Q101 / 150 · MHT CET 2026 (18 April Shift 1) · Physics

If two resistors of resistances have values $R_1 = (350 \pm 3)\ \Omega$ and $R_2 = (140 \pm 4)\ \Omega$. The percentage error for the sum and difference of $R_1$ and $R_2$ are respectively

  • Correct answer
    A
    $1.43\%$ , $3.33\%$
  • B
    $7\%$ , $1\%$
  • C
    $1\%$ , $7\%$
  • D
    $3.33\%$ , $1.43\%$

Solution

Given $R_1 = (350 \pm 3)\ \Omega$ and $R_2 = (140 \pm 4)\ \Omega$.

For the sum of resistances:
$R_s = R_1 + R_2 = 350 + 140 = 490\ \Omega$
The absolute error in the sum is the sum of absolute errors:
$\Delta R_s = \Delta R_1 + \Delta R_2 = 3 + 4 = 7\ \Omega$
Percentage error in sum = $\dfrac{\Delta R_s}{R_s} \times 100 = \dfrac{7}{490} \times 100 = \dfrac{100}{70} \approx 1.43\%$

For the difference of resistances:
$R_d = R_1 - R_2 = 350 - 140 = 210\ \Omega$
The absolute error in the difference is also the sum of absolute errors:
$\Delta R_d = \Delta R_1 + \Delta R_2 = 3 + 4 = 7\ \Omega$
Percentage error in difference = $\dfrac{\Delta R_d}{R_d} \times 100 = \dfrac{7}{210} \times 100 = \dfrac{100}{30} \approx 3.33\%$

The percentage errors are $1.43\%$ and $3.33\%$ respectively.

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

Relevant Questions