Solution:
Terminal velocity \(v_T \propto r^2(\sigma - \rho)\). For iron, \(v_1 \propto 1^2(7.6 - 13.6) = -6\). For lead, \(v_2 \propto 2^2(11.6 - 13.6) = -8\). Thus, the ratio is \(v_1 : v_2 = -6 / -8 = 0.75\).
Terminal velocity of iron ball of radius \(1\text{ mm}\) in glycerine is \(v_1\) and \(v_2\) is the terminal velocity of lead ball of radius \(2\text{ mm}\) in glycerine then \(v_1 : v_2\) is : [\(\rho_{\text{glycerine}} = 13.6\text{ g/cc}\), \(\sigma_{\text{Fe}} = 7.6\text{ g/cc}\), \(\sigma_{\text{Pb}} = 11.6\text{ g/cc}\)]
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