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JEE Main 20246 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
6 April
Shift
Morning Shift
Subject
Physics
Chapter
Current Electricity

JEE Main 2024 · 6 April · Morning Shift · Physics · Current Electricity

A wire of resistance R and radius r is stretched till its radius...

A wire of resistance RR and radius rr is stretched till its radius became r/2r / 2. If new resistance of the stretched wire is xRx R, then value of xx is _______

Correct answer

0

Step-by-step solution

We know R=ρlA,R∝lr2\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 Vi=Vfπr2l=πr24lflf=4lRnew Rold =(4lr24)r2l=16Rnew =16R∴x=16\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}
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