A) 51.2%
B) 62.6%
C) 15%
D) 21.2%
Correct Answer: A
Solution :
Fraction of the left \[\frac{m{{v}^{2}}}{\sqrt{2g}}\] Here, \[m(2g{{h}^{3}})\] Now, \[mg\left[ \frac{b}{2} \right]\] \[mg\left[ a+\frac{b}{2} \right]\] \[mg\left[ \frac{b-a}{2} \right]\] \[mg\left[ \frac{b+a}{2} \right]\] \[i={{i}_{1}}\cos \omega t+{{i}_{2}}\sin \omega t\] \[\frac{{{i}_{1}}+{{i}_{2}}}{2}\] If T is half period \[\frac{{{({{i}_{1}}+{{i}_{2}})}^{2}}}{\sqrt{2}}\] \[\frac{1}{\sqrt{2}}\sqrt{i_{1}^{2}+i_{2}^{2}}\] Again \[\frac{i_{1}^{2}+i_{2}^{2}}{2}\] \[\Omega \] \[\frac{30.8}{\sqrt{T}}\overset{0}{\mathop{A}}\,\]
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