Air of density 1.2 kg m$$^{-3}$$ is blowing across the horizontal wings of an aeroplane in such a way that its speeds above and below the wings are 150 ms$$^{-1}$$ and 100 ms$$^{-1}$$, respectively. The pressure difference between the upper and lower sides of the wings, is :

To solve this problem, we need to find the pressure difference between the upper and lower sides of the aeroplane wings. We are given that the density of air, ρ, is 1.2 kg m⁻³. The speed of air above the wing is 150 m s⁻¹, and below the wing is 100 m s⁻¹.

We use Bernoulli's theorem, which states that for an ideal fluid in steady flow, the sum of pressure energy, kinetic energy per unit volume, and potential energy per unit volume remains constant along a streamline. Since the wings are horizontal, the height difference is negligible, so we can ignore the potential energy term. Therefore, Bernoulli's equation simplifies to:

$$ P + \frac{1}{2} \rho v^2 = \text{constant} $$

Applying this equation to points above and below the wing:

$$ P_{\text{above}} + \frac{1}{2} \rho v_{\text{above}}^2 = P_{\text{below}} + \frac{1}{2} \rho v_{\text{below}}^2 $$

We rearrange this equation to find the pressure difference, $$ P_{\text{below}} - P_{\text{above}} $$:

$$ P_{\text{below}} - P_{\text{above}} = \frac{1}{2} \rho v_{\text{above}}^2 - \frac{1}{2} \rho v_{\text{below}}^2 $$

Factor out $$\frac{1}{2} \rho$$:

$$ P_{\text{below}} - P_{\text{above}} = \frac{1}{2} \rho \left( v_{\text{above}}^2 - v_{\text{below}}^2 \right) $$

Now substitute the given values: ρ = 1.2 kg m⁻³, $$ v_{\text{above}} = 150 $$ m s⁻¹, and $$ v_{\text{below}} = 100 $$ m s⁻¹.

First, calculate $$ v_{\text{above}}^2 $$:

$$ v_{\text{above}}^2 = (150)^2 = 22500 $$

Next, calculate $$ v_{\text{below}}^2 $$:

$$ v_{\text{below}}^2 = (100)^2 = 10000 $$

Now find the difference:

$$ v_{\text{above}}^2 - v_{\text{below}}^2 = 22500 - 10000 = 12500 $$

Substitute into the equation:

$$ P_{\text{below}} - P_{\text{above}} = \frac{1}{2} \times 1.2 \times 12500 $$

First, multiply 1.2 and 12500:

$$ 1.2 \times 12500 = 1.2 \times (125 \times 100) = (1.2 \times 125) \times 100 $$

$$ 1.2 \times 125 = 1.2 \times (100 + 25) = (1.2 \times 100) + (1.2 \times 25) = 120 + 30 = 150 $$

Then, $$ 150 \times 100 = 15000 $$, so:

$$ 1.2 \times 12500 = 15000 $$

Now multiply by $$\frac{1}{2}$$:

$$ \frac{1}{2} \times 15000 = 7500 $$

Therefore, the pressure difference is 7500 N m⁻².

Comparing with the options:

A. 60 Nm⁻²

B. 180 Nm⁻²

C. 7500 Nm⁻²

D. 12500 Nm⁻²

Hence, the correct answer is Option C.