When the load on a wire is slowly increased from `3kgwt` to `5 kg wt`, the elongation increases from `0.61` to `1.02 mm`. The work done during the extension of wire is

Options

A. `16 xx10^(-3)J`
B. `8xx10^(-2)J`
C. `20xx10^(-2)J`
D. `11 xx 10^(-3)J`

Answer

A

Explanation

Work done in the stretching the wire through 0.61 mm under the load of 3 kg
`W_(1) =(1)/(2)` stretching force `xx` extension `=(1)/(2) xx3xx9.8 xx xx 0.61 xx 10^(-3)= = 8.967 xx 10^(-3)J`.
Work done stretching the wire through 1.02 mm.
`W_(2) = (1)/(2) xx59.8 xx 1.02 xx 10^(-3) = 24.99 xx 10^(-3)J`.
Hence the work done is streching the wire form 0.61 mm to 1.02 m
`DeltaW = W_(2) -W_(1) =(24.99-8.961) xx 10^(-3) ~= 16 xx 10^(-3)J`.