Use app×
QUIZARD
QUIZARD
JEE MAIN 2026 Crash Course
NEET 2026 Crash Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
107 views
in Physics by (90.2k points)
closed by
A metal string is fixed between rigid supports. It is initially at negligible tensin. Its Young modulus is `Y`, density `rho` and coefficient of thermal expansion is `alpha`. If it is now cooled through a temperature `=t`, transverse waves will move along it with speed
A. `Ysqrt(alphat)/rho`
B. `alphatsqrt(Y/rho)`
C. `sqrt((Yalphat)/rho)`
D. `tsqrt((Yalpha)/rho)`

1 Answer

0 votes
by (90.9k points)
selected by
 
Best answer
c) From, `Y = (F//A)/(DeltaL//L)=(T.L)/(ADeltaL)`
where, `F=T="tension"`.
On cooling, `DeltaL = alphaL(Deltatheta)= alphaLt`
`thereforeY = (TL)/(AalphaLt)`
`T=YAalphaT`
Also, mass per unit length of string
`m=Arho`
As wave velocity = `=sqrt(T/M)` `therefore` v=`sqrt((YAalphat)/(Arho)) = sqrt((Yalphat)/rho)`

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...