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

Let the velocity of splash be $v$. The problem states that $v$ depends on weight of car $W$, power delivered by engine $P$, and thickness of water layer $h$. We assume a relationship of the form $v = k W^a P^b h^c$, where $k$ is a dimensionless constant and $a, b, c$ are exponents.

The dimensions of the quantities are:

Velocity $v$: $[L T^{-1}]$

Weight $W$: $[M L T^{-2}]$

Power $P$: $[M L^2 T^{-3}]$

Thickness $h$: $[L]$

Substituting these dimensions into the 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: $0 = a + b$ (1)

For L: $1 = a + 2b + c$ (2)

For T: $-1 = -2a - 3b$ (3)

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

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

$-1 = -2(-b) - 3b \implies -1 = 2b - 3b \implies -1 = -b \implies b = 1$.

Since $a = -b$, $a = -1$.

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

$1 = (-1) + 2(1) + c \implies 1 = -1 + 2 + c \implies 1 = 1 + c \implies c = 0$.

The relationship is $v = k W^{-1} P^1 h^0 = k \frac{P}{W}$.

The velocity of the splash is proportional to the power and inversely proportional to the weight, and it does not depend on the thickness of the water layer.

Let $v_1, W_1, P_1$ be the values in the first experiment and $v_2, W_2, P_2$ be the values in the second case.

$v_1 = k \frac{P_1}{W_1}$

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

Taking the ratio:

$\frac{v_2}{v_1} = \frac{k \frac{P_2}{W_2}}{k \frac{P_1}{W_1}} = \frac{P_2}{W_2} \times \frac{W_1}{P_1} = \frac{P_2}{P_1} \times \frac{W_1}{W_2}$

Given values:

$v_1 = 3$ m/sec

$W_1 = 1000$ N

$P_1 = 800$ kW

$W_2 = 500$ N

$P_2 = 400$ kW

Substitute the values into the ratio equation:

$\frac{v_2}{3 \text{ m/sec}} = \frac{400 \text{ kW}}{800 \text{ kW}} \times \frac{1000 \text{ N}}{500 \text{ N}}$

$\frac{v_2}{3} = \frac{1}{2} \times \frac{2}{1}$

$\frac{v_2}{3} = 1$

$v_2 = 3 \times 1 = 3$ m/sec.

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