Given;
$Springmakesanangle={37}^{0}\phantom{\rule{0ex}{0ex}}Itsnaturallengthlandspringcons\mathrm{tan}tkisfixedontheground\phantom{\rule{0ex}{0ex}}andtheotherisfittedwithasmoothringofmassmwhichslides\phantom{\rule{0ex}{0ex}}onahorizontalrodfixed.\phantom{\rule{0ex}{0ex}}\mathit{D}\mathit{e}\mathit{t}\mathit{e}\mathit{r}\mathit{m}\mathit{i}\mathit{n}\mathit{e}\mathbf{}\mathit{t}\mathit{h}\mathit{e}\mathbf{}\mathit{s}\mathit{p}\mathit{e}\mathit{e}\mathit{d}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{t}\mathit{h}\mathit{e}\mathbf{}\mathit{r}\mathit{i}\mathit{n}\mathit{g}\mathbf{}\mathit{w}\mathit{h}\mathit{e}\mathit{n}\mathbf{}\mathit{t}\mathit{h}\mathit{e}\mathbf{}\mathit{s}\mathit{p}\mathit{r}\mathit{i}\mathit{n}\mathit{g}\mathbf{}\mathit{b}\mathit{e}\mathit{c}\mathit{o}\mathit{m}\mathit{e}\mathit{s}\mathbf{}\mathit{v}\mathit{e}\mathit{r}\mathit{t}\mathit{i}\mathit{c}\mathit{a}\mathit{l}\mathbf{}\mathbf{?}$
We know that;
From the given figure,
$lengthofthespringis;\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{l}{\mathrm{cos}{37}^{0}}\phantom{\rule{0ex}{0ex}}or,\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{l}{4/5}=\frac{5l}{4}\phantom{\rule{0ex}{0ex}}Therefore,\phantom{\rule{0ex}{0ex}}Extensionofthespringis;\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{5l}{4}-l=\frac{l}{4}$
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