A) 1s
B) 0.5s
C) 2s
D) 1.5s
Correct Answer: A
Solution :
At a distance y from bottom end consider a string element of length dy. Tension at this height is, \[T=\frac{m}{l}\times yg\] So, \[\frac{dy}{dt}=\] velocity of wave \[=\sqrt{\frac{T}{\mu }=}\sqrt{gy}\] \[\Rightarrow \]\[\int\limits_{0}^{2.45}{\frac{dy}{\sqrt{gy}}=\int\limits_{0}^{t}{dt}}\]\[\Rightarrow \]\[t=1s\]
You need to login to perform this action.
You will be redirected in
3 sec
