JEE Mains · Physics · STD 12 - Current electricity
- JEE Mains 2024
- Hard
- Numerical
A wire of resistance \(R\) and radius \(r\) is stretched till its radius became \(r / 2\). If new resistance of the stretched wire is \(xR\), then value of \(x\) is _______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(16\)
Step-by-step Solution
Detailed explanation
We know \(\mathrm{R}=\frac{\rho l}{\mathrm{~A}}, \mathrm{R} \propto \frac{l}{\mathrm{r}^2}\)
As we starch the wire, its length will increase but its radius will decrease keeping the volume constant
\(\mathrm{V}_{\mathrm{i}}=\mathrm{V}_{\mathrm{f}}\)
\(\pi^2 l=\pi \frac{\mathrm{r}^2}{4} l_{\mathrm{f}}\)
\(l_{\mathrm{f}}=4 l\)
\(\frac{\mathrm{R}_{\text {nelv }}}{\mathrm{R}_{\text {old }}}=\left(\frac{4 l}{\frac{\mathrm{r}^2}{4}}\right) \frac{\mathrm{r}^2}{l}=16\)
\(\mathrm{R}_{\text {neer }}=16 \mathrm{R}\)
\(\therefore \mathrm{x}=16\)
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