A) 0.3 s
B) 0.05 s
C) 0.03 s
D) 0.5 s
Correct Answer: B
Solution :
Given, mass per unit length (L) \[\mu =0.25\,g\text{/}cm\] \[\Rightarrow \] \[\frac{10}{L}=0.25\] \[\Rightarrow \] \[L=40\,cm\] Stiffness of spring, \[k=160\,N{{m}^{-1}}\] deformation, \[x=1\text{ }cm=0.01\text{ }m\] Now, force (restoring),\[T=kx=160\times 0.01\] \[=16N=16\times {{10}^{4}}\text{dyne}\] But, \[v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{16\times {{10}^{4}}}{0.25}}\] \[=8\times {{10}^{2}}cm\text{/}s\] The time taken by pulse to reach the spring \[t=\frac{40}{800}=\frac{1}{20}=0.05\,s\]You need to login to perform this action.
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