A spring of force constant 15 N/m is cut int | vedclass

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(C) The force constant $K$ of a spring is inversely proportional to its length $\ell$,i.e.,$K \ell = 	ext{constant}$.Let the original length be $\ell

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(C) The force constant $K$ of a spring is inversely proportional to its length $\ell$,i.e.,$K \ell = 	ext{constant}$.Let the original length be $\ell$ and the original force constant be $K = 15 \ N/m$.The spring is cut into two pieces with lengths in the ratio $1 : 3$. Thus,the lengths of the pieces are $\ell_1 = \frac{\ell}{4}$ and $\ell_2 = \frac{3\ell}{4}$.The smaller piece has length $\ell_1 = \frac{\ell}{4}$.Using the relation $K \ell = K^{\prime} \ell_1$,we get:$15 	imes \ell = K^{\prime} 	imes (\frac{\ell}{4})$$K^{\prime} = 15 	imes 4 = 60 \ N/m$.

$A$ spring of force constant $15 \ N/m$ is cut into two pieces. If the ratio of their lengths is $1 : 3$,then the force constant of the smaller piece is . . . . . . $N/m$.

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