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. $0.2\text{ m}$
• B. $0.8\text{ m}$
• C. $0.6\text{ m}$
• D. $0.5\text{ m}$

Hint:

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 $l_Q = 0.8\text{ m}$.

Answer 1

The Correct Option is B

Step 1: Track how stretching changes P's resistance.
Let the original length of P be $l$ and of Q be $(1-l)$ in metres. Stretching P to double its length keeps its volume fixed, so its area halves. Since $R = \rho l/A$ and $A' = A/2$ while $l' = 2l$, the new resistance is \[ R_P' = \rho\frac{2l}{A/2} = 4\left(\rho\frac{l}{A}\right) = 4R_P \]
Step 2: Write Q's (unchanged) resistance in the same units. \[ R_Q = \rho\frac{1-l}{A} \]
Step 3: Set the stretched P equal to Q, as given. \[ 4\left(\rho\frac{l}{A}\right) = \rho\frac{1-l}{A} \quad\Rightarrow\quad 4l = 1-l \quad\Rightarrow\quad l = 0.2\text{ m} \]
Step 4: Get the length of Q. \[ l_Q = 1 - 0.2 \] \[ \boxed{l_Q = 0.8\text{ m}} \]