A) 27 h
B) 8.3 h
C) 11h
D) 16h
Correct Answer: D
Solution :
\[{{e}^{{{a}_{1}}+{{a}_{2}}+...+{{a}_{n}}}}\] Equivalent of \[\frac{{{a}_{1}}+{{a}_{2}}+...+{{a}_{n}}}{n}\] According to first law of faraday electrolysis \[{{a}_{1}}{{a}_{2}}...{{a}_{n}}\] \[\int_{{}}^{{}}{\frac{{{\sec }^{2}}x}{{{(\sec x+\tan x)}^{9/2}}}dx}\] \[-\frac{1}{{{(\sec x+\tan x)}^{11/2}}}\left\{ \frac{1}{11}-\frac{1}{7}{{(\sec x+\tan x)}^{2}} \right\}+K\] \[\frac{1}{{{(\sec x+\tan x)}^{11/2}}}\left\{ \frac{1}{11}-\frac{1}{7}{{(\sec x+\tan x)}^{2}} \right\}+K\]\[-\frac{1}{{{(\sec x+\tan x)}^{11/2}}}\left\{ \frac{1}{11}+\frac{1}{7}(\sec x+\tan x) \right\}+K\]
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