Question8. 11. The conductance of a wire is $2,5 \mathrm{~S}$ . Another wire of the same material and at the same temperature has a diameter one-forth as great and the length twice as great. Find the conductance of the second wire. Ans. $78,1 \mathrm{mS}$ .

Studdy Solution

STEP 1

1. Conductance is inversely proportional to the length of the wire.
2. Conductance is directly proportional to the cross-sectional area of the wire.
3. The wires are made of the same material and are at the same temperature, so the resistivity remains constant.
4. The diameter of the second wire is one-fourth that of the first wire.
5. The length of the second wire is twice that of the first wire.

STEP 2

1. Define the relationship between conductance, length, and cross-sectional area.
2. Calculate the change in cross-sectional area due to the change in diameter.
3. Calculate the change in conductance due to changes in length and cross-sectional area.
4. Calculate the conductance of the second wire.

STEP 3

Define the relationship between conductance, length, and cross-sectional area.
Conductance $G$ is given by the formula: $G = \frac{1}{R} = \frac{\sigma A}{L}$ where $\sigma$ is the conductivity, $A$ is the cross-sectional area, and $L$ is the length of the wire.

STEP 4

Calculate the change in cross-sectional area due to the change in diameter.
The cross-sectional area $A$ of a wire is proportional to the square of its diameter $d$ . If the diameter of the second wire is one-fourth of the first, then: $A_2 = \left(\frac{1}{4}\right)^2 A_1 = \frac{1}{16} A_1$

STEP 5

Calculate the change in conductance due to changes in length and cross-sectional area.
The conductance of the second wire $G_2$ can be expressed in terms of the conductance of the first wire $G_1$ as: $G_2 = G_1 \times \frac{A_2}{A_1} \times \frac{L_1}{L_2}$
Substituting the values: $G_2 = 2.5 \, \mathrm{S} \times \frac{1}{16} \times \frac{1}{2}$

STEP 6

Calculate the conductance of the second wire.
$G_2 = 2.5 \, \mathrm{S} \times \frac{1}{32}$ $G_2 = \frac{2.5}{32} \, \mathrm{S}$ $G_2 = 0.078125 \, \mathrm{S}$ $G_2 = 78.125 \, \mathrm{mS}$
Rounding to one decimal place, the conductance of the second wire is: $\boxed{78.1 \, \mathrm{mS}}$

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