Question

Two wires A and B with circular Cross-sections having identical lengths and are made of the same material. Yet, wire A has four times the resistance of wire B. How many times greater is the diameter of wire B than wire A?​

Answer 1

Answer:

Explanation:using the formula rho(l)/area we have to solve

Here rho is resistivity of material

L is length of wire

A is area of cross section of wire

And it is given that rho is same length is same . Hence solve it as follows

Answer 2

Answer:

The diameter of wire B is 2 times greater than the diameter of wire B.  

Explanation:

Given:

$L_{A}=L_{B}$

$\rho_{A}=\rho_{B}$

Because both wires are made of the same material.

Here,

The length of wire A is denoted by $L_{A}$.

The length of wire B is denoted by $L_{B}$.

The area of cross-section of the wire A is denoted by $A_{A}$.

The area of cross-section of the wire B is denoted by $A_{B}$.

The resistivity of wire A is denoted by $\rho_{A}$.

The resistivity of wire B is denoted by $\rho_{B}$.

The resistance of wire A is denoted by $R_{A}$.

The resistance of wire B is denoted by $R_{B}$.

The diameter of wire A is denoted by $D_{A}$.

The diameter of wire B is denoted by $D_{B}$.

Now,

By the equation,

$R_{A} = \frac{\rho_{A} L_{A} }{A_{A} }$

$R_{A} = \frac{\rho_{A} L_{A} }{\frac{\pi \times D_{A} ^{2} }{4} }$

$R_{A} = \frac{4\rho_{A} L_{A} }{\pi \times D_{A} ^{2} }$                             .......(1)

Then,

By the equation,

$R_{B} = \frac{\rho_{B} L_{B} }{A_{B} }$

$R_{B} = \frac{\rho_{B} L_{B} }{\frac{\pi \times D_{B} ^{2} }{4} }$

$R_{B} = \frac{4\rho_{B} L_{B} }{\pi \times D_{B} ^{2} }$                            .......(2)

Then,

According to the question,

$R_{A} =4R_{B}$

Putting Equation one and two,    

$\frac{4\rho_{A} L_{A} }{\pi \times D_{A} ^{2} }=4 \frac{4\rho_{B} L_{B} }{\pi \times D_{B} ^{2} }$

$\frac{4\rho_{A} L_{A} }{\pi \times D_{A} ^{2} }=4 \frac{4\rho_{A} L_{A} }{\pi \times D_{B} ^{2} }$                     ∴ $L_{A}=L_{B}$ and $\rho_{A}=\rho_{B}$    

$\frac{D_{A} }{D_{B} } =\frac{1}{2}$

So, the diameter of wire B is 2 times greater than the diameter of wire B.