Q191 / 200 · MHT CET - Medical 2026 (25 April Shift 2) · Physics
The heat generated in a circuit is given by $Q = I^2 R t$. If the percentage error in the measurement of $Q$ is $1\%$, $R$ is $2\%$, and $t$ is $1\%$, then the percentage error in the measurement of current $I$ is
- A$3\%$
- Correct answerB$2\%$
- C$4\%$
- D$5\%$
Solution
Given $Q = I^2 R t$
Rearranging the formula for $I$, we get:
$I = \left(\dfrac{Q}{R t}\right)^{1/2}$
The maximum percentage error in $I$ is given by:
$\dfrac{\Delta I}{I} \times 100 = \dfrac{1}{2} \left( \dfrac{\Delta Q}{Q} \times 100 \right) + \dfrac{1}{2} \left( \dfrac{\Delta R}{R} \times 100 \right) + \dfrac{1}{2} \left( \dfrac{\Delta t}{t} \times 100 \right)$
Substituting the given percentage errors:
$\dfrac{\Delta I}{I} \times 100 = \dfrac{1}{2}(1\%) + \dfrac{1}{2}(2\%) + \dfrac{1}{2}(1\%)$
$\dfrac{\Delta I}{I} \times 100 = 0.5\% + 1\% + 0.5\% = 2\%$
Answer: $2\%$
Rearranging the formula for $I$, we get:
$I = \left(\dfrac{Q}{R t}\right)^{1/2}$
The maximum percentage error in $I$ is given by:
$\dfrac{\Delta I}{I} \times 100 = \dfrac{1}{2} \left( \dfrac{\Delta Q}{Q} \times 100 \right) + \dfrac{1}{2} \left( \dfrac{\Delta R}{R} \times 100 \right) + \dfrac{1}{2} \left( \dfrac{\Delta t}{t} \times 100 \right)$
Substituting the given percentage errors:
$\dfrac{\Delta I}{I} \times 100 = \dfrac{1}{2}(1\%) + \dfrac{1}{2}(2\%) + \dfrac{1}{2}(1\%)$
$\dfrac{\Delta I}{I} \times 100 = 0.5\% + 1\% + 0.5\% = 2\%$
Answer: $2\%$
Hence, the correct answer is option: 'B'.