Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

A steel wire of length 1 m , mass 0.1 kg and uniform cross sectional area 10−6 m2 is rigidly fixed at both ends. The temperature of the wire is lowered by 20∘ C. If the transverse waves are set up plucking the string in the middle, calculate the frequency of the fundamental mode of vibration. (Y=2×1011  n/m2,  a=1.21×10−5 / ∘ C)

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya

a

22 Hz

b

33 Hz

c

44 Hz

d

11 Hz

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

n=12lTμ =12lYAα(Δt)ml    (∵T=YAα△t)     =12l(YAα(Δt)lm)       =12(1)2×1011×10−6×1.21×10−5×20×10.1     =124×121     =11 Hz
Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

  • promsvg

    live classes

  • promsvg

    progress tracking

  • promsvg

    24x7 mentored guidance

  • promsvg

    study plan analysis

download the app

gplay
mentor

Download the App

gplay
whats app icon
personalised 1:1 online tutoring
A steel wire of length 1 m , mass 0.1 kg and uniform cross sectional area 10−6 m2 is rigidly fixed at both ends. The temperature of the wire is lowered by 20∘ C. If the transverse waves are set up plucking the string in the middle, calculate the frequency of the fundamental mode of vibration. (Y=2×1011  n/m2,  a=1.21×10−5 / ∘ C)