Wire of glass!

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 V V volts, find the sum of digits of V V .

Assume :

  • Assume that the Earth is a perfect sphere with a radius of 6 , 372 k m 6,372km

  • ρ g l a s s = 1 0 13 m \rho_{glass}=10^{13}Ω m

  • e = 1.6 × 1 0 19 C e=1.6\times10^{-19}C

  • Assume a day to be of 24 24 hours

Note:

  • Enter 1 -1 if you think that the scenario is not possible


The answer is 11.

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1 solution

Zakir Husain
Jul 12, 2020

Length of glass cylinder = = Circumference of the Earth = 2 π 6372 × 1 0 3 m =2\pi6372\times10^{3}m

Cross section area of glass cylinder = π ( 1 0 9 ) 2 = π 1 0 18 m 2 =\pi(10^-9)^2=\pi10^{-18}m^2

Total resistance over the glass cylinder = ρ l A = 1 0 13 2 π 6372 × 1 0 3 m π 1 0 18 m 2 = ρ × 2 × 6372 × 1 0 21 m 1 =\rho \dfrac{l}{A}=10^{13}\dfrac{2\cancel{\pi}6372\times10^{3}m}{\cancel{\pi}10^{-18}m^2}={\rho\times2\times6372\times10^{21}}m^{-1}

= 2 × 6372 × 1 0 24 =2\times6372\times10^{24}Ω

Electric current = e 60 × 60 × 24 s = 1.6 × 1 0 19 86400 A = 1 54 × 1 0 22 A =\dfrac{e}{60\times60\times24s}=\dfrac{1.6\times10^{-19}}{86400}A=\dfrac{1}{54}\times10^{-22}A

From ohm's law: V = I R = 2 × 6372 × 1 0 24 × 1 54 × 1 0 22 V = 118 × 2 × 1 0 12 V = 236 × 1 0 12 V V=IR=2\times6372\times10^{24}\times\dfrac{1}{54}\times10^{-22}\red{V}=118\times2\times10^{12}\red{V}=236\times10^{12}\red{V}

Sum of digits of V = 2 + 3 + 6 = 8 + 3 = 11 =2+3+6=8+3=\boxed{11}

Note :

  • V \red{V} indicates 'volts'

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