Physics · Moving Charges and Magnetic Field

JEE Main 2026 — 28 January, Evening Shift — Question 44

The time period of a simple harmonic oscillator is $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$ . The measured value of mass (m) of the object is 10 g with an accuracy of 10 mg , and time for 50 oscillations of the spring is found to be 60 s using a watch of 2 s resolution. Percentage error in determination of spring constant(k) is $\_\_\_\_$ %.

Option A:
3.43
Option B:
3.35
Option C:
7.6
Option D:
6.76
Correct

Answer: D

Step-by-step solution

$\frac{\Delta \mathrm{K}}{\mathrm{K}}=\frac{2 \Delta \mathrm{~T}}{\mathrm{~T}}+\frac{\Delta \mathrm{m}}{\mathrm{m}}$ $\mathrm{T}=\frac{60}{50}=1.2 \mathrm{sec}$ $\Delta \mathrm{T}=\frac{2}{50}$ $\therefore \frac{\Delta \mathrm{K}}{\mathrm{K}}=\frac{2 \times 2}{50 \times 1.2}+\frac{10 \times 10^{-3}}{10}=0.0676$ ∴ % Error $=6.76 \%$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 28 January, Evening Shift
Subject: Physics
Chapter: Moving Charges and Magnetic Field
Topic: Magnetic Field Due to Current-Carrying Wire - Biot-Savart Law