In an organ pipe closed at one end produces a fundamental note of frequency '$\nu$'. The pipe is cut into two pipes of equal length. The fundamental frequencies produced in the two pipes are

Let the length of the original closed organ pipe be $L$.

The fundamental frequency of the closed organ pipe is given by $\nu = \dfrac{v}{4L}$.

When the pipe is cut into two equal halves, the length of each half becomes $L' = \dfrac{L}{2}$.

One of the halves will be a closed organ pipe (having the original closed end) and the other half will be an open organ pipe (having both ends open).

For the closed half, the fundamental frequency is $\nu_1 = \dfrac{v}{4L'} = \dfrac{v}{4(L/2)} = \dfrac{v}{2L} = 2\left(\dfrac{v}{4L}\right) = 2\nu$.

For the open half, the fundamental frequency is $\nu_2 = \dfrac{v}{2L'} = \dfrac{v}{2(L/2)} = \dfrac{v}{L} = 4\left(\dfrac{v}{4L}\right) = 4\nu$.

Thus, the fundamental frequencies produced in the two pipes are $2\nu$ and $4\nu$.

Answer: $2\nu$ , $4\nu$