wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A resonance tube is resonated with tuning fork of frequency 256 Hz. If the length of first and second resonating air columns are 32 cm and 100 cm, then end correction will be

A
1 cm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
2 cm
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
4 cm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
6 cm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 2 cm
First resonating length:

λ4=L1+e

Here,

λ is wavelength

L1 is length of first resonating column

e is the end correction

λ4=32+e ...........(1)



Second resonating length:

3λ4=L2+e

Here,

λ is wavelength

L2 is length of second resonating column

e is the end correction

3λ4=100+e ...........(2)



Solving the equation (1) and (2),
(2) - (1) gives,

λ2=10032

λ2=68

λ=136cm

Hence,

λ4=32+e

e=136432

=3432=2cm

Therefore the end correction is equal to 2 cm



flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Pascal's Law
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon