Question

two different wires have specific resistivity lengths, area of cross section are in the ratio 4:3,9:2 and 27:8.then the ratio of resistance of two wires​

Answer 1

Given that

$\boxed{ \frac{\rho_1}{\rho_2} = \frac{4}{3}} \\ \\ \\ \boxed{ \frac{l_1}{l_2} = \frac{9}{2}} \\ \\ \\ \boxed{ \frac{a_1}{a_2} = \frac{27}{8}}$

Where,

$\rho$ represents the Resistivity of two wires, l represents the length of two wires and a represents the area of cross section of two wires.

Now,

We already know that :-

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

So,

Let the resistance of first wire be R1 and second wire be R2.

So,

R1/R2 is equal to :-

$\frac{r1}{r2} = \frac{ \frac{\rho1 \times l1}{a1} }{ \frac{\rho2 \times l2}{a2} } \\ \\ \\ \frac{r1}{r2} = \frac{\rho1 \times l1 \times a2}{\rho2 \times l2 \times a1} \\ \\ \\ = \sf{\frac{4}{3} \times \frac{9}{2} \times \frac{8}{27}} \\ \\ \\ = \sf{\frac{16}{9}}$

So,

The ratio of their resistances is :-

$\boxed{ \sf{16:9}}$

Answer 2

Answer :-

Given :-

Ratio of two Wires Resistivity = 4 : 3.

=> Resistivity of Wire1 & Wire2 is 4 & 3.

Ratio of Length = 9 : 2

=> Length of two wires are 9 & 2 respt.

& Ratio of Cross-Section Area = 27 :8.

=> Cross Section Area of two wires is 27 & 8 respt.

Now,

By the Resistance Formula. we get,

Resistance of Wire1 / Resistance of Wire2

=> (4 × 9/27) / (3×2/8)

=> 8*(4×9) / 27*(3×2)

=> 16 : 9.

Hence,

The Ratio of Resistance of the wires 16 : 9.