Question

In an experiment to measure the velocity of water splash when a car passes over a water layer on the road, it is found that the velocity depends on weight of car, power delivered by its engine and thickness of water layer. If velocity of splash was found to be 3 m/sec when a car weighing 1000N passes over a layer of thickness 3 mm and its engine is supplying a power of 800 kW. Find the velocity (in m/sec) of splash when car weighing 500 N passes over a layer of thickness 4mm & its engine is supplying a power of just 400 kW.

Answer 1

3

Answer 2

The velocity of the water splash $v$ is stated to depend on the weight of the car $W$, the power delivered by the engine $P$, and the thickness of the water layer $t$. We can use dimensional analysis to find the relationship between these quantities.

Let the relationship be $v \propto W^a P^b t^c$.

The dimensions of the quantities are:

• Velocity $v$: $[L T^{-1}]$
• Weight $W$: $[M L T^{-2}]$ (Weight is a force)
• Power $P$: $[M L^2 T^{-3}]$ (Power = Work/Time = Force $\times$ Distance / Time)
• Thickness $t$: $[L]$

Equating the dimensions on both sides of the proportional relationship:

$[L T^{-1}] = [M L T^{-2}]^a [M L^2 T^{-3}]^b [L]^c$ $[L T^{-1}] = [M^{a+b} L^{a+2b+c} T^{-2a-3b}]$

Equating the exponents of M, L, and T on both sides:

For M: $0 = a + b \quad (1)$ For L: $1 = a + 2b + c \quad (2)$ For T: $-1 = -2a - 3b \quad (3)$

From equation (1), $a = -b$.

Substitute $a = -b$ into equation (3):

$-1 = -2(-b) - 3b$ $-1 = 2b - 3b$ $-1 = -b$ $b = 1$

Substitute $b = 1$ into $a = -b$:

$a = -1$

Substitute $a = -1$ and $b = 1$ into equation (2):

$1 = (-1) + 2(1) + c$ $1 = -1 + 2 + c$ $1 = 1 + c$ $c = 0$

So, the relationship is $v \propto W^{-1} P^1 t^0$.

This means $v \propto \frac{P}{W}$.

The thickness $t$ does not affect the velocity of the splash according to this dimensional analysis based on the given parameters.

We can write the relationship as $v = K \frac{P}{W}$, where $K$ is a dimensionless constant.

We are given the first set of values:

$v_1 = 3$ m/sec $W_1 = 1000$ N $P_1 = 800$ kW = $800 \times 10^3$ W $t_1 = 3$ mm (not used in the formula)

Using these values, we can find the constant $K$:

$v_1 = K \frac{P_1}{W_1}$ $3 = K \frac{800 \times 10^3}{1000}$ $3 = K \times 800$ $K = \frac{3}{800}$

Now, we need to find the velocity $v_2$ for the second set of values:

$W_2 = 500$ N $P_2 = 400$ kW = $400 \times 10^3$ W $t_2 = 4$ mm (not used in the formula)

Using the same relationship $v = K \frac{P}{W}$:

$v_2 = K \frac{P_2}{W_2}$

Substitute the value of $K$ and the second set of values:

$v_2 = \frac{3}{800} \times \frac{400 \times 10^3}{500}$ $v_2 = \frac{3}{800} \times \frac{400000}{500}$ $v_2 = \frac{3}{800} \times \frac{4000}{5}$ $v_2 = \frac{3}{800} \times 800$ $v_2 = 3$

The velocity of the splash in the second case is 3 m/sec.