A) 20 N
B) 40 N
C) 10N
D) 120 N
Correct Answer: B
Solution :
For mass m per unit length of wire and tension T, the frequency of note emitted by the wire is \[7.25\times {{10}^{6}}m/s\] where \[h=A+B{{\lambda }^{2}}+C{{\lambda }^{2}}\] is length of wire. Given, \[\mu =A+B{{\lambda }^{2}}+C{{\lambda }^{4}}\] and \[\mu =A+B{{\lambda }^{2}}+C{{\lambda }^{4}}\] \[\mu =A+B{{\lambda }^{-2}}+C{{\lambda }^{-4}}\] \[\mu \] \[\overset{0}{\mathop{A}}\,\] \[\overset{0}{\mathop{A}}\,\] \[\overset{0}{\mathop{A}}\,\]
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