A conductor with rectangular cross-section has dimensions (a × 2 a × 4 a) as shown in the figure. Resistance across AB is x , across CD is y and across EF is z . Then - .

Question

A conductor with rectangular cross-section has dimensions (a × 2 a × 4 a) as shown in the figure. Resistance across AB is x , across CD is y and across EF is z . Then - . <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-a6wmnhoolwcc.png" /> (A) x=y=z (B) x>y>z (C) y>z>x (D) x>z>y

Tardigrade Admin · Accepted Answer

We know that R=(ρ l/A) x=(ρ (4 a)/a × 2 a)=(2 ρ /a) y=(ρ (a)/2 a × 4 a)=(ρ /8 a) z=(ρ (2 a)/a × 4 a)=(ρ /2 a) ∴ x>z>y

Q. A conductor with rectangular cross-section has dimensions $(a \times 2 a \times 4 a)$ as shown in the figure. Resistance across $A B$ is $x$ , across $C D$ is $y$ and across $EF$ is $z$ . Then -

.

$x = y = z$

$x > y > z$

$y > z > x$

$x > z > y$

We know that $R = \frac{ρl}{A}$
$x = \frac{ρ ( 4 a )}{a \times 2 a} = \frac{2 ρ}{a}$
$y = \frac{ρ ( a )}{2 a \times 4 a} = \frac{ρ}{8 a}$
$z = \frac{ρ ( 2 a )}{a \times 4 a} = \frac{ρ}{2 a}$
$∴ x > z > y$

Q. A conductor with rectangular cross-section has dimensions (a×2a×4a) as shown in the figure. Resistance across AB is x , across CD is y and across EF is z . Then - .

x=y=z

x>y>z

y>z>x

x>z>y