Please answer the practice it and exercise parts only. 4:20Access WebAssignEXAMPLE 10.4Rings and RodsGOAL Apply the equation of area expansion.PROBLEM (a) A circular copper ring at 20.0°C has a hole with an area of 9.980 cm?. What minimum temperature mustit have so that it can be slipped onto a steel metal rod having a cross-sectional area of 10.000 em?? (b) Suppose the ringand the rod are heated simultaneously. What minimum change in temperature of both will allow the ring to be slipped ontothe end of the rod? (Assume no significant change in the coefficients of linear expansion over this temperature range.)STRATEGY In part (a), finding the necessary temperature change is just a matter of substituting given values into theequation of area expansion. Remember tthat y - 20. Part (b) is a little harder because now the rod is also expanding. Ifthe ring is to slip onto the rod, however, the final cross-sectional areas of both ring and rod must be equal, Write thiscondition in mathematical tems, and solve for AT.SOLUTION(A) Find the temperature of the ring that will allow it to slip onto the rod.Write the equation of area expansion andAA- YAAT0.020 cm? = [34 x 10 ("cy'9.980 cm?)(AT)substitute known values, leaving AT as thesole unknownSolve for AT, then add this change to theAT = 59°Cinitial temperature to get the finalT= Tg +AT = 20.0°C + 59°C = 79'Ctemperature.(B) If both ring and rod are heated, find the minimum change in temperature that will allow the ring to be slipped onto therod.Set the final areas of the copper ring andAc + AAc = Ag + AASsteel rod equal to each other.خي ي + A ي الم +AA وAATي = - آش هY(YAc - YAg) AT Ag - AcSubstitute for each change in area, AA.Rearrange terms to get AT on one side only,factor it out, and solve.As - AcAT =10.000 cm2 - 9.980 cm?(34 x 10-6 "c(9.980 cm) - (22 x 10-6-c10.000 cm)AT - 170°CLEARN MOREREMARKS Warming and cooling strategies are sometimes useful for separating glass parts in a chemistry lab, such asthe glass stopper in a bottle of reagent.QUESTION If instead of heating the copper ring in part (a) the steel rod is cooled, would the magnitude of the requiredtemperature change be larger, smaller, or the same? Why? (Dont calculate it')O smaller, because steel has a larger coefficient of area expansionOlarger, because the steel rod has a smaler radiusO larger, because steel has a smaller coefficient of area expansionOsmaller, because the steel rod has a smaller radiusPRACTICE ITUse the worked example above to help you solve this problem.(a) A circular copper ring at 22.0°C has a hole with an area of 9.95 cm. What minimum temperature must it haveso that it can be slipped onto a steel metal rod having a cross-sectional area of 10.0 cm??5."C(b) Suppose the ring and the rod are heated simultaneously. What change in temperature of both will allow thering to be slipped onto the end of the rod? (Assume no significant change in the coefficients of linear expansionover this temperature range.)"CEXERCISEHINTS: GETTING STARTED I IM STUCK!A steel ring with a hole having area of 7.970 cm2 is to be placed on an aluminum rod with cross-sectional area of 8.000cm. Both rod and ring are initially at a temperature of 25.0°C. At what common temperature can the steel ring be slippedonto one end of the aluminum rod? (The coefficient of linear expansion of steel is11 x 10-6 ("c)-1, and the coefficient oflinear expansion aof aluminum is24 - 10 ("C)-1)"CII

Question

Please answer the practice it and exercise parts only. 4:20Access WebAssignEXAMPLE 10.4Rings and RodsGOAL Apply the equation of area expansion.PROBLEM (a) A circular copper ring at 20.0°C has a hole with an area of 9.980 cm?. What minimum temperature mustit have so that it can be slipped onto a steel metal rod having a cross-sectional area of 10.000 em?? (b) Suppose the ringand the rod are heated simultaneously. What minimum change in temperature of both will allow the ring to be slipped ontothe end of the rod? (Assume no significant change in the coefficients of linear expansion over this temperature range.)STRATEGY In part (a), finding the necessary temperature change is just a matter of substituting given values into theequation of area expansion. Remember tthat y - 20. Part (b) is a little harder because now the rod is also expanding. Ifthe ring is to slip onto the rod, however, the final cross-sectional areas of both ring and rod must be equal, Write thiscondition in mathematical tems, and solve for AT.SOLUTION(A) Find the temperature of the ring that will allow it to slip onto the rod.Write the equation of area expansion andAA- YAAT0.020 cm? = [34 x 10 (&#34;cy'9.980 cm?)(AT)substitute known values, leaving AT as thesole unknownSolve for AT, then add this change to theAT = 59°Cinitial temperature to get the finalT= Tg +AT = 20.0°C + 59°C = 79'Ctemperature.(B) If both ring and rod are heated, find the minimum change in temperature that will allow the ring to be slipped onto therod.Set the final areas of the copper ring andAc + AAc = Ag + AASsteel rod equal to each other.خي ي + A ي الم +AA وAATي = - آش هY(YAc - YAg) AT Ag - AcSubstitute for each change in area, AA.Rearrange terms to get AT on one side only,factor it out, and solve.As - AcAT =10.000 cm2 - 9.980 cm?(34 x 10-6 &#34;c(9.980 cm) - (22 x 10-6-c10.000 cm)AT - 170°CLEARN MOREREMARKS Warming and cooling strategies are sometimes useful for separating glass parts in a chemistry lab, such asthe glass stopper in a bottle of reagent.QUESTION If instead of heating the copper ring in part (a) the steel rod is cooled, would the magnitude of the requiredtemperature change be larger, smaller, or the same? Why? (Dont calculate it')O smaller, because steel has a larger coefficient of area expansionOlarger, because the steel rod has a smaler radiusO larger, because steel has a smaller coefficient of area expansionOsmaller, because the steel rod has a smaller radiusPRACTICE ITUse the worked example above to help you solve this problem.(a) A circular copper ring at 22.0°C has a hole with an area of 9.95 cm. What minimum temperature must it haveso that it can be slipped onto a steel metal rod having a cross-sectional area of 10.0 cm??5.&#34;C(b) Suppose the ring and the rod are heated simultaneously. What change in temperature of both will allow thering to be slipped onto the end of the rod? (Assume no significant change in the coefficients of linear expansionover this temperature range.)&#34;CEXERCISEHINTS: GETTING STARTED I IM STUCK!A steel ring with a hole having area of 7.970 cm2 is to be placed on an aluminum rod with cross-sectional area of 8.000cm. Both rod and ring are initially at a temperature of 25.0°C. At what common temperature can the steel ring be slippedonto one end of the aluminum rod? (The coefficient of linear expansion of steel is11 x 10-6 (&#34;c)-1, and the coefficient oflinear expansion aof aluminum is24 - 10 (&#34;C)-1)&#34;CII

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Since this question has multiple parts, i am just solving the last part. Please post another part…