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A rod of metal - 1 of length 50.0 cm elongates by 0.10 cm when it is heated from 0^(@)C to 100^(@)C. Another rod of metal - 2 of length 80.0. cm elongates by 0.08 cm for the same rise in temperature. A third rod of length 50.0 cm, made by welding pieces of rod 1 rad and 2 rad placed end to end, elongates by 0.03 cm when it is heated from 0^(@)C to 50^(@)C. Then what is the length of metal - 1 in the third rod at 0^(@)C? |
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Answer» SOLUTION :`"For META - 1,For metal - 2,"` `alpha_(1)=(Deltal)/(L Deltat)=(0.10)/(50.0xx100)""alpha_(2)=(Deltal)/(lDeltat)=(0.08)/(80.0xx100)` `alpha_(1)=2.0xx10^(-5)//^(@)C""alpha_(2)=1.0xx10^(-5)//^(@)C` LET the lengths of metal - 1 and metal - 2 in the third rod at `0^(@)C` be `l_(1) and l_(2)`, respectively. Then `""l_(1)+l_(2)=50.0` When this rod is heated to `50^(@)C`, then `(l._(1)=l_(1)(l+alpha_(1)50), l._(2)=l_(2)(l+alpha_(2)50)` `and""l._(1)+l._(2)=l_(1)+l_(2)+(l_(1)alpha_(1)+l_(2)alpha_(2))50` `50.03=50.0+(2l_(1)+l_(2))xx50xx10^(-5)` `2l_(1)+l_(2)=60"...(2)"` `"SOLVING (1) and (2),"l_(1)=10.0cm, l_(2)=40.0cm` |
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