Four wires of the same material are stretched by the same load. Which one of them will elongate most if their dimensions are as follows

Options

A. L = 100 cm , r = 1mm
B. L = 200 cm , r = 3 mm
C. L = 300 , r = 3 mm
D. L = 400, r = 4mm

Answer

A

Explanation

`DeltaL = (FL)/(AY)`
Because , wires of the same material are stretched by the same load . So , F and Y will be constant.
`therefore Delta L prop (L)/(pi r^(2))`
`therefore DeltaL_(1) = (100)/(pi xx (1 xx 10^(-3))^(2))`
`= (100)/(pi xx 10^(-6)) = (100)/(pi) xx 10^(-6)`
` therefore " " DeltaL_(2) = (200)/(pi xx (3 xx 10^(-3))^(2))`
`= (200)/(pi xx 9 xx 10^-6) = (22.2)/(pi) xx 10^(6)`
`therefore DeltaL_(3) = (300)/(pi xx (3 xx 10^(-3))^(2))`
`= (300)/(pi xx 9 xx 10^(-6)) = (33.3)/(pi) xx 10^(6)`
`Delta L_(4) = (400)/(pi xx (4 xx 10^(-3))^(2))`
`= (400)/(pi xx 16 xx 10^(-6)) = (25)/(pi) xx 10^(6)`
We can see that , L = 100cm and r = 1 mm will elongate most.