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 two supports 40.05 cm apart. In its fundamental mode, it vibrates with a frequency 100064 Hz. Find the Young's modulus of the wire.

A
1×109 N-m2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
2×109 N-m2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
3×109 N-m2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
4×109 N-m2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is A 1×109 N-m2As linear mass density is μ=3.2 gm40.05 cm=3.2×10−340.05×10−2≈324000 kg/m For fundamental mode, l=λ2 ⇒λ=2l We use, fundamental frequency n0=Vλ=12l√Tμ ⇒100064=12×40.05×10−2  ⎷T324000 ⇒[100064×2×40.05×10−2]2324000=T ⇒(100064)2×4×1600×10−4×324000≈T ⇒T≈108 N From Hooke's law, Young's modulus can be calculated as Y=T/AΔL/L=10×0.48×10−6×0.05×10−2=109 N/m2

Suggest Corrections
10
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program