MHTCET 25 Apr 2025 Shift 1 Paper

Concept:For a closed organ pipe, only odd harmonics are produced.The frequency of the $n^{\text{th}}$ mode is $f_n = (2n-1)\dfrac{v}{4L}$.Here, the “2nd harmonic” is actually the 3rd harmonic of the closed pipe.Explanation:Let the original length of the closed pipe be $L$ and speed of sound be $v$.The fundamental frequency is $f_1 = \dfrac{v}{4L} = 400\ \text{Hz}$.Therefore, $v = 1600L$.When $\dfrac{1}{3}$ of the pipe is filled with water, the length of the air column becomes $L' = L - \dfrac{L}{3} = \dfrac{2L}{3}$.For the new air column, the fundamental frequency is $f_1' = \dfrac{v}{4L'}$.The 2nd harmonic means the 3rd harmonic of the closed pipe, so $f_2' = 3\dfrac{v}{4L'}$.Substitute $L' = \dfrac{2L}{3}$ and $v = 1600L$:$f_2' = \dfrac{3 \times 1600L}{4 \times \dfrac{2L}{3}} = \dfrac{4800L}{\dfrac{8L}{3}} = 1800\ \text{Hz}$.Answer:The frequency of the 2nd harmonic is $1800\ \text{Hz}$.Correct option: D. 1800 Hz.