MHTCET 15 Apr 2026 Shift 2 Paper

Concept:When a pipe open at both ends is dipped in water, the water surface acts as a closed end.The air column becomes a pipe closed at one end and open at the other.This changes its fundamental frequency.Explanation:Let the full length of the open pipe be $L$.For an open pipe, the fundamental frequency is given by:$$f = \frac{v}{2L}$$Here, $v$ is the speed of sound in air.Since $\frac{2}{3}$ of the pipe is dipped in water, the length of the air column is:$$L_{\text{air}} = L - \frac{2L}{3} = \frac{L}{3}$$The water surface acts as a closed end, so the air column behaves like a closed pipe.For a closed pipe, the fundamental frequency is:$$f' = \frac{v}{4L_{\text{air}}}$$Substitute $L_{\text{air}} = \frac{L}{3}$:$$f' = \frac{v}{4\left(\frac{L}{3}\right)} = \frac{3v}{4L}$$Now compare the new frequency with the original frequency:$$\frac{f'}{f} = \frac{\frac{3v}{4L}}{\frac{v}{2L}} = \frac{3v}{4L} \times \frac{2L}{v} = \frac{3}{2}$$So,$$f' = \frac{3}{2} f$$Answer:The new fundamental frequency is $\frac{3}{2} f$.Therefore, the correct option is Option B: $\frac{3}{2} f$.