Resonance

In a resonance tube closed at one end, the fundamental frequency corresponds to a resonance at the first harmonic (where the tube behaves like a pipe with a closed end). The lengths and correspond to successive resonances, and we use this information to calculate the frequency of sound. The difference in lengths between two resonant positions corresponds to half of the wavelength of the sound. Thus, we have: This represents half the wavelength, so the full wavelength is: Now, we use the wave equation , where is the speed of sound and is the frequency of the sound. Solving for : Thus, the frequency of the sound is 1000 Hz. Therefore, the correct answer is (B) 1000 Hz.

Step 1: Use the resonance condition for an LCR circuit

For maximum current in a series LCR circuit, resonance must occur. At resonance, the angular frequency is given by:

Step 2: Convert all quantities to SI units

• Inductance:
• Capacitance:

Step 3: Substitute values into the resonance formula

Step 4: Simplify the square root

Step 5: Calculate the final value of

Final Answer: The angular frequency at resonance is .