A metal rod A of 50 cm length expands by 0.10 cm when its temperature is raised from 0^(0) C " to " 100^(0) C . Another rod B of a different metal of length 60 cm expands by 0.06 emf or the same rise in temperature. A third rod C of 50 cm length is made up of pieces of rods A and B placed end to end expands by 0.03 cm on heating from 0^(0)C " to" 50^(0)C. Find the length of each portion of composite rod C.

A metal rod A of 50 cm length expands by 0.10 cm when its temperature

1.	A metal rod A of 50 cm length expands by 0.10 cm when its temperature is raised from 0^(0) C " to " 100^(0) C . Another rod B of a different metal of length 60 cm expands by 0.06 emf or the same rise in temperature. A third rod C of 50 cm length is made up of pieces of rods A and B placed end to end expands by 0.03 cm on heating from 0^(0)C " to" 50^(0)C. Find the length of each portion of composite rod C.
Answer» Solution :From the data for rod A, we have `Delta` L = `alpha_(A) L Delta T ` or `alpha_(A) = (Delta L)/(L Delta T) = (0.10)/(50 xx 100) = 2 xx 10^(-5)""^(@)C^(-1)` For rod B, we have `Delta L = a_(B) L Delta T ` If rod C is made of segments of rod A and B of lengths `l_(1) and l_(2)` RESPECTIVELY then we have at `0 " "^(@)C` . `l_(1) + l_(2) = 50` cm At T = `50^(@)C "" l_(1). + l_(2). = 50.03` cm Thus `alpha_(A) l_(1) Delta T + alpha_(B) l_(2) Delta T = 0.03 ` cm or `2 xx 10^(-5) xx l_(1) xx 50 + 10^(-5) xx l_(2) xx 50 = 0.03 cm` or `2l_(1) + l_(2) = (0.03)/(50) xx 10^(5) = 60 cm "" `..... (b) SOLVING (a) and (b) we get `l_(1) = 10 cm and l_(2)` = 40 cm