Applying Bernoulli’s theorem: \(P_1 + \rho g h + \frac{1}{2} \rho v^2 = P_2 + \frac{1}{2} \rho (2v)^2\)
Putting the values, \(4100=800(\frac{3}{2}v^2−10)\)
\(⇒v=\frac{\sqrt{363}}{6}m/s\)
If \[ \frac{dy}{dx} + 2y \sec^2 x = 2 \sec^2 x + 3 \tan x \cdot \sec^2 x \] and
and \( f(0) = \frac{5}{4} \), then the value of \[ 12 \left( y \left( \frac{\pi}{4} \right) - \frac{1}{e^2} \right) \] equals to: