Physics - Fluid Question with Solution | TestHub

Required lift force, $F=P_{2}A-P_{1}A=5.4\times {10}^5\times g$

$\Rightarrow P_2-P_1=\frac{5.4\times {10}^6}{500}=5.4\times 2\times {10}^3$

$=10.8\times {10}^3$

Now applying Bernoulli's Theorem we get

$P_2+0+\frac{1}{2}\rho V_2^2=P_1+0+\frac{1}{2}\rho V_1^2$

$$\begin{gathered} \Rightarrow P_2-P_1=\frac{1}{2}\rho \left(V_1^2-V_2^2\right)=\frac{1}{2}\rho \left(V_1-V_2\right)\left(V_1+V_2\right) \\ \left[\text{Since for small change}, V_1=V_2=V=1080 \mathrm{km} {\mathrm{h}}^{-1}=300 \mathrm{m} {\mathrm{s}}^{-1}\right] \end{gathered}$$

$\Rightarrow 10.8\times {10}^3=\frac{1}{2}\times 1.2\left(V_1-V_2\right)\times 2\times 300$

$\Rightarrow 10.8\times 10=3.6\left(V_1-V_2\right)$

$\Rightarrow V_1-V_2=30$

Therefore, fractional increase will be$\left(\frac{V_1-V_2}{V}\right)\times 100=\frac{30}{300}\times 100=10\%$