Question 49 of 380
A wire of resistance and radius is stretched till its radius became . If new resistance of the stretched wire is , then value of is _______
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Answer: 16
We know
As we starch the wire, its length will increase but its radius will decrease keeping the volume constant
$\begin{aligned}
& \mathrm{V}_{\mathrm{i}}=\mathrm{V}_{\mathrm{f}} \\
& \pi \mathrm{r}^2 l=\pi \frac{\mathrm{r}^2}{4} l_{\mathrm{f}} \\
& l_{\mathrm{f}}=4 l \\
& \frac{\mathrm{R}_{\text {new }}}{\mathrm{R}_{\text {old }}}=\left(\frac{4 l}{\frac{\mathrm{r}^2}{4}}\right) \frac{\mathrm{r}^2}{l}=16 \\
& \mathrm{R}_{\text {new }}=16 \mathrm{R} \\
\therefore & \mathrm{x}=16
\end{aligned}$
As we starch the wire, its length will increase but its radius will decrease keeping the volume constant
$\begin{aligned}
& \mathrm{V}_{\mathrm{i}}=\mathrm{V}_{\mathrm{f}} \\
& \pi \mathrm{r}^2 l=\pi \frac{\mathrm{r}^2}{4} l_{\mathrm{f}} \\
& l_{\mathrm{f}}=4 l \\
& \frac{\mathrm{R}_{\text {new }}}{\mathrm{R}_{\text {old }}}=\left(\frac{4 l}{\frac{\mathrm{r}^2}{4}}\right) \frac{\mathrm{r}^2}{l}=16 \\
& \mathrm{R}_{\text {new }}=16 \mathrm{R} \\
\therefore & \mathrm{x}=16
\end{aligned}$