Ring just slip over rod when area of hole is equals to area of cross-section of rod. ⇒Aring at T2 temperature =Arod at T2 temperature ⇒[A0(1+β∆T)]ring =[A0(1+β∆T)]rod Here, A0( ring )=9.98cm2 A0(rod)=10cm2 βring =2×17×10−6C−1(∵β=2α) βrod =2×11×10−6C−1 So, we have from Eq. (i),
