\\\\(C = 5 \\mu F\\\\)minimum capacity is obtained when all are in a series,\\\\(\ herefore \\dfrac{1}{c_{s}}=\\dfrac{1}{c_1}+\\dfrac{1}{c_2}.....+\\dfrac{1}{c_{10}}\\\\)\\\\( \\dfrac{1}{c_{s}}=\\dfrac{1}{5}+\\dfrac{1}{5}+.....+\\dfrac{1}{5}=\\dfrac{10}{5}\\\\)\\\\(\ herefore c_{s}=\\dfrac{1}{2}\\\\)maximum capacity is obtained when they are parallety connected\\\\(\ herefore c_{eff}=c_1+c_2+.......+c_{10}\\\\)= 5 + 5 + ......+5\\\\(c_{p}{=10(5)}\\\\)\\\\(c_{p}=50\\\\)\\\\(\ herefore\\\\) Ratio of min to max capacity, \\\\(\\dfrac{c_s}{c_p}= \\dfrac{0.5}{50}=1 : 100\\\\)