# A three quarter inch wire has 12 ohms resistance. How much resistance has the same length of...

## Question:

A three quarter inch wire has {eq}12 {/eq} ohms resistance. How much resistance has the same length of half-inch wire, if resistance varies inversely as the square of the diameter?

## Variation:

(i) If x varies directly as y then {eq}x=my {/eq}, where 'm' is a proportionality constant.

(ii) If x varies inversely as y then {eq}x= \dfrac{n}{y} {/eq}, where 'n' is a proportionality constant.

Let us assume that the diameter and the resistance are denoted by {eq}d {/eq} and {eq}R {/eq} respectively.

The problem says, "resistance varies inversely as the square of the diameter".

So by the definition of inverse variation:

$$R = \dfrac{k}{d^2} \,\,\,\,\,\, \rightarrow (1)$$

The problem says {eq}R=12 {/eq} when {eq}d= \dfrac{3}{4} {/eq}.

Substitute these values in (1):

$$12 = \dfrac{k}{ \left( \dfrac{3}{4} \right)^2} \\ 12 = \dfrac{k}{ \dfrac{9}{16}} \\ \text{Multiply both sides by } \dfrac{9}{16}, \\ k= 12 \times \dfrac{9}{16} = \dfrac{27}{4}$$

Substitute this and {eq}d= \dfrac{1}{2} {/eq} in (1) now:

\begin{align} R&= \dfrac{ \left( \dfrac{27}{4}\right)}{ \left( \dfrac{1}{2} \right)^2} \\ &= \dfrac{27}{4} \times \dfrac{4}{1} \\ &= \boxed{\mathbf{27}} \end{align}

Therefore, the required resistance is {eq}\boxed{\mathbf{27}} {/eq} ohms. 