A) 70%
B) 65.4%
C) 94.4%
D) 25.5%
Correct Answer: C
Solution :
\[{{P}_{sb}}={{P}_{c}}\,{{\left( \frac{{{m}_{a}}}{2} \right)}^{2}}={{P}_{c}}\frac{{{(0.5)}^{2}}}{4}=0.0625\,{{P}_{c}}\] Also \[P={{P}_{c}}\,\left( 1+\frac{m_{a}^{2}}{2} \right)={{P}_{c}}\,\left( 1+\frac{{{(0.5)}^{2}}}{2} \right)=1.125{{P}_{c}}\] \ % saving\[=\frac{(1.125{{P}_{c}}-0.0625{{P}_{c}})}{1.125{{P}_{c}}}\times 100=94.4%.\]
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