Hiii shrikant answer to your question is
=1μ∈−−−√v=1μ∈=1(μoμr∈o∈r)−−−−−−−−−√=1(μoμr∈o∈r)
=cμr∈r−−−−√=cμr∈r
⇒v2=2μr∈r⇒v2=2μr∈r
⇒∈r=c2μrv2⇒∈r=c2μrv2=(3×108)2(2×108)2=(3×108)2(2×108)2=2.25
Final answer is 2.25
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