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Physics · Units, Dimensions & Error Analysis

JEE Main 2026 — 21 January, Morning Shift — Question 29

In an experiment the values of two spring constants were measured as k1=(10±0.2)N/m\mathrm{k}_{1}=(10 \pm 0.2) \mathrm{N} / \mathrm{m} and k2=(20±0.3)N/m\mathrm{k}_{2}=(20 \pm 0.3) \mathrm{N} / \mathrm{m}. If these springs are connected in parallel, then the percentage error in equivalent spring constant is :

  1. Option A:

    2.67%2.67 \%

  2. Option B:

    2.33%2.33 \%

  3. Option C:

    1.33%1.33 \%

  4. Option D:

    1.67%1.67 \%

    Correct

Answer: D

Step-by-step solution

For parallel combination of spring,

Keq=K1+K2=30 N/mΔ Keq=ΔK1+ΔK2=0.2+0.3=0.5 N/m∴% Error in K=0.530×100=1.67%\begin{aligned} & \mathrm{K}_{\mathrm{eq}}=\mathrm{K}_{1}+\mathrm{K}_{2}=30 \mathrm{~N} / \mathrm{m} & \Delta \mathrm{~K}_{\mathrm{eq}}=\Delta \mathrm{K}_{1}+\Delta \mathrm{K}_{2}=0.2+0.3=0.5 \mathrm{~N} / \mathrm{m} & \therefore \quad \% \text { Error in } \mathrm{K}=\frac{0.5}{30} \times 100=1.67 \% \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
21 January, Morning Shift
Subject
Physics
Chapter
Units, Dimensions & Error Analysis
Topic
Significant Figures and Error Analysis

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