GIVEN as,\(K = \frac{{{F_2} - {F_1}}}{{{L_1} - {L_2}}}\)where, F1 = weight of pan, F2 = total weight of pan and extra mass, L1 = initial length of spring, L2 = final length of springThe un-deformed length of spring is, Deflection of spring due to 2 kg pan + ORIGINAL length of springCalculation:Given:\(k= \frac{{\left( {22 - 2} \right) \times 9.81}}{{\left( {200 - 100} \right) \times {{10}^{ - 3}}}} = 1962~N/m\)Deflection of spring due to 2 kg pan\(\delta= \frac{{2 \times 9.81}}{{1962}} = 0.01~m = 10\;mm\)∴ Unperformed length = 200 + 10= 210 mm