Q.A standing wave is produced on a string fixed at one end with the other end free. The length of the string (A) must be an odd integral multiple of $\lambda/4$.
(B) must be an odd integral multiple of $\lambda/2$.
(C) must be an odd integral multiple of $\lambda$.
(D) must be an even integral multiple of $\lambda$.

[!TLDR] A string fixed at one end (node) and free at the other (antinode) needs an odd number of quarter-wavelengths to fit -- exactly the pipe-closed-at-one-end condition, $L=(2p …

A string fixed at one end has a NODE there (the fixed end cannot move); a free end always forms an ANTINODE (maximum displacement, since nothing constrains its motion there) -- this is exactly the same pair of boundary conditions as a pipe closed at one end (node at the closed end, antinode at the open end, section 6.7.2). By that same derivation, the allowed lengths are $L=\lambda/4,\,3\lambda/4,\,5\lambda/4,\ldots=(2p+1)\lambda/4$ for $p=0,1,2,\ldots$ -- i.e. the string's length must always be an ODD integral multiple of …

Recognise a fixed-end/free-end string as having the same node-at-fixed-end, antinode-at-free-end boundary conditions as a pipe closed at one en …

Treating the string as if BOTH ends were fixed (the usual string case), which instead gives $L$ as an integral multiple of $\lambda/2$ -- forgetting that one end here is explicitly FREE, n …