Question 233: Properties of Matter

Question

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 $\_\_\_\_$ .

Accepted Answer

Dimensions of Each Quantity
We first write the dimensions of every symbol in the formula.
Time $[\mathrm{t}]=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^1ight]$
Density $[ho]=[\sigma]=\left[\frac{	ext { Mass }}{	ext { Volume }}ight]=\left[\mathrm{ML}^{-3}ight]$
Radius $[\mathrm{r}]=[\mathrm{L}]$
From Stokes' Law ( $F=6 \pi \eta$ rv), we have $\eta=$ F/rv.
Using StokesÃ¢ÂÂ law, $\eta$ depends on force, radius, and velocity, so its dimensions come from $F$, $r$, and $v$.
Dimension $=\left[\mathrm{MLT}^{-2}ight][\mathrm{L^{-1}}]\left[\mathrm{L^{-1}T}ight]=\left[\mathrm{ML}^{-1}{ \mathrm{~T}^{-1}}ight]$
Constant (A) is dimensionless
According to the given formula $t=A ho^a r^b \eta^c \sigma^d$ :
$$ [\mathrm{T}]=\left[\mathrm{ML}^{-3}ight]^{\mathrm{a}}[\mathrm{~L}]^{\mathrm{b}}\left[\mathrm{ML}^{-1}{ \mathrm{~T}^{-1}}ight]^{\mathrm{c}}\left[\mathrm{ML}^{-3}ight]^{\mathrm{d}} $$
$$\Rightarrow $$ $\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^1ight]=\left[\mathrm{M}^{\mathrm{a}+\mathrm{c}+\mathrm{d}}ight]\left[\mathrm{L}^{-3 \mathrm{a}+\mathrm{b}-\mathrm{c}-3 \mathrm{~d}}ight]\left[\mathrm{T}^{-\mathrm{c}}ight]$
Comparing the exponents for $\mathrm{M}, \mathrm{L}$, and T on both sides :
For $\mathrm{T}:-\mathrm{c}=1 \Rightarrow \mathrm{c}=-1$
For $\mathrm{M}: \mathrm{a}+\mathrm{c}+\mathrm{d}=0 \Rightarrow \mathrm{a}-1+\mathrm{d}=0 \Rightarrow \mathrm{a}+\mathrm{d}=1$
For $\mathrm{L}:-3 \mathrm{a}+\mathrm{b}-\mathrm{c}-3 \mathrm{~d}=0$
Substituting $\mathrm{c}=-1$ :
$$ -3 a+b+1-3 d=0 $$
$$\Rightarrow $$ $$ b+1-3(a+d)=0 $$
Substituting $(a+d)=1$
$$ b+1-3(1)=0 $$
$$\Rightarrow $$ $$ b-2=0 \Rightarrow b=2 $$
We need to find the value of $\frac{b+c}{a+d}$ :
$$ \frac{b+c}{a+d}=\frac{2+(-1)}{1}=1 $$
Therefore, the correct answer is $\mathbf{1}$.

Properties of Matter

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 \_\_\_\_ .

Integer / Numeric Answer Type