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Q.A rod has λ = k sqrt(x). The potential at one end is

a
2kL3/23×4πϵ0\dfrac{2 k L^{3/2}}{3 \times 4\pi \epsilon_0}
b
kL4πϵ0\dfrac{k L}{4\pi \epsilon_0}
c
kL24πϵ0\dfrac{k L^2}{4\pi \epsilon_0}
d
none

Correct Answer: Option A

The correct solution involves applying the fundamental concept to derive the final value step by step...

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