Q191 / 200 · MHT CET - Medical 2026 (25 April Shift 1) · Physics
The least count of voltmeter is 0.2 V, and least count of ammeter is 1 mA. If the potential difference is 2 V, and the current is 10 mA, what is the percentage error in measurement of resistance?
- A$2\%$
- Correct answerB$20\%$
- C$1\%$
- D$10\%$
Solution
Resistance is given by Ohm's law:
$R = \dfrac{V}{I}$
The maximum percentage error in resistance is the sum of the percentage errors in voltage and current:
$\dfrac{\Delta R}{R} \times 100 = \left( \dfrac{\Delta V}{V} + \dfrac{\Delta I}{I} \right) \times 100$
Given values are:
$V = 2 \text{ V}$, $\Delta V = 0.2 \text{ V}$
$I = 10 \text{ mA}$, $\Delta I = 1 \text{ mA}$
Substituting these values:
$\dfrac{\Delta R}{R} \times 100 = \left( \dfrac{0.2}{2} + \dfrac{1}{10} \right) \times 100$
$\dfrac{\Delta R}{R} \times 100 = (0.1 + 0.1) \times 100$
$\dfrac{\Delta R}{R} \times 100 = 0.2 \times 100 = 20\%$
Answer: $20\%$
$R = \dfrac{V}{I}$
The maximum percentage error in resistance is the sum of the percentage errors in voltage and current:
$\dfrac{\Delta R}{R} \times 100 = \left( \dfrac{\Delta V}{V} + \dfrac{\Delta I}{I} \right) \times 100$
Given values are:
$V = 2 \text{ V}$, $\Delta V = 0.2 \text{ V}$
$I = 10 \text{ mA}$, $\Delta I = 1 \text{ mA}$
Substituting these values:
$\dfrac{\Delta R}{R} \times 100 = \left( \dfrac{0.2}{2} + \dfrac{1}{10} \right) \times 100$
$\dfrac{\Delta R}{R} \times 100 = (0.1 + 0.1) \times 100$
$\dfrac{\Delta R}{R} \times 100 = 0.2 \times 100 = 20\%$
Answer: $20\%$
Hence, the correct answer is option: 'B'.