The largest mass (m) that can be moved by a flowing river depends on velocity (v), density (ρ) of river water and acceleration due to gravity (g). The correct relation is

Correct Answer: m ∝ ρ v⁶ /g³

To find the correct relationship between the largest mass ($m$) that can be moved by a flowing river, the velocity of the water ($v$), the density of water ($\rho$), and the acceleration due to gravity ($g$), we can use the method of dimensional analysis.

Let the relationship be represented as:
$m\propto {\rho }^a{v}^b{g}^c$
where $a$, $b$, and $c$ are dimensionless constants to be determined.

First, let us write the dimensional formulas for each of the physical quantities involved:
1. Mass ($m$): $\left[m\right]=\left[M^1{L}^0{T}^0\right]$
2. Density ($\rho$): $\left[\rho \right]=\left[M^1{L}^{-3}{T}^0\right]$
3. Velocity ($v$): $\left[v\right]=\left[M^0{L}^1{T}^{-1}\right]$
4. Acceleration due to gravity ($g$): $\left[g\right]=\left[M^0{L}^1{T}^{-2}\right]$

Substituting these dimensions into the proportionality equation, we get:
$\left[M^1{L}^0{T}^0\right]={\left[M^1{L}^{-3}{T}^0\right]}^a{\left[M^0{L}^1{T}^{-1}\right]}^b{\left[M^0{L}^1{T}^{-2}\right]}^c$

Combining the exponents of the fundamental units on the right-hand side:
$\left[M^1{L}^0{T}^0\right]=\left[M^a{L}^{-3a+b+c}{T}^{-b-2c}\right]$

By equating the exponents of $M$, $L$, and $T$ from both sides, we get the following system of linear equations:
1. For $M$: $a=1$
2. For $T$: $-b-2c=0\Rightarrow b=-2c$
3. For $L$: $-3a+b+c=0$

Substitute $a=1$ and $b=-2c$ into the equation for $L$:
$-3\left(1\right)+(-2c)+c=0$
$-3-c=0$
$c=-3$

Now, calculate the value of $b$ using $b=-2c$:
$b=-2(-3)=6$

Substituting the values $a=1$, $b=6$, and $c=-3$ back into the original proportionality relation, we obtain:
$m\propto {\rho }^1{v}^6{g}^{-3}$
which can be rewritten as:
$m\propto \frac{\rho v^6}{g^3}$