(b) $\frac{I_{centre}}{I_{edge}} = \frac{\left( r^{2} + h^{2} \right)^{3/2}}{h^{3}} = \frac{\left( 1 + \frac{1}{4} \right)^{3/2}}{1^{3}} = \left( \frac{5}{4} \right)^{3/2}$
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.