JEE Mains · Physics · STD 12 - 3. current electricity
The variation of applied potential and current flowing through a given wire is shown in figure. The length of wire is \(31.4 \,cm\). The diameter of wire is measured as \(2.4 \,cm\). The resistivity of the given wire is measured as \(x \times 10^{-3} \,\Omega cm\). The value of \(x\) is_______ [Take \(\pi=3.14]\)
- A \(134\)
- B \(143\)
- C \(150\)
- D \(144\)
Answer & Solution
Correct Answer
(D) \(144\)
Step-by-step Solution
Detailed explanation
\(1=\rho \frac{\ell}{ A }\) \(1=\frac{\rho \times 31.4}{\frac{\pi(2.4)^{2}}{4}}\) \(\frac{\pi(2.4)^{2}}{4}=\rho \times 314\) \(\frac{2.4 \times 2.4}{4}=\rho \times 10\) \(\frac{0.6 \times 2.4}{10}=\rho\) \(\frac{1.44}{10}=\rho\) \(0.144=\rho\) \(144 \times 10^{-3}=\rho\)
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