Question

Taking the resistivity of platinoid as $3.3\times {10}^{-7}\Omega m$, find the resistance of $7.0m$ of platinoid wire of average diameter $0.14cm$.

• A. $25\Omega$
• B. $1.5\Omega$
• C. $2.5\Omega$
• D. $15\Omega$

Answer 1

The correct option is B $1.5\Omega$

Let the platinoid wire of uniform cross-section has a resistance of $R$.
Let the diameter be $d$.
Let the length of the wire be $l$.
According to question:
$l=7m$
$d=0.14cm$
Cross sectional Area be $A$:
$A=\pi \frac{d^2}{4}$ = $\pi \times \frac{{.0014}^2}{4}$
$\therefore$ $A=15.4$$\times {10}^{-}{}^7{m}^2$
Now the resistance of a wire can be expressed as:
$R=\rho \frac{l}{A}$

Where resistivity is $\rho$.
$\rho =3.3\times {10}^{-}{}^7$ $\Omega m$
Putting the values of $l,A$ and $\rho$ in Eq.1
$R=$ $\frac{3.3\times {10}^{-7}}{15.4\times {10}^{-7}}$ $\times 7$ $=1.5$ ohm