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

A 40 cm long wire having a mass 3.2 gm and area of cross section 1 mm2 is stretched between the support 40.05 cm apart. In its fundamental mode, it vibrates with a frequency of 100064 Hz. The young's modulus of the wire is 10x N/m2, then x is.

A
9.0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
9
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
9.00
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

μ=3.2 gm40 cm=3.2×10340×102=3.2400=324000 kg/m
l=λ2
λ=2l.......(1)

f=vλ=12lTμ
100064=12×40×102T32/4000
[100064×2×40×102]2324000=T
1000064×324000=TT=108 N
y=10/8106.05×10240×102=107840(.05)=109 N/m2.

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