Answer
Choice A.
Work Step by Step
In a tube of length L, open at both ends, the wavelength of the lowest vibration mode is 2L.
In a tube of length L that is closed at one end, the wavelength of the lowest vibration mode is 4L.
We see that the wavelength doubles when one end of the pipe is suddenly closed off.
We know that $v = \lambda f$, so because the speed of sound v remains constant, doubling the wavelength must cut the oscillation frequency in half. This corresponds to a lower pitch note.
