Question

when the tension on a wire is 4N its length is l1. when the tension on the wire is 5N its length is l2. Find its natural length.

Answer 1

let x1 and x2 be the extension with 4N and 5N respectively.
then the equations would be
4=k*×1 and 5=k*×2
and let l be the natural length of the wire .
Hence,
x1+l=l1
and
x2+2=l2
put x1 and x2 values in tensions after dividing them respectively.
therefore,
4/5=x1/x2=>(l1-l)/(l2-l)
4l2 -4l=5l1-51
l=5l1-4l2
hence,would be the natural length.

Answer 2

Answer:let x1 and x2 be the extension with 4N and 5N respectively.

then the equations would be

4=k*×1 and 5=k*×2

and let l be the natural length of the wire .

Hence,

x1+l=l1

and

x2+2=l2

put x1 and x2 values in tensions after dividing them respectively.

therefore,

4/5=x1/x2=>(l1-l)/(l2-l)

4l2 -4l=5l1-51

l=5l1-4l2

hence,would be the natural length.