A ball of radius r and density $$\rho$$ dropped through a viscous liquid of density $$\sigma$$ and viscosity $$\eta$$ attains its terminal velocity at time t, given by $$t= A \rho^{a}r^{b}\eta^{c}\sigma^{d}$$, where A is a constant and a, b, c and d are integers. The value of $$\frac{b+c}{a+d}$$ is _________.

We need to find $$\frac{b+c}{a+d}$$ where the terminal velocity time $$t = A\rho^a r^b \eta^c \sigma^d$$. Since we use dimensional analysis, we note the dimensions of each quantity: $$[t] = T$$, $$[\rho] = ML^{-3}$$ (density of the ball), $$[r] = L$$ (radius), $$[\eta] = ML^{-1}T^{-1}$$ (viscosity), and $$[\sigma] = ML^{-3}$$ (density of the liquid).

Writing the dimensional equation gives $$T = (ML^{-3})^a \cdot L^b \cdot (ML^{-1}T^{-1})^c \cdot (ML^{-3})^d$$ which simplifies to $$T = M^{a+c+d} \cdot L^{-3a+b-c-3d} \cdot T^{-c}$$.

Equating the exponents of M, L, and T on both sides leads to the system $$a + c + d = 0$$ ...(i), $$-3a + b - c - 3d = 0$$ ...(ii), and $$-c = 1 \implies c = -1$$ ...(iii).

Substituting $$c = -1$$ into equation (i) gives $$a - 1 + d = 0 \implies a + d = 1$$ ...(iv).

Substituting $$c = -1$$ into equation (ii) yields $$-3a + b - (-1) - 3d = 0 \implies -3a + b + 1 - 3d = 0$$ so that $$b = 3a + 3d - 1 = 3(a + d) - 1 = 3(1) - 1 = 2$$ ...(v).

This gives the required ratio as $$\frac{b + c}{a + d} = \frac{2 + (-1)}{1} = \frac{1}{1} = 1$$.

Option 1