If the wire has resistivity ρ per unit area of cross-section, the resistance between A and B is

Question

If the wire has resistivity ρ per unit area of cross-section, the resistance between A and B is (A) (ρ l/√2) (B) √2 ρ l (C) 2 ρ l (D) none of these

Tardigrade Admin · Accepted Answer

The figure can be redrawn as: <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/7f9640ea38d50d8b9ef1a0e3d08eeb1e-.png" /> R=(ρ l/2) R1=(ρ l/√2) R eq =(1/2)[(2 R R1/2 R+R1)+2 R]=(ρ l/√2)

Q. If the wire has resistivity $ρ$ per unit area of cross-section, the resistance between $A$ and $B$ is

$\frac{ρl}{2}$

$2 ρl$

$2 ρl$

none of these

The figure can be redrawn as:

$R = \frac{ρl}{2}$
$R_{1} = \frac{ρl}{2}$
$R_{e q} = \frac{1}{2} [\frac{2 R R _{1}}{2 R + R _{1}} + 2 R] = \frac{ρl}{2}$

Q. If the wire has resistivity ρ per unit area of cross-section, the resistance between A and B is

2​ρl​

2​ρl

2ρl

none of these

The figure can be redrawn as: R=2ρl​ R1​=2​ρl​ Req​=21​[2R+R1​2RR1​​+2R]=2​ρl​

Q. If the wire has resistivity $ρ$ per unit area of cross-section, the resistance between $A$ and $B$ is

$\frac{ρl}{2}$

$2 ρl$

$2 ρl$

The figure can be redrawn as:

$R = \frac{ρl}{2}$
$R_{1} = \frac{ρl}{2}$
$R_{e q} = \frac{1}{2} [\frac{2 R R _{1}}{2 R + R _{1}} + 2 R] = \frac{ρl}{2}$