A) \[=-48.78\]
B) \[G=\frac{\text{Current sensitivity}}{\text{Voltage sensitivity}}\]
C) \[=\frac{10}{2}=5\Omega \]
D) None of the above
Correct Answer: B
Solution :
\[^{47}{{C}_{4}}+\sum\limits_{r=1}^{5}{{{C}_{3}}{{=}^{47}}{{C}_{4}}{{+}^{51}}{{C}_{3}}{{+}^{50}}{{C}_{3}}{{+}^{49}}{{C}_{3}}}\] \[{{+}^{48}}{{C}_{3}}{{+}^{47}}{{C}_{3}}\] \[{{=}^{51}}{{C}_{3}}{{+}^{50}}{{C}_{3}}{{+}^{49}}{{C}_{3}}{{+}^{48}}{{C}_{3}}+{{(}^{47}}{{C}_{3}}{{+}^{47}}{{C}_{4}})\] \[{{=}^{51}}{{C}_{3}}{{+}^{50}}{{C}_{3}}{{+}^{49}}{{C}_{3}}+{{(}^{48}}{{C}_{3}}{{+}^{480}}{{C}_{4}})\] \[[{{\because }^{n}}{{C}_{r}}{{+}^{n}}{{C}_{r-1}}{{=}^{n+1}}{{C}_{r}}]\] \[{{=}^{52}}{{C}_{4}}\]
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