MHTCET 15 May 2024 Shift 1 Paper

Concept:The beat frequency is the difference between the fundamental frequencies of two open organ pipes of different lengths.Explanation:For an open organ pipe, the fundamental frequency is $f = \dfrac{V}{2L}$, where $V$ is the speed of sound and $L$ is the pipe length.For the first pipe of length $l$, the frequency is:$$f_1 = \frac{V}{2l}$$For the second pipe of length $l + l_1$, the frequency is:$$f_2 = \frac{V}{2(l + l_1)}$$Since $l + l_1 > l$, we have $f_2 < f_1$. Therefore, the beat frequency is:$$f_b = f_1 - f_2 = \frac{V}{2l} - \frac{V}{2(l + l_1)}$$Taking a common denominator:$$f_b = \frac{V(l + l_1 - l)}{l(l + l_1)} \times \frac{1}{2} = \frac{V l_1}{2l(l + l_1)}$$Given that $l_1 \ll l$, the term $l + l_1 \approx l$. Hence:$$f_b = \frac{V l_1}{2l^2}$$Answer:$$\frac{V l_1}{2 l^2}$$Therefore, the correct option is B.