A simple spring has length 'l' and force cons | vedclass

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(B) The force constant of a spring is inversely proportional to its length,i.e.,$k \propto \frac{1}{l}$,which implies $kl = 	ext{constant}$.Let $k_1$

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$\frac{K(n+1)}{n}$

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(B) The force constant of a spring is inversely proportional to its length,i.e.,$k \propto \frac{1}{l}$,which implies $kl = 	ext{constant}$.Let $k_1$ and $k_2$ be the force constants of the two springs of lengths $l_1$ and $l_2$ respectively.Given $l = l_1 + l_2$ and $l_1 = n l_2$.From the relation $kl = k_1 l_1 = k_2 l_2$,we have:$k_1 = \frac{kl}{l_1}$ and $k_2 = \frac{kl}{l_2}$.Since $l_1 = n l_2$,we have $l = n l_2 + l_2 = l_2(n+1)$.Substituting $l$ in the expression for $k_1$:$k_1 = \frac{k \cdot l_2(n+1)}{n l_2} = \frac{K(n+1)}{n}$.

$A$ simple spring has length '$l$' and force constant '$K$'. It is cut into two springs of length '$l_1$' and '$l_2$' such that $l_1 = n l_2$ ($n$ is an integer). The force constant of the spring of length '$l_1$' is:

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