TG EAMCET 2 May 2025 Shift 2 Paper

Step 1: Write the formula for frequency in an open pipeThe frequency of the $n$th harmonic in an open organ pipe is:$f_0=\frac{n v}{2 l_0}$Step 2: Find the frequency for the 3rd harmonic in the open pipeFor the third harmonic, $n=3$ :$f_0'=\frac{3 v}{2 l_0}$Step 3: Write the formula for frequency in a closed pipeThe frequency of the $n$th harmonic in a closed pipe is:$f_c=\frac{n v}{4 l_c}$Step 4: Find the frequency for the 5th harmonic in the closed pipeFor the fifth harmonic, $n=5$ :$f_c'=\frac{5 v}{4 l_c}$Step 5: Write the ratio of the two frequenciesWe want to find the ratio $\frac{f_0'}{f_c'}$.$\frac{f_0'}{f_c'}=\frac{\frac{3 v}{2 l_0}}{\frac{5 v}{4 l_c}}$Step 6: Simplify the ratioFirst, simplify by dividing:$\frac{3 v}{2 l_0} \times \frac{4 l_c}{5 v} = \frac{12 l_c}{10 l_0}$Step 7: Use the ratio of lengthsIt is given that the ratio of lengths of the open and closed pipes is $2: 3$, so $\frac{l_0}{l_c} = \frac{2}{3}$.Step 8: Substitute the value for $\frac{l_0}{l_c}$$\frac{12}{10} \times \frac{1}{\frac{l_0}{l_c}} = \frac{12}{10} \times \frac{1}{\frac{2}{3}} = \frac{12}{10} \times \frac{3}{2} = \frac{36}{20} = \frac{9}{5}$Step 9: State the final ratioThe ratio of the frequencies is $9: 5$.