Waves

JEE Main level, numerical answer, medium

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Question 13 of 142

The mass per unit length of a uniform wire is 0.135gcm−10.135g{cm}^{-1}. A transverse wave of the form y=−0.21sin⁡(x+30t)y=-0.21\sin \left(x+30t\right) is produced in it, where xx is in meter and tt is in second. Then, the expected value of tension in the wire is x×10−2Nx\times {10}^{-2}N. Value of xx is (Round-off to the nearest integer)

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Answer: 1215

μ=0.135gmcm−1=0.0135kgm−1\mu =0.135gm{cm}^{-1}=0.0135kg{m}^{-1}
y=−0.21sin⁡(x+30t)y=-0.21\sin \left(x+30t\right)
(xx in meter & tt in sec)
v=ωk=301=30ms−1v=\dfrac{\omega }{k}=\dfrac{30}{1}=30m{s}^{-1}
v=Tμ⇒T=v2μ=(30)2(0.0135)v=\sqrt{\dfrac{T}{\mu }}\Rightarrow T={v}^{2}\mu ={\left(30\right)}^{2}\left(0.0135\right)
=12.15=12.15
=x×10−2N=x\times {10}^{-2}N
⇒x=1215\Rightarrow x=1215

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