Question

a lead wire and an iron wire are having same dimensions.their respective specific resistances are in the ratio 49:24.the ratio of their resistances are
a)49:24
b)3:5
c)9:4
d)7:4​

Answer 1

Given:

Ratio of specific resistances of the lead wire and the iron wire = 49:24

To find:

The ratio of the resistances of the lead wire and the iron wire.

Solution:

Tge resistance of a wire is given by:

$r = \rho \frac{l}{a}$

Where r is the resistance of the wire,

$\rho$

is the specific resistance of the wire,

l is the length of the wire,

And a is the cross-sectional area of the wire.

If the dimensions of the wire are same, then ratio of resistances will be equal to ratio of specific resistances.

Hence the required ratio is 49:24.

The correct ootion is (a).

Answer 2

Given:

A lead wire and an iron wire are having same dimensions.their respective specific resistances are in the ratio 49:24.

To find:

Ratio of resistance?

Calculation:

General expression of resistance of a wire is:

$\boxed{R = \rho \times \dfrac{l}{a} }$

• $\rho$ is specific resistance (or resistivity) , 'l' is length and 'a' is area of cross-section.

Now, required ratio will be :

$\therefore \: \dfrac{R1}{R2} = \dfrac{ \bigg \{\dfrac{ (\rho1)l}{a} \bigg \} }{\bigg \{\dfrac{ (\rho2)l}{a} \bigg \}}$

$\implies \: \dfrac{R1}{R2} = \dfrac{ \rho1}{ \rho2}$

$\implies \: R1 : R2 = \rho1 :\rho2$

$\implies \: R1 : R2 = 49:24$

So, the ratio of resistance is 49 : 24.