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(A) Given,drag force $F_d$ depends on air density $\sigma$,velocity of ball $v$,and cross-sectional area $A$.$F_d \propto \sigma^a v^b A^c$Using dimen

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$2^{1/6}$

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(A) Given,drag force $F_d$ depends on air density $\sigma$,velocity of ball $v$,and cross-sectional area $A$.$F_d \propto \sigma^a v^b A^c$Using dimensional analysis: $[MLT^{-2}] = [ML^{-3}]^a [LT^{-1}]^b [L^2]^c$.Equating exponents: $a=1$,$-3a+b+2c=1$,and $-b=-2 \Rightarrow b=2$.Substituting $a=1, b=2$ into the second equation: $-3(1)+2+2c=1 \Rightarrow 2c=2 \Rightarrow c=1$.Thus,$F_d = k \sigma v^2 A$.At terminal velocity $v_T$,$mg = F_d = k \sigma v_T^2 A$.Since $A = \pi R^2$ and $m = \frac{4}{3} \pi R^3 \rho_{ball}$,we have $R^2 \propto m^{2/3}$.Substituting $A \propto m^{2/3}$ into the force equation: $mg \propto v_T^2 m^{2/3} \Rightarrow v_T^2 \propto m^{1/3} \Rightarrow v_T \propto m^{1/6}$.Therefore,the ratio of terminal velocities is $\frac{v_1}{v_2} = \left(\frac{m_1}{m_2}\right)^{1/6} = \left(\frac{250}{125}\right)^{1/6} = 2^{1/6}$.

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