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JEE Mains · Physics · STD 12 - Current electricity
  • JEE Mains 2017
  • Hard
  • Numerical

A uniform wire of length \(l\) and radius \(r\) has a resistance of \(100\, \Omega \). It is recast into a wire of radius \(\frac{r}{2}\). The resistance of new wire will be ............... \(\Omega\)

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Answer & Solution

Correct Answer

(A) \(1600\)

Step-by-step Solution

Detailed explanation

Given, \({R_1} = 100\,\Omega ,{r^\prime } = r/2,{R_2} = ?\)

Resistivity of wire, \(\mathrm{R}=\frac{\rho l}{\mathrm{A}}\)

\(\because\) Area \(\times\) length \(=\) volume

Hence, \(R=\frac{\rho V}{A^{2}}\)

since, \(\rho \rightarrow\) constant, \(V \rightarrow\) constant

\(\mathrm{R} \propto \frac{1}{\mathrm{A}^{2}}\)

or \(\quad \mathrm{R} \propto \frac{1}{\mathrm{r}^{4}} \quad \because \quad \mathrm{A}=\pi \mathrm{r}^{2}\)

\(\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=16 \Rightarrow \mathrm{R}_{2}=16 \times 100=1600 \,\Omega,\)

Resistance of new wire.