Q108 / 200 · NEET 2024 (Re-NEET) · Physics

In an electrical circuit, the voltage is measured as $V=(200 \pm 4)$ volt and the current is measured as $I=(20 \pm 0.2) \mathrm{A}$. The value of the resistance is:

  • A
    $(10 \pm 4.2) \Omega$
  • Correct answer
    B
    $(10 \pm 0.3) \Omega$
  • C
    $(10 \pm 0.1) \Omega$
  • D
    $(10 \pm 0.8) \Omega$

Solution

$R=\frac{V}{l}=\frac{200}{20}=10 \Omega$ $\frac{\Delta R}{R}=\frac{\Delta V}{V}+\frac{\Delta l}{l}=\frac{4}{200}+\frac{0.2}{20}=\frac{4+2}{200}=\frac{6}{200}$ $\Rightarrow \Delta R=\frac{6}{200} \times R=\frac{6}{20}=\frac{3}{10}=0.3 \Omega$ $\therefore$ Resistance $=R \pm \Delta R=(10 \pm 0.3) \Omega$

Hence, the correct answer is option: 'B'.

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