wiz-icon
MyQuestionIcon
MyQuestionIcon
7
You visited us 7 times! Enjoying our articles? Unlock Full Access!
Question

An object of density ρ=500 kg/m3 is hanging from a thin steel wire. The fundamental frequency of transverse standing wave on the wire is 200 Hz. The set up is now immersed in water, so that one third of the object's volume is submerged. The new fundamental frequency (in Hz) is

A
2003
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
200
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2003
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
100
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 2003
Density of water (ρw)=1000 kg/m3



Fundamental frequency on a stretched thin wire
f=12lTμ

In air : T=mg=Vρg [ V is the volume of object]
f=12lVρgμ .........(1)

When object is one third immersed in water, then
T=mgupthrust =VρgV3ρwg=V3g(3ρρw)
The new fundamental frequency is
f=12lTμ=12l  Vg3(3ρρw)μ .......ii)

From (1) and (2)

ff=  Vg3(3ρρw)Vρg=(1ρw3ρ)

From the data given,

f=(110003×500)×200=2003 Hz

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon