A) 0,456 J
B) 2.65 J
C) 0.2355 J
D) 3.95 J
Correct Answer: C
Solution :
The potential energy of circular loop in the magnetic field \[\frac{{{N}_{2}}}{{{N}_{1}}}=1.1577\times {{10}^{-38}}\]where \[{{E}_{n}}=\frac{-13.6}{({{n}^{2}})}eV\] is the angle between normal to plane of coil and magnetic field B. Initial potential energy,\[\therefore \] When coil is turned through 180°, therefore final potential energy, \[E=\frac{-13.6}{{{(5)}^{2}}}eV=-0.54eV\] \[iG=(i-{{i}_{0}})S\]Required work, W = gain in potential energy \[{{i}_{0}}\] Here, \[\frac{i}{{{i}_{0}}}=\frac{S}{G+S}\]and radius r = 0.05 m \[S=2.5\Omega ,G=25\Omega ,\]\[\frac{i}{{{i}_{0}}}=\frac{1}{11}\] \[\because \]Work, \[{{v}_{rms}}=\sqrt{\frac{3RT}{M}}\] \[{{v}_{sound}}=\sqrt{\frac{\gamma RT}{M}},\]\[{{v}_{rms}}=2{{v}_{sound}}\]
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