Row, Row, Row Your Boat

Power delivered by the rowers is lost due to drag force in steady state and there is no net increment in velocity.

Therefore, $n \times P_{p} = F_{d} \times v_{boat}$

So, $\rho A_{boat} v_{boat}^{3} = n \times P_{p}$ -Equation 1

Puting $A_{boat}=(n V_{p})^{2/3}$ in Equation , we get

$\rho (n V_{p})^{2/3} v_{boat}^{3} = n \times P_{p}$

On simplifying and taking $v_{boat}$ on Left hand side,

$v_{boat}=\rho^{-1/3} P_{p}^{1/3} n^{1/9} V_{p}^{-2/9}$

Therefore, $\alpha=1/9$ , $\beta=1/3$ , $\gamma=-2/9$ , $\delta=-1/3$

$\alpha \beta \gamma \delta= 2/729$

so, $a=2$ and $b=729$

Therefore, $a+b= \boxed{731}$

To get a Steady State Velocity, Clearly,

$P_{input} = P_{drag}$

Thus,

$nP_p = F_d v_{boat}$
$\Rightarrow nP_p = \rho A_{boat} V_{boat}^3$

Simplifying,

$v_{boat} ~ n^{1/9} P_p^{1/3} V_p^{-2/9} \rho^{-1/3}$

Thus,

$\alpha\beta\gamma\delta = \frac {2}{729}$

Finally,

$a+b = 731$

In steady state, energy lost/second is same as the power supplied,

Clearly, energy lost/sec = work done by drag force/sec = $k \rho A_{boat} v_{boat}^2 \times v_{boat}$

Hence, $n \times P_{p} = k \rho A_{boat} v_{boat}^3$ , $A_{boat} = k' (n V_{p})^{\frac{2}{3}}$

Hence, ve get $v_{boat} \propto n^{1/9} P_{p}^{1/3} \rho^{-1/3} V_{p}^{-2/9}$

Hence, $\alpha \beta \gamma \delta = \frac{2}{729}$

sum of power of persons=power of drag force

using the simple fact that Power = Force . Velocity,

we get $n \times P_{p}$ = $F_{p} \times v_{p}$ plugging in the values using any constants can get the answer as we only need the proportionality relation $\boxed{731}$

This question is actually a simple dimensions problem. We start with what we are given as : The drag force, $F_{d}$ = $\rho$ $v^{2}_{boat}$ $n^{\frac {2}{3}}$ $V^{\frac {2}{3}}_{p}$ Now we know that $Power = Force \times velocity$ And we are also given that each man exerts a power of $P_{p}$ . So, total power exerted by all men $= n \times P_{p}$ . Therefore substituting in our original equation : $nP_{p} = F_{d} \times v_{boat}$ $\Rightarrow nP_{p} = \rho$ $v^{2}_{boat}$ $n^{\frac {2}{3}}$ $V^{\frac {2}{3}}_{p}$ $\Rightarrow v_{boat} = n^{\frac {1}{9}}P^{\frac {1}{3}}_{p}V^{\frac {-2}{9}}_{p}\rho^{\frac {-1}{3}}$ Now I am sure you can do the rest and enjoy the mathematics. Thank you.

Note that for a power $P$ , it is defined as the work done per unit time. Thus, $P = \frac{Work Done}{t}$ . As work done is defined as the product of force and distance, we have $P = \frac{(F)(x)}{t} = (F)(v)$ , where $v$ represents the speed.

As each person provides a power P, the net power provided is defined as $nP_p$ . At the steady state velocity, $v$ is constant, implying that $F_{net} = 0 N$ . Hence, as the applied force = drag force. , at a constant speed, the power applied = power loss due to drag force. Thus $nP_p = P_d = (F_d)(v)$ .

Hence, $nP_p = (F_d)(v)$

= $pAv^3$ , by substituting the expression for $F_d$ .

Thus, $nP_p = pn^{\frac{2}{3}}V_p^{\frac{2}{3}}v^3$ , by substituting the expression for $A_{boat}$

Hence, $v = n^{\frac{1}{9}}P_p^{\frac{1}{3}}V_p^{-\frac{2}{9}}p^{-\frac{1}{3}}$

Thus, the product of the exponents is given by $\frac{1}{9}$ x $\frac{1}{3}$ x $(-\frac{2}{9})$ x $(-\frac{1}{3})$ = $\frac{2}{729}$

Hence, $2 + 729 = \boxed{731}$