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Ans:- Given that the magnatic force varies inversly with the square of the distance between the magnets.Let x be the distance between magnets and F be the force then we thaveF=\frac{k}{x^{2}} \quad(k \text { is a constant) }Given that when x=0.06 meter then F=3 \mathrm{~N} Hence from (1) we get\begin{array}{l}3=\frac{K}{(0.06)^{2}} \\\Rightarrow \quad K=0.0108 \\\end{array}Therefore (1) becomesF=\frac{0.0108}{x^{2}}Now-the work by the force F(x) to move the magnets from a disfance of a meter apart to a distance of b meter apant is given bythe formulaW=\int_{a}^{b} f(x) d xHene a=0.01 and b=0.11Hence W=\int_{0.01}^{0.11} \frac{0.0108}{x^{2}} d x=0.0108 \int_{0.01}^{0.11} x^{-2} d x\begin{array}{l}=0.0108\left[-\frac{1}{x}\right]_{0.01}^{0.11} \\=0.0108\left[-\frac{1}{0.11}+\frac{1}{0.01}\right] \\=0.0108 \times 90.9091\end{array}=0.982 (Rounded to 3 decimal place)Therefore the required works is 0.982 Joules ...