Question:
A wire of length $1\text{ m}$ is broken into two unequal parts $P$ and $Q$. Part $P$ is extended to double its length so that its resistance become equal to resistance of $Q$. The length of $Q$ part is:
A wire of length $1\text{ m}$ is broken into two unequal parts $P$ and $Q$. Part $P$ is extended to double its length so that its resistance become equal to resistance of $Q$. The length of $Q$ part is:
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When a wire is stretched to $n$ times its length, its resistance increases by a factor of $n^2$.
Here, $n = 2$, so the resistance of $P$ increases by $2^2 = 4$ times.
This leads to the direct equation: $4 \times l_P = l_Q$.
Since $l_P + l_Q = 1$, we have $5 l_P = 1 \implies l_P = 0.2\text{ m}$ and $l_Q = 0.8\text{ m}$.
Here, $n = 2$, so the resistance of $P$ increases by $2^2 = 4$ times.
This leads to the direct equation: $4 \times l_P = l_Q$.
Since $l_P + l_Q = 1$, we have $5 l_P = 1 \implies l_P = 0.2\text{ m}$ and $l_Q = 0.8\text{ m}$.
Updated On: Jul 22, 2026
- $0.2\text{ m}$
- $0.8\text{ m}$
- $0.6\text{ m}$
- $0.5\text{ m}$
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Question:
A $1\text{ m}$ wire is cut into two pieces, $P$ and $Q$.
Piece $P$ is stretched to double its original length, causing its resistance to increase.
After stretching, the resistance of $P$ equals that of $Q$. We need to find the original length of piece $Q$.
Step 2: Key Formula and Approach:
The resistance of a wire is given by:
\[ R = \rho \frac{l}{A} \] When a wire is stretched, its volume $V = A \cdot l$ remains constant.
If length is doubled ($l' = 2l$), the cross-sectional area must be halved ($A' = \frac{A}{2}$), making the new resistance $R' = 4R$.
Step 3: Detailed Explanation:
• Let variables for initial lengths:
Let the initial length of part $P$ be $l_P = x\text{ m}$.
Since the total length is $1\text{ m}$, the initial length of part $Q$ is:
\[ l_Q = (1 - x)\text{ m} \]
• Calculate Initial Resistances:
Let $\rho$ be the resistivity and $A$ be the initial cross-sectional area:
\[ R_P = \rho \frac{x}{A} \] \[ R_Q = \rho \frac{1 - x}{A} \]
• Calculate Resistance of P after stretching:
When $P$ is stretched to double its length ($l_P' = 2x$), its area becomes $A_P' = \frac{A}{2}$:
\[ R_P' = \rho \frac{l_P'}{A_P'} = \rho \frac{2x}{A/2} = 4 \left( \rho \frac{x}{A} \right) = 4 R_P \]
• Equate resistances:
We are given $R_P' = R_Q$:
\[ 4 \left(\rho \frac{x}{A}\right) = \rho \frac{1 - x}{A} \] \[ 4x = 1 - x \] \[ 5x = 1 \implies x = 0.2\text{ m} \]
• Calculate the length of part $Q$:
\[ l_Q = 1 - x = 1 - 0.2 = 0.8\text{ m} \]
Step 4: Final Answer:
The length of the $Q$ part is $0.8\text{ m}$, which corresponds to Option (B).
A $1\text{ m}$ wire is cut into two pieces, $P$ and $Q$.
Piece $P$ is stretched to double its original length, causing its resistance to increase.
After stretching, the resistance of $P$ equals that of $Q$. We need to find the original length of piece $Q$.
Step 2: Key Formula and Approach:
The resistance of a wire is given by:
\[ R = \rho \frac{l}{A} \] When a wire is stretched, its volume $V = A \cdot l$ remains constant.
If length is doubled ($l' = 2l$), the cross-sectional area must be halved ($A' = \frac{A}{2}$), making the new resistance $R' = 4R$.
Step 3: Detailed Explanation:
• Let variables for initial lengths:
Let the initial length of part $P$ be $l_P = x\text{ m}$.
Since the total length is $1\text{ m}$, the initial length of part $Q$ is:
\[ l_Q = (1 - x)\text{ m} \]
• Calculate Initial Resistances:
Let $\rho$ be the resistivity and $A$ be the initial cross-sectional area:
\[ R_P = \rho \frac{x}{A} \] \[ R_Q = \rho \frac{1 - x}{A} \]
• Calculate Resistance of P after stretching:
When $P$ is stretched to double its length ($l_P' = 2x$), its area becomes $A_P' = \frac{A}{2}$:
\[ R_P' = \rho \frac{l_P'}{A_P'} = \rho \frac{2x}{A/2} = 4 \left( \rho \frac{x}{A} \right) = 4 R_P \]
• Equate resistances:
We are given $R_P' = R_Q$:
\[ 4 \left(\rho \frac{x}{A}\right) = \rho \frac{1 - x}{A} \] \[ 4x = 1 - x \] \[ 5x = 1 \implies x = 0.2\text{ m} \]
• Calculate the length of part $Q$:
\[ l_Q = 1 - x = 1 - 0.2 = 0.8\text{ m} \]
Step 4: Final Answer:
The length of the $Q$ part is $0.8\text{ m}$, which corresponds to Option (B).
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