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Grade: 12

                        

errors chapter a given wire is stretched with force F1 and final length is x. when stretching force is F2 final length is y. the original length of the wire is a. f2y-f1x/f2-f1 b. f2x-f1y/f2-f1 c. f1y-f2x/f2-f1 d. f2y-f1x/f1-f2 correct answer is b.

3 years ago

Answers : (1)

Harshit Goyal
13 Points
							Let when constant of the wore be `k` and original length be `l`When force F1 is applied extension = (x-l) as final length was x and original length was l Similarly on application of force F2 extension was (y-l)We know than when a force F is applied on a spring an equal amount of restoring force -(kx) is applied by it (Newton`s Third law ) and hence-k(x-l) = F1 and  -k(y-l) =F2-k= F1/(x-l) and -k = F2/(y-l)Since wire in both cases was same k would be same in both cases and hence F1/(x-l) = F2/(y-l)(x-l)/F1 = (y-l)/F2F1y -F1l = F2x - F2ll(F2-F1) = F2x -F1yHence l = (F2x - F1y)/(F2-F1)Correct answer = Option B
						
3 years ago
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