InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
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 |
|
Answer» Here, `F_(1)=3kg f=3 xx 9.8 N, Delta l_(1) = 0.61 xx 10^(-3)m` `F_(2)=5kg f=5 xx 9.8 N, Delta l_(2) = 1.02 xx 10^(-3)m` Work done, `W=1.2 F_(2) Delta l_(2) - 1/2 F_(1)Delta l_(1)` `=1/2 [(5 xx 9.8) xx (1.02 xx 10^(-3))` `-(3 xx 9.8) xx (0.61 xx 10^(-3))]` `=16.023 xx 10^(-3)J`. |
|