When a load of 10 ,kg is suspended on a metal | vedclass

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(C) The force constant $K$ of a wire is defined as the ratio of the applied force $F$ to the elongation $\Delta x$ produced in the wire.$K = \frac{F}{

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$5 	imes 10^4$

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(C) The force constant $K$ of a wire is defined as the ratio of the applied force $F$ to the elongation $\Delta x$ produced in the wire.$K = \frac{F}{\Delta x}$Given:Mass $m = 10 \,kg$,so the force $F = mg = 10 	imes 10 = 100 \,N$ (taking $g = 10 \,m/s^2$).Elongation $\Delta x = 2 \,mm = 2 	imes 10^{-3} \,m = 0.002 \,m$.Substituting these values into the formula:$K = \frac{100}{0.002} = \frac{100000}{2} = 5 	imes 10^4 \,N/m$.Therefore,the correct option is $C$.

When a load of $10 \,kg$ is suspended on a metallic wire,its length increases by $2 \,mm$. The force constant of the wire is ........ $N/m$.

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