JEE Main 2026 — Jan 21, Shift 1 — Physics solution

JEE Main2026Jan 21, Shift 1Physics

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

A
2.67%
B
2.33%
C
1.33%
D
1.67%

Solution

Springs in parallel add: $K_{e q} = k_{1} + k_{2} = 30$ N/m, with absolute errors adding directly: $Δ K_{e q} = 0.2 + 0.3 = 0.5$ N/m.

Percentage error $= \frac{0.5}{30} \times 100 \approx 1.67%$ .

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