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JEE Main 202621 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
21 January
Shift
Morning Shift
Subject
Physics
Chapter
Mathematics in Physics

JEE Main 2026 · 21 January · Morning Shift · Physics · Mathematics in Physics

In an experiment the values of two spring constants were measured...

In an experiment the values of two spring constants were measured as k1=(10±0.2)N/mk_{1}=(10 \pm 0.2) \mathrm{N} / \mathrm{m} and k2=(20±0.3)N/mk_{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 :

Options

  1. A1.33%
  2. B1.67%
  3. C2.67%
  4. D2.33%

Correct answer

B. 1.67%

Step-by-step solution

For parallel combination: keq=k1+k2=10+20=30k_{eq} = k_1 + k_2 = 10 + 20 = 30 N/m
For sum, absolute errors add:
Δkeq=Δk1+Δk2=0.2+0.3=0.5\Delta k_{eq} = \Delta k_1 + \Delta k_2 = 0.2 + 0.3 = 0.5 N/m
Percentage error =Δkeqkeq×100=0.530×100=1.67%= \frac{\Delta k_{eq}}{k_{eq}} \times 100 = \frac{0.5}{30} \times 100 = 1.67\%
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