A) 77/80
B) 71/80
C) 31/32
D) 15/16 By using \[\Delta l.\] \[\Delta T\]\[\Delta l\] \[\Delta T\]\[\rho \] \[100\Omega \]\[{{f}_{1}}\] Hence,\[{{f}_{2}}\] \[n=3,{{f}_{2}}=\frac{3}{4}{{f}_{1}}\]\[n=5,{{f}_{2}}=\frac{3}{4}{{f}_{1}}\]\[n=3,{{f}_{2}}=\frac{5}{4}{{f}_{1}}\]\[n=5,{{f}_{2}}=\frac{5}{4}{{f}_{1}}\] So, fraction that decays \[{{T}_{1}}\]
Correct Answer: C
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