The fundamental frequency of a closed pipe is 400 Hz . If $\left(\frac{1}{3}\right)^{\mathrm{rd}}$ length of the pipe is filled with water, the frequency of the $2^{\text {nd }}$ harmonic of the pipe will be (Neglect end correction)

The fundamental frequency of a closed pipe is given by $f_1 = \frac{v}{4L}$, where $v$ is the speed of sound and $L$ is the pipe length.

Given $f_1 = 400$ Hz, we have $\frac{v}{4L} = 400$ Hz.

When $\frac{1}{3}$ of the pipe is filled with water, the effective air column length becomes $L' = \frac{2}{3}L$. The new fundamental frequency is $f_1' = \frac{v}{4L'} = \frac{v}{4(\frac{2}{3}L)} = \frac{3v}{8L}$.

Substituting $\frac{v}{4L} = 400$ Hz into the expression for $f_1'$ yields $f_1' = \frac{3}{2}(400) = 600$ Hz.

For a closed pipe, the harmonic frequencies are the odd multiples of the fundamental frequency. The second harmonic corresponds to $3f_1' = 3 \times 600 = 1800$ Hz.

Final answer: $\boxed{1800\ \text{Hz}}$