A wire has an initial length of 7.5 cm, an initial diameter of 20 mm, and an initial resistance of 0.024 ohms. The wire is pulled so that its diameter is reduced to 10 mm. We need to find the new resistance of the wire.

1. Problem Description

A wire has an initial length of 7.5 cm, an initial diameter of 20 mm, and an initial resistance of 0.024 ohms. The wire is pulled so that its diameter is reduced to 10 mm. We need to find the new resistance of the wire.

2. Solution Steps

Since the volume of the wire remains constant during the pulling process, we have:

$V_1 = V_2$

where $V_1$ and $V_2$ are the initial and final volumes of the wire, respectively.

The volume of a cylindrical wire is given by:

$V = A \times L = \pi r^2 L$

where $A$ is the cross-sectional area, $L$ is the length, and $r$ is the radius of the wire.

Therefore, we have:

$\pi r_1^2 L_1 = \pi r_2^2 L_2$

where $r_1$ and $r_2$ are the initial and final radii, and $L_1$ and $L_2$ are the initial and final lengths, respectively.

We are given that the initial diameter $d_1 = 20$ mm and the final diameter $d_2 = 10$ mm. Therefore, $r_1 = d_1/2 = 10$ mm and $r_2 = d_2/2 = 5$ mm. Also, $L_1 = 7.5$ cm. Substituting these values, we get:

$(10)^2 \times 7.5 = (5)^2 \times L_2$

$100 \times 7.5 = 25 \times L_2$

$750 = 25 L_2$

$L_2 = 750/25 = 30$ cm

So, the new length of the wire is $L_2 = 30$ cm.

The resistance of a wire is given by:

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

where $\rho$ is the resistivity of the material, $L$ is the length, and $A$ is the cross-sectional area.

Thus we can write:

$R_1 = \rho \frac{L_1}{A_1} = \rho \frac{L_1}{\pi r_1^2}$

$R_2 = \rho \frac{L_2}{A_2} = \rho \frac{L_2}{\pi r_2^2}$

Dividing the two equations gives:

$\frac{R_2}{R_1} = \frac{L_2}{L_1} \times \frac{r_1^2}{r_2^2}$

$R_2 = R_1 \times \frac{L_2}{L_1} \times \frac{r_1^2}{r_2^2}$

$R_2 = 0.024 \times \frac{30}{7.5} \times \frac{(10)^2}{(5)^2}$

$R_2 = 0.024 \times 4 \times \frac{100}{25}$

$R_2 = 0.024 \times 4 \times 4$

$R_2 = 0.024 \times 16$

$R_2 = 0.384 \, \Omega$

3. Final Answer

The new resistance of the wire is 0.384 $\Omega$.