A) \[1.641\times {{10}^{-19}}C\]
B) \[1.630\times {{10}^{-19}}C\]
C) \[1.648\times {{10}^{-19}}C\]
D) \[1.602\times {{10}^{-19}}C\]
Correct Answer: A
Solution :
Any charge in the universe is given by \[q=ne\Rightarrow e=\frac{q}{n}\] (where n is an integer) \[{{q}_{1}}:{{q}_{2}}:{{q}_{3}}:{{q}_{4}}:{{q}_{5}}:{{q}_{6}}::{{n}_{1}}:{{n}_{2}}:{{n}_{3}}:{{n}_{4}}:{{n}_{5}}:{{n}_{6}}\] \[6.563:8.204:11.5:13.13:16.48:18.09\]\[::{{n}_{1}}:{{n}_{2}}:{{n}_{3}}:{{n}_{4}}:{{n}_{5}}:{{n}_{6}}\] Divide by 6.563 \[1:1.25:1.75:2.0:2.5:2.75\]\[::{{n}_{1}}:{{n}_{2}}:{{n}_{3}}:{{n}_{4}}:{{n}_{5}}:{{n}_{6}}\] Multiplied by 4 \[4:5:7:8:10:11\]\[::{{n}_{1}}:{{n}_{2}}:{{n}_{3}}:{{n}_{4}}:{{n}_{5}}:{{n}_{6}}\] \[e=\frac{{{q}_{1}}+{{q}_{2}}+{{q}_{3}}+{{q}_{4}}+{{q}_{5}}+{{q}_{6}}}{{{n}_{1}}+{{n}_{2}}+{{n}_{3}}+{{n}_{4}}+{{n}_{5}}+{{n}_{6}}}=\frac{73.967\times {{10}^{-19}}}{45}\] \[=1.641\times {{10}^{-19}}C\] (Note : If you take 45.0743 in place of 45, you will get the exact value)
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