A ball of radius r and density ρ dropped through a viscous liquid of density σ and viscosity η attains its terminal velocity at time t, given by t = A ρ^a r^b η^c σ^d, where A is a constant and a, b, c and d are integers. The value of (b + c)/(a + d) is ___

A ball of radius r and density ρ dropped through a viscous liquid of density σ and viscosity η attains its terminal velocity at time t

Q. 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 ____.

Correct Answer: 1

Explanation

The constant $A$ is dimensionless.

Dimensions of the given quantities are:

$$ [t] = [T] $$

$$ [\rho] = [\sigma] = [ML^{-3}] $$

$$ [r] = [L] $$

From Stokes’ law, the dimension of coefficient of viscosity is:

$$ [\eta] = [ML^{-1}T^{-1}] $$

Using the given relation:

$$ t = A\rho^a r^b \eta^c \sigma^d $$

Writing dimensions on both sides:

$$ [T] = [ML^{-3}]^a [L]^b [ML^{-1}T^{-1}]^c [ML^{-3}]^d $$

$$ [T] = [M^{a+c+d} L^{-3a+b-c-3d} T^{-c}] $$

Comparing powers of $M$, $L$ and $T$:

For time $T$:

$$ -c = 1 \Rightarrow c = -1 $$

For mass $M$:

$$ a + c + d = 0 \Rightarrow a - 1 + d = 0 \Rightarrow a + d = 1 $$

For length $L$:

$$ -3a + b - c - 3d = 0 $$

Substituting $c = -1$:

$$ -3a + b + 1 - 3d = 0 $$

$$ b + 1 - 3(a+d) = 0 $$

Using $a + d = 1$:

$$ b + 1 - 3 = 0 \Rightarrow b = 2 $$

Now,

$$ \frac{b+c}{a+d} = \frac{2 + (-1)}{1} = 1 $$

Final Answer: $$ \boxed{1} $$

Explanation

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