There is a cylindrical piece of glass of radius 1 nanometre and length equal to the circumference of the Earth (along the equator).
This glass cylinder is then connected to an electric battery, due to which a current flow such that through a given point (on the glass cylinder) only one electron passes per day.
Now if the potential difference between the two endpoints of the glass cylinder is volts, find the sum of digits of .
Assume :
Assume that the Earth is a perfect sphere with a radius of
Assume a day to be of hours
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Length of glass cylinder = Circumference of the Earth = 2 π 6 3 7 2 × 1 0 3 m
Cross section area of glass cylinder = π ( 1 0 − 9 ) 2 = π 1 0 − 1 8 m 2
Total resistance over the glass cylinder = ρ A l = 1 0 1 3 π 1 0 − 1 8 m 2 2 π 6 3 7 2 × 1 0 3 m = ρ × 2 × 6 3 7 2 × 1 0 2 1 m − 1
= 2 × 6 3 7 2 × 1 0 2 4 Ω
Electric current = 6 0 × 6 0 × 2 4 s e = 8 6 4 0 0 1 . 6 × 1 0 − 1 9 A = 5 4 1 × 1 0 − 2 2 A
From ohm's law: V = I R = 2 × 6 3 7 2 × 1 0 2 4 × 5 4 1 × 1 0 − 2 2 V = 1 1 8 × 2 × 1 0 1 2 V = 2 3 6 × 1 0 1 2 V
Sum of digits of V = 2 + 3 + 6 = 8 + 3 = 1 1
Note :