Question

A wire is stretched to double its length. New resistance becomes:

• A. $R/2$
• B. $R$
• C. $2R$
• D. $4R$

Hint:

For any wire stretched by a factor of $n$ (where $l' = n \cdot l$), the new resistance is given by the shortcut formula:
\[ R' = n^2 \cdot R \] Since the length is doubled ($n = 2$), the new resistance is $2^2 \cdot R = 4R$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The electrical resistance (\(R\)) of a wire is a measure of the opposition it offers to the flow of electric current. According to Ohm's law and the properties of materials, resistance is directly proportional to the length of the conductor and inversely proportional to its cross-sectional area. However, a key factor in "stretching" problems is the conservation of mass and volume. When a wire is pulled or stretched to increase its length, its material doesn't change, and no mass is added or removed. This implies that the total volume of the wire remains constant. As the length increases, the wire must become thinner, meaning the cross-sectional area decreases simultaneously. Both of these changes contribute to an increase in the final resistance.
Step 2: Key Formula or Approach:
The resistance of a uniform conductor is given by:
\[ R = \rho \frac{l}{A} \]
Where \(\rho\) is the resistivity (constant for a given material), \(l\) is the length, and \(A\) is the cross-sectional area.
Since volume (\(V\)) is constant:
\[ V = A_{1}l_{1} = A_{2}l_{2} \]
Where the subscripts 1 and 2 represent the initial and final states, respectively.
Step 3: Detailed Explanation:
Let the initial resistance be \(R_{1}\), initial length be \(l_{1}\), and initial area be \(A_{1}\). We are given:
\[ R_{1} = \rho \frac{l_{1}}{A_{1}} \]
The wire is stretched to double its length, so the new length \(l_{2}\) is:
\[ l_{2} = 2l_{1} \]
Using the conservation of volume (\(A_{1}l_{1} = A_{2}l_{2}\)), we can determine the new cross-sectional area \(A_{2}\):
\[ A_{2} = \frac{A_{1}l_{1}}{l_{2}} = \frac{A_{1}l_{1}}{2l_{1}} = \frac{A_{1}}{2} \]
Now, let's calculate the new resistance \(R_{2}\) using the new dimensions:
\[ R_{2} = \rho \frac{l_{2}}{A_{2}} \]
Substitute the values of \(l_{2}\) and \(A_{2}\) in terms of the initial parameters:
\[ R_{2} = \rho \frac{(2l_{1})}{(A_{1}/2)} \]
Using simple algebraic fraction rules (dividing by a fraction is multiplying by its reciprocal):
\[ R_{2} = \rho \frac{2l_{1} \cdot 2}{A_{1}} = 4 \left( \rho \frac{l_{1}}{A_{1}} \right) \]
Since the term in the parenthesis is the original resistance \(R_{1}\):
\[ R_{2} = 4R_{1} \]
Physical justification: When the length doubles, the electrons have to travel twice the distance, facing twice the number of collisions (doubling the resistance). At the same time, the area halves, meaning there is only half the space for current to flow (doubling the resistance again). Together, these two factors result in a four-fold increase (\(2 \times 2 = 4\)).
Step 4: Final Answer:
If a wire is stretched to double its length, its resistance increases by four times, becoming \(4R\).