**Text:**The drawing shows a solid cylindrical rod made from a center cylinder of lead and an outer concentric jacket of copper. Except for its ends, the rod is insulated (not shown), so that the loss of heat from the curved surface is negligible. When a temperature difference is maintained between its ends, this rod conducts 1/7 the amount of heat that it would conduct if it were solid copper. Determine the ratio of the radii \( r_1/r_2 \).**Diagram Explanation:**The diagram displays a cross-sectional view of a cylindrical rod. The inner portion of the rod, labeled "Lead," is solid and cylindrical in shape, denoted by the radius \( r_1 \). Surrounding this lead core is an outer layer labeled "Copper," which forms a concentric jacket with the radius \( r_2 \). The copper jacket encloses the lead core completely. The image is used to aid in understanding how the differences in material and geometry affect thermal conduction properties.

Given that: Let the temperature at the both conducting end is maintained as T1 and T2 respectively as shown. Also let the length of the conductor is L.Formula used:Rate of heat transfer through a conductor of conductivity k, area of cross section A,length L and temperature difference ∆T is given by:Q˙=kA∆TL∆T=LkAQ˙∆T=RTQ˙Here RT is known as thermal resistance of the conductor, RT=LkA We know that if the voltage across each resistance (equal and unequal) in a ideal current circuit is same, then they are connected in parallel. Comparing to ideal current circuits, if temperature difference across each conductor in a system of heat conductors are same, then they are connected in parallel.Thermal resistance of lead:RL=LKLAL=LπKLr12Thermal resistance of Copper:RCu=LKCuACu=LπKCu(r22-r12)If the Net thermal resistance=RN, then1RN=1RL+1RCu1RN=πL[KLr12+KCu(r22-r12)] Net rate heat transfer in this case:Q1˙=(T1-T2)RNQ1˙=(T1-T2)πL[KLr12+KCu(r22-r12)] If the material was only cupper:R'Cu=LKCuA'Cu=LπKCur22Net rate heat transfer in this case:Q2˙=(T1-T2)R'CuQ2˙=(T1-T2)πLKCur22Q1˙=(T1-T2)πL[KLr12+KCu(r22-r12)] Q2˙=(T1-T2)πLKCur22As per the given condition:Q1˙=17Q2˙(T1-T2)πL[KLr12+KCu(r22-r12)] =17[(T1-T2)πLKCur22]KLr12+KCu(r22-r12)=KCur227Using standard values:KL=35 W/mKKCu=386 W/mK35r12+386(r22-r12)=386r22735r12+386(r22-r12)=55.14r22330.8r22=351r12r12r22=330.8351r1r2=0.97 (ans)(Note that: the ratio found can vary a little if as different sites claim different standard values of thermal conductivities)

Answered: The drawing shows a solid cylindrical…