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 ............... \(\Omega\)
Pick an option to check your answer.
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.
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