Basic Maths and Units and DimensionsEasy
Question
If the capacitance of a nanocapacitor is measured in terms of a unit ′u′ made by combining the electronic charge ′e′ Bohr radius′a0′, Planck′s constant ′h′ and speed of light ′c′ then.
Options
A.
B.
C.
D.
Solution
[u] = [e]a[a0]b[h]c[c]d
[M-1L-2T+4A+2] = [A1T1]a [L]b[ML2T-1]c[LT-1]d
[M-1L-2T+4A+2] = [MCLb+2c+d Ta-c-dAa]
a = 2, b = 1, c = - 1, d = - 1
u =
[M-1L-2T+4A+2] = [A1T1]a [L]b[ML2T-1]c[LT-1]d
[M-1L-2T+4A+2] = [MCLb+2c+d Ta-c-dAa]
a = 2, b = 1, c = - 1, d = - 1
u =
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