A) 500
B) 50
C) 5
D) 10
Correct Answer: B
Solution :
Drop is stationary,\[ie\], weight of drop = electric force \[ie\], \[mg=qE\] or \[mg=q\frac{V}{d}\] or \[mg=ne\frac{V}{d}\] \[\therefore \] \[n=\frac{mgd}{eV}\] \[=\frac{1.8\times {{10}^{-13}}\times 10\times 0.9\times {{10}^{-2}}}{1.6\times {{10}^{-19}}\times 2000}\] \[n=50\]You need to login to perform this action.
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