A faulty resonance tube

Classical Mechanics Level 5

A 3.6 m 3.6 \text{ m} long vertical pipe is filled completely with a liquid. A small hole is drilled at the base of the pipe due to which the liquid starts leaking out. This pipe resonates with a tuning fork. The first two resonances occur when height of water column is 3.22 m 3.22 \text{ m} and 2.34 m 2.34 \text{ m} respectively. What is the area of cross-section of the pipe?

Take exact value of π \pi and state the answer in cm 2 \text{cm}^{2}


The answer is 314.15.

Tanishq Varshney by Tanishq Varshney , 23, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Mvs Saketh
Apr 19, 2015

use the fact that 0.6 R is the correction term (apparent extra length) of the tube

5 Helpful 0 Interesting 0 Brilliant 0 Confused

yes as simple as that ⌣ ¨ \ddot \smile

Tanishq Varshney - 6 years, 1 month ago

https://brilliant.org/problems/projectile-velocity/?group=AJ66ctqqSEx5&ref_id=648396 hey just try this problem created by me and feel free to comment on any error.

Athul Nambolan - 6 years, 1 month ago
Aditya Chauhan
May 30, 2016

The end correction of the tube is equal to e = 0.6 r e=0.6r

Acc. to ques 38 + e 126 + e = 1 3 \dfrac{38+e}{126+e}=\dfrac{1}{3}

e = 6 e=6

0.6 r = 6 0.6r=6

r = 10 r=10

Area= 100 π \boxed{100\pi}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...