Step 1: Formula for frequency in an open organ pipe.- Fundamental frequency f1 = $\frac{v}{2L}$, where v = 360m/s, L = 0.4m.- Frequency of second harmonic f2 = 2f1.
Step 2: Calculate the frequency.
f1 = $\frac{360}{2 \times 0.4}$ = 450Hz.
f2 = 2 × 450 = 900Hz.
Final Answer: The frequency of the second harmonic is 900Hz

The heat generated in 1 minute between points A and B in the given circuit, when a battery of 9 V with internal resistance of 1 \(\Omega\) is connected across these points is ______ J. 
The given circuit works as: 
Let the lines $L_1 : \vec r = \hat i + 2\hat j + 3\hat k + \lambda(2\hat i + 3\hat j + 4\hat k)$, $\lambda \in \mathbb{R}$ and $L_2 : \vec r = (4\hat i + \hat j) + \mu(5\hat i + + 2\hat j + \hat k)$, $\mu \in \mathbb{R}$ intersect at the point $R$. Let $P$ and $Q$ be the points lying on lines $L_1$ and $L_2$, respectively, such that $|PR|=\sqrt{29}$ and $|PQ|=\sqrt{\frac{47}{3}}$. If the point $P$ lies in the first octant, then $27(QR)^2$ is equal to}