(b) IA = IB
(c) IA < IB
(d) depends on the actual values of t and r.digAnsr: cAns : (c) IA < IB
Moment of inertia of circular disc of radius r:
I = `` \frac{1}{2}m{r}^{2}``
Mass = Volume × Density
Volume of disc = `` \,\mathrm{\,\pi \,}{r}^{2}t``
Here, t is the thickness of the disc.
As density is same for both the rods, we have:
Moment of inertia, `` I\propto \,\mathrm{\,thickness\,}\times {\left(\,\mathrm{\,radius\,}\right)}^{4}``
`` \frac{{I}_{A}}{{I}_{B}}=\frac{t.{\left(r\right)}^{4}}{{\displaystyle \frac{t}{4}}{\left(4r\right)}^{4}}<1``
`` \Rightarrow \frac{{I}_{A}}{{I}_{B}}<1``
`` \Rightarrow {I}_{\,\mathrm{\,A\,}}<{I}_{\,\mathrm{\,B\,}}``
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