A) 4/3
B) 16/9
C) 15/12
D) 32/27
Correct Answer: D
Solution :
\[R:h\]\[6\times {{10}^{15}}Hz.\] \[1.324\times {{10}^{15}}Hz\] \[2.295\times {{10}^{16}}Hz\] \[3.678\times {{10}^{18}}z\]\[2.7\times {{10}^{14}}Hz\] \[{{y}_{1}}=\alpha \sin kx\cos \omega t\]\[{{y}_{2}}=\alpha \sin (\omega t-kx).\]\[{{x}_{1}}=\frac{\pi }{3k}\]\[{{x}_{2}}=\frac{3\pi }{2k}\] Similarly m second case,\[{{\phi }_{1}}\] \[{{\phi }_{2}}\]\[\frac{{{\phi }_{1}}}{{{\phi }_{1}}}\] Here, \[{{a}_{0}}=g/2.\]specific gravity (say S) \[50\mu A\]\[1.1924\times {{10}^{-8}}N\]\[2.1\times {{10}^{-8}}N\]\[1.6\times {{10}^{-8}}N\]
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