1.

In above question, the work done in the two wire is

Answer»

0.5 J, 0.03 J
0.25 J, 0 J
0.03 J, 0.25 J
0J, 0j

Solution :As work DONE, `W = (F^(2)I)/(2((pi D^(2))/(4))y)`
Where, Y,I and F are constants.
`implies W prop (1)/(D^(2))` or `(W_(1))/(W_(2)) = (D_(2)^(2))/(D_(1)^(2)) = 16`
Now, `W_(1) = (1)/(2) xx 10^(3 xx 1 xx 10^(-3) = 0.5 J`
`:. W_(2) = (1)/(2) xx 10^(3) xx (10^(-3))/(16) = (1)/(32) = 0.03125`
`(w_(1))/(W_(2)) = (0.5)/(0.3125) = 16`


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