Chapter 7 - Circular Motion and Gravitation - Learning Path Questions and Exercises - Exercises - Page 261: 16

a). $301.4rad$ b). $90.43m$

a). Since diameter of the wire $=0.5cm$, No of turns$=\frac{24}{0.5}=48$. So, radians covered $=48\times2pi=301.4rad$ b). Perimeter$=2pi\times r=188.4cm$ Length of the wire $=48\times 188.4=9043cm=90.43m$