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Units and Measurements: JEE Main Physics Question with Solution

Two resistances are given as

\mathrm{R}_1 = (10 \pm 0.5)\Omega

and

\mathrm{R}_2 = (15 \pm 0.5)\Omega

. The percentage error in the measurement of equivalent resistance when they are connected in parallel is

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6.33

2.33

4.33

5.33

Step-by-step solutionView

Correct answer

The percentage error in the equivalent parallel resistance is $4.33\%$ .

Option analysis

Why each option works or fails

A ·

6.33

Believing that percentage errors add directly for parallel combinations, i.e., $(0.5/10 + 0.5/15)\times 100\% = 5\% + 3.33\% = 8.33\%$ or miscalculating the differential weighting. Use the differential form $\frac{\Delta R_{eq}}{R_{eq}^2} = \frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2}$ to combine errors for parallel resistors.

B ·

2.33

Omitting the factor of equivalent resistance $R_{eq}$ when converting the absolute error equation into fractional error. Remember that $\frac{\Delta R_{eq}}{R_{eq}} = R_{eq}\left(\frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2}\right)$ .

C ·

4.33

None. This is the correct calculation. Correctly applying $\frac{\Delta R_{eq}}{R_{eq}} \times 100 = R_{eq}\left(\frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2}\right) \times 100$ gives $6\left(\frac{0.5}{100} + \frac{0.5}{225}\right) \times 100 = 4.33\%$ .

D ·

5.33

Making an arithmetic mistake when summing the fraction terms $\frac{0.5}{100} + \frac{0.5}{225}$ . Carefully compute $\frac{0.5}{100} = 0.005$ and $\frac{0.5}{225} \approx 0.00222$ , summing to $0.00722$ before multiplying by $600$ .

Reviewed route

Solution

StepWorking

01given

$R_1 = 10\ \Omega$ , $\Delta R_1 = 0.5\ \Omega$ , $R_2 = 15\ \Omega$ , $\Delta R_2 = 0.5\ \Omega$ .

02find

The percentage error in equivalent parallel resistance, $\frac{\Delta R}{R} \times 100$ .

03strategise

For parallel combination, $\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$ . Differentiating gives $\frac{\Delta R}{R^2} = \frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2}$ , which leads to $\frac{\Delta R}{R} = R \left(\frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2}\right)$ , where $R = \frac{R_1 R_2}{R_1 + R_2}$ .

04execute

First, calculate $R = \frac{10 \times 15}{10 + 15} = \frac{150}{25} = 6\ \Omega$ . Then, $\frac{\Delta R}{R} \times 100 = 6 \times 100 \times \left(\frac{0.5}{10^2} + \frac{0.5}{15^2}\right) = 600 \times 0.5 \times \left(\frac{1}{100} + \frac{1}{225}\right) = 300 \times \left(\frac{9 + 4}{900}\right) = 300 \times \frac{13}{900} = \frac{13}{3} \approx 4.33\%$ .

✓verify

Percentage error of $R_1$ is $5\%$ and of $R_2$ is $3.33\%$ . The equivalent percentage error in parallel must lie between $3.33\%$ and $5\%$ . $4.33\%$ falls right in this valid interval.

Hints that build this answer step by step

What is the equivalent resistance $R_{eq}$ for $R_1 = 10\,\Omega$ and $R_2 = 15\,\Omega$ connected in parallel?
$R_{eq} = 6\,\Omega$
Which relation correctly expresses the propagation of error for resistors in parallel, $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$ ?
$\frac{\Delta R_{eq}}{R_{eq}^2} = \frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2}$
Substitute the values to find the percentage error $\frac{\Delta R_{eq}}{R_{eq}} \times 100\%$ .
$4.33\%$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Physics
Concepts assessed: Physics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 9 September 2026

Quick checks

Students also ask

Why isn't ΔR/R simply equal to ΔR1/R1 + ΔR2/R2?

Because R is not given by R = R1 * R2. The relation is 1/R = 1/R1 + 1/R2. Differentiating 1/R gives -dR/R^2, so each term carries a square in the denominator.

Answer

The percentage error in the equivalent parallel resistance is $4.33\%$ .

Why each option works or fails

A: $6.33$ - Believing that percentage errors add directly for parallel combinations, i.e., $(0.5/10 + 0.5/15)\times 100\% = 5\% + 3.33\% = 8.33\%$ or miscalculating the differential weighting. Use the differential form $\frac{\Delta R_{eq}}{R_{eq}^2} = \frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2}$ to combine errors for parallel resistors.
B: $2.33$ - Omitting the factor of equivalent resistance $R_{eq}$ when converting the absolute error equation into fractional error. Remember that $\frac{\Delta R_{eq}}{R_{eq}} = R_{eq}\left(\frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2}\right)$ .
C · correct: $4.33$ - None. This is the correct calculation. Correctly applying $\frac{\Delta R_{eq}}{R_{eq}} \times 100 = R_{eq}\left(\frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2}\right) \times 100$ gives $6\left(\frac{0.5}{100} + \frac{0.5}{225}\right) \times 100 = 4.33\%$ .
D: $5.33$ - Making an arithmetic mistake when summing the fraction terms $\frac{0.5}{100} + \frac{0.5}{225}$ . Carefully compute $\frac{0.5}{100} = 0.005$ and $\frac{0.5}{225} \approx 0.00222$ , summing to $0.00722$ before multiplying by $600$ .

Step-by-step solution

given: $R_1 = 10\ \Omega$ , $\Delta R_1 = 0.5\ \Omega$ , $R_2 = 15\ \Omega$ , $\Delta R_2 = 0.5\ \Omega$ .
find: The percentage error in equivalent parallel resistance, $\frac{\Delta R}{R} \times 100$ .
strategise: For parallel combination, $\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}$ . Differentiating gives $\frac{\Delta R}{R^2} = \frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2}$ , which leads to $\frac{\Delta R}{R} = R \left(\frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2}\right)$ , where $R = \frac{R_1 R_2}{R_1 + R_2}$ .
execute: First, calculate $R = \frac{10 \times 15}{10 + 15} = \frac{150}{25} = 6\ \Omega$ . Then, $\frac{\Delta R}{R} \times 100 = 6 \times 100 \times \left(\frac{0.5}{10^2} + \frac{0.5}{15^2}\right) = 600 \times 0.5 \times \left(\frac{1}{100} + \frac{1}{225}\right) = 300 \times \left(\frac{9 + 4}{900}\right) = 300 \times \frac{13}{900} = \frac{13}{3} \approx 4.33\%$ .
verify: Percentage error of $R_1$ is $5\%$ and of $R_2$ is $3.33\%$ . The equivalent percentage error in parallel must lie between $3.33\%$ and $5\%$ . $4.33\%$ falls right in this valid interval.

Units and Measurements: JEE Main Physics Question with Solution

The percentage error in the equivalent parallel resistance is 4.33%4.33\%4.33%.

01given

02find

03strategise

04execute

✓verify

Students also ask

The percentage error in the equivalent parallel resistance is $4.33\%$ .