Search any question & find its solution
Question:
Answered & Verified by Expert
An open organ pipe having fundamental frequency (n) is in unison with a vibrating string. If the tube is dipped in water so that $75 \%$ of the length of the tube is inside the water then the ratio of fundamental frequency of the air column of dipped tube with that of string will be (Neglect end corrections)
Options:
Solution:
2584 Upvotes
Verified Answer
The correct answer is:
$2: 1$
$\mathrm{n}_{\text {open }}=\frac{\mathrm{v}}{2 \mathrm{~L}}.... (i)$
When dipped in water, pipe becomes closed at one end and open at the other.
Length available for resonance is
$\begin{aligned}
l_1 & =25 \% \times \mathrm{L} \\
& =\mathrm{L} \times \frac{25}{100} \\
& =\mathrm{L} / 4
\end{aligned}$
$\therefore \quad \mathrm{n}_{\text {closed }}=\frac{\mathrm{v}}{4 l_1}=\frac{\mathrm{v}}{4 \times \frac{\mathrm{L}}{4}}=\frac{\mathrm{v}}{\mathrm{L}}.... (ii)$
Comparing (i) and (ii),
$\therefore \quad \frac{\mathrm{n}_{\text {closed }}}{\mathrm{n}_{\text {open }}}=\frac{\left(\frac{\mathrm{v}}{\mathrm{L}}\right)^{\prime}}{\left(\frac{\mathrm{v}}{2 \mathrm{~L}}\right)}=\frac{2}{1}$
When dipped in water, pipe becomes closed at one end and open at the other.
Length available for resonance is
$\begin{aligned}
l_1 & =25 \% \times \mathrm{L} \\
& =\mathrm{L} \times \frac{25}{100} \\
& =\mathrm{L} / 4
\end{aligned}$
$\therefore \quad \mathrm{n}_{\text {closed }}=\frac{\mathrm{v}}{4 l_1}=\frac{\mathrm{v}}{4 \times \frac{\mathrm{L}}{4}}=\frac{\mathrm{v}}{\mathrm{L}}.... (ii)$
Comparing (i) and (ii),
$\therefore \quad \frac{\mathrm{n}_{\text {closed }}}{\mathrm{n}_{\text {open }}}=\frac{\left(\frac{\mathrm{v}}{\mathrm{L}}\right)^{\prime}}{\left(\frac{\mathrm{v}}{2 \mathrm{~L}}\right)}=\frac{2}{1}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.