A) 1
B) 10
C) 100
D) 1000
Correct Answer: A
Solution :
From the relation, \[q\,E=mg\] \[\left( \text{Since,}\,E=\frac{V}{d} \right)\] \[\left( E=\frac{100}{0.016} \right)\] \[\therefore \] \[q=\frac{mg}{E}\] ?(i) \[\left( \begin{align} & \text{Given: m =1}{{\text{0}}^{-13}}\,g \\ & ={{10}^{-16}}\,\text{kg} \\ & d=0.016 \\ & V=100\,\text{volt} \\ & e\text{ }=\text{ }1.6\times {{10}^{-19}}\text{coulomb} \\ & g\text{ }=\text{ }10\text{ }m/{{s}^{2}} \\ \end{align} \right)\] Now, putting the given value and value of E then form (ii) in Eq. (i) \[q=\frac{{{10}^{-6}}\times 10\times 0.016}{100}\,\text{coulomb}\] Number of charges are given as \[n=\frac{q}{e}\] \[=\frac{{{10}^{-16}}\times 10\times 0.016}{100\times 1.6\times {{10}^{-19}}}\] = 1You need to login to perform this action.
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