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Two wires — one steel, one brass — are connected in series and support three equal masses (each of mass m) hanging from their lower ends, with the steel wire attached to the ceiling. The ratios L_steel/L_brass = a, r_steel/r_brass = b, and Y_steel/Y_brass = c. Find the ratio of the elongation of the steel wire to that of the brass wire.
- 2a²c/b
- 3a/(2b²c)
- 2ac/b²
- 3c/(2ab²)
Correct answer: 3a/(2b²c)
Solution
The steel wire (top) bears the weight of all three masses (3mg) while the brass wire (middle) bears two masses (2mg). Substituting the given ratios into delta_L = FL/(pi*r²*Y) gives the ratio (3/2)*(a)*(1/b²)*(1/c) = 3a/(2b²*c).
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