Length of a spring is α when a force of 4,N is applied on it and length is β when 5,N force is app

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5.Work, Energy, Power and Collision

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The length of a spring is $\alpha $ when a force of $4\,N$ is applied on it and the length is $\beta $ when $5\,N$ force is applied. Then the length of spring when $9\,N$ force is applied is

$5\beta -4\alpha $

$\beta -\alpha $

$5\alpha -4\beta $

$9(\beta - \alpha )$

Solution

Let the natural length of the spring $=\ell_{0}$

From figure

$4=\mathrm{k}\left(\alpha-\ell_{0}\right) \ldots(\mathrm{i})$

$5=\mathrm{k}\left(\beta-\ell_{0}\right) \dots(\mathrm{ii})$

$9=\mathrm{k}\left(\gamma-\ell_{0}\right) \ldots(\mathrm{iii})$

eq. $\frac{(\mathrm{iit})-(\mathrm{i})}{(\mathrm{ii})-(\mathrm{i})} \Rightarrow \frac{5}{4}=\frac{\mathrm{k}(\gamma-\alpha)}{\mathrm{k}(\gamma-\beta)}$

$\gamma=5 \beta-4 \alpha$

Standard 11

Physics

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The length of a spring is $\alpha $ when a force of $4\,N$ is applied on it and the length is $\beta $ when $5\,N$ force is applied. Then the length of spring when $9\,N$ force is applied is

Solution

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