Question

The ratio of masses of two aluminium wires is 2 : 1 and their corresponding ratio of lengths is 1 : 2. Then, the ratio of their resistances is​

Answer 1

Answer:

Ratio of their resistances is 1 : 8.

Explanation:

Given:

Ratio of masses of two aluminium wires is 2 : 1.

Ratio of lengths of two aluminium wires is 1 : 2.

To find:

Ratio of their resistances = ?

Solution:

Resistance of a wire, R is given as:

$R = \rho \dfrac{L}A$

Where $\rho$ is the resistivity of the material.

L is the length of wire and

A is the cross sectional Area.

Now, we know that, Density of any material is given as:

$D =\dfrac{m}{V}$

Where m is the mass and

V is the volume

Volume of any wire is given as:

$V = A \times L$

Putting above:

$D =\dfrac{m}{A \times L}\\\Rightarrow A = \dfrac{m}{D \times L}$

Putting value of A in the main equation, the resistance of material is given as:

$R = \rho \dfrac{L}A \\\Rightarrow R = \rho \dfrac{L\times D \times L}{M}\\\Rightarrow R = \rho \dfrac{L^2\times D }{M}$

For the first wire, Resistance:

$R_1 = \rho_1 \dfrac{L_1^2\times D_1 }{M_1}$

For the second wire, Resistance:

$R_2 = \rho_2 \dfrac{L_2^2\times D_2 }{M_2}$

Here, material of both the wires is same so:

$\rho_1=\rho_2\\D_1=D_2$

Now, given that the ratio are:

$m_1:m_2=2:1\\L_1:L_2=1:2$

Now, the ratio of resistances:

$R_1:R_2 = \rho_1\dfrac{L_1^2 \times D_1}{M_1}: \rho_2\dfrac{L_2^2 \times D_2}{M_2}\\\Rightarrow R_1:R_2 = \dfrac{L_1^2}{M_1}: \dfrac{L_2^2}{M_2}\\\Rightarrow R_1:R_2 = \dfrac{1^2}{2}: \dfrac{2^2}{1}\\\Rightarrow R_1:R_2 = \dfrac{1}{2}: \dfrac{4}{1}\\\Rightarrow R_1:R_2 = 1:8$

So, the answer is:

Ratio of their resistances is 1 : 8.