If along a uniform rod of length l carrying current I , the voltage V changes with position x along the length of the rod such that dV/dx=-k , where k is a positive number, then the resistance of the rod is

Question

If along a uniform rod of length l carrying current I , the voltage V changes with position x along the length of the rod such that dV/dx=-k , where k is a positive number, then the resistance of the rod is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-zqjtq2j2g46iqexm.jpg" /> (A) kl/2I (B) kl/I (C) I/kl (D) klI

Tardigrade Admin · Accepted Answer

dV=-IdR=-Iρ (d x/A) (d V/d x)=-I(ρ /A)=-k ⇒ R=kl/I II Method: V2-V1= displaystyle ∫ ((d V/d ℓ ))dl=kL=IR

Q. If along a uniform rod of length $l$ carrying current $I$ , the voltage $V$ changes with position $x$ along the length of the rod such that $d V / d x = - k$ , where $k$ is a positive number, then the resistance of the rod is

$k l /2 I$

$k l / I$

$I / k l$

$k l I$

$d V = - I d R = - I ρ \frac{d x}{A}$
$\frac{d V}{d x} = - I \frac{ρ}{A} = - k$
$\Rightarrow R = k l / I$
II Method:
$V_{2} - V_{1} = \int (\frac{d V}{d ℓ}) d l = k L = I R$

Q. If along a uniform rod of length l carrying current I , the voltage V changes with position x along the length of the rod such that dV/dx=−k , where k is a positive number, then the resistance of the rod is

kl/2I

kl/I

I/kl

klI