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
Electrolysis of 600 mL aqueous solution of NaCl for 5 min changes the pH of the solution to 12. The current in Amperes used for the given electrolysis is ….. (Nearest integer).