A stretched string of length 0.50 m, linear density 0.010 kg m^-1 and tension 100 N is fixed at both ends. It is sounded near a pipe of length 0.85 m…

String: $v=\sqrt{T/\mu}=100\ \mathrm{m\,s^{-1}}$, so $f_n=n\,v/2L=100n\ \mathrm{Hz}$. Closed pipe: $f=(2k+1)v_a/4L_p=(2k+1)\times100\ \mathrm{Hz}$. (A) Both fundamentals are $100\ \mathrm{Hz}$: true. (B) The string's second harmonic is $200\ \mathrm{Hz}$, an even multiple of $100$, which a closed pipe cannot sustain: false. (C) With $4T$ the wave speed doubles and $f_n=200n\ \mathrm{Hz}$, always an even multiple of $100\ \mathrm{Hz}$, while the pipe supports only odd multiples: no match, true. (D) The pipe fundamental becomes $346.8/3.4=102\ \mathrm{Hz}$, giving $2\ \mathrm{Hz}$ beats with the string's $100\ \mathrm{Hz}$: true. The key idea is that a closed pipe supports only odd harmonics of its fundamental while a string supports all harmonics, so coincidences require the string frequencies to include odd multiples.