Question

two copper wires have their masses in the ratio 2:3 and length in the ratio 3:4 find the ratio of their resistances

Answer 1

GIVEN,

Masses of a two copper wire are in ratio = 2:3

Lengths are in ratio = 3:4

So for masses let the common ratio be x,

So for first wire  its mass is = 2x.

For second wire its mass is 3x.

For length let the common ratio be l.

So for first wire  its mass is = 3l.

For second wire its mass is 4l.

We know,

$Resistance = \rho\dfrac{l}{A}$

$Since,\\\\Volume = Area \times length\\\\\dfrac{1}{A}= \dfrac{Length}{Volume}$

$Resistance = \rho \times\dfrac{l^{2}}{volume}$

Since,

$Volume=\dfrac{Mass}{Density}$

So,

$Resistance = \rho \times\dfrac{l^{2} \times density}{mass}$

So ,

For First wire,

$Resistance R_1 = \rho \times\dfrac{3l^{2} \times density}{2x}$

$Resistance R_1 = \rho \times\dfrac{ 9l^{2}\times density}{2x}$

For First wire,

$Resistance R_2= \rho \times\dfrac{4l^{2} \times density}{3x}$

$Resistance R_2 = \rho \times\dfrac{ 16l^{2}\times density}{3x}$

On dividing resistance of first wire by second,

$\dfrac{R_1}{R_2} = \rho \times\dfrac{ 9l^{2}\times density}{2x} \div\rho \times\dfrac{ 16l^{2}\times density}{3x}$

$\dfrac{R_1}{R_2} =\dfrac{ \rho \times 9l^{2}\times density}{2x} \times \dfrac{3x}{ \rho \times 16l^{2}\times density}$

on solving we get,

$\dfrac{R_1}{R_2} =\dfrac{9\times 3}{ 2 \times 16}$

$\dfrac{R_1}{R_2} =\dfrac{27}{ 32}$

So, the ratio of their resistances will be  27:32.

Answer 2

Answer:

Required rasistance ratio is 27:32.

Explanation:

Given ratio of masses of two copper wires is 2:3.

And ratio of their lengths is 3:4.

Let mass of first copper wire be 2x and mass of second copper wire be 3x.

Again let length of first copper wire be 3y and length of Second copper wire be 4x.

We know,

Resistance R=$\frac{pl}{Al} = \frac{p {l}^{2} }{Al} =\frac{p {l}^{2} }{v} = \frac{p {l}^{2} }{ \frac{m}{d} }$

$= \frac{pd {l}^{2} }{m}$

R₁ : R₂$= {(\frac{3y}{4y}) }^{2} \times \frac{3x}{2x} = \frac{27}{32}$

Required rasistance ratio is 27:32.