JEE PYQ: Units & Measurements - Question ID cb8723255ab8 (JEE Main 2024)

ID: cb8723255ab8JEE Main 2024Single Correct MCQ

To find the spring constant (k)(k) of a spring experimentally, a student commits 2%2 \% positive error in the measurement of time and 1%1 \% negative error in measurement of mass. The percentage error in determining value of kk is :

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Step-by-step Explanation

Core Formula & Concept:

In simple harmonic motion (SHM) of a mass-spring system, the time period TT of oscillation is given by: T=2πmkT = 2\pi \sqrt{\frac{m}{k}} where

  • mm is the mass attached to the spring,
  • kk is the spring constant.
To find the spring constant experimentally, we rearrange the formula: k=4π2mT2k = \frac{4\pi^2 m}{T^2} This expression shows that kk depends on mm and T2T^2. Any errors in measuring mm or TT will propagate to the calculated value of kk.

Step-by-Step Derivation:

Step 1: Express the relative error in kk
Since k=4π2mT2k = \frac{4\pi^2 m}{T^2}, we take the natural logarithm on both sides to find the relative error: lnk=ln(4π2)+lnm2lnT\ln k = \ln(4\pi^2) + \ln m - 2 \ln T Differentiating both sides with respect to the measured quantities: dkk=dmm2dTT\frac{dk}{k} = \frac{dm}{m} - 2 \frac{dT}{T} The relative error in kk is: Δkk=Δmm2ΔTT\frac{\Delta k}{k} = \frac{\Delta m}{m} - 2 \frac{\Delta T}{T}

Step 2: Substitute the given percentage errors
The student commits:

  • 1%1\% negative error in mass, so Δmm=1%\frac{\Delta m}{m} = -1\%,
  • 2%2\% positive error in time, so ΔTT=+2%\frac{\Delta T}{T} = +2\%.
Substituting these values: Δkk=(1%)2(+2%)=1%4%=5%\frac{\Delta k}{k} = (-1\%) - 2(+2\%) = -1\% - 4\% = -5\% The negative sign indicates the direction of error, but the percentage error is the magnitude, so 5%5\%.

Step 3: Conclusion
The percentage error in determining kk is 5%5\%. Hence, the correct option is A: 5%.

Common Traps & Exam Tip:

  • Sign Confusion: Students often forget that a negative error in mass (Δm/m=1%\Delta m/m = -1\%) and a positive error in time (ΔT/T=+2%\Delta T/T = +2\%) combine differently. The negative sign in the formula Δkk=Δmm2ΔTT\frac{\Delta k}{k} = \frac{\Delta m}{m} - 2 \frac{\Delta T}{T} is crucial.
  • Squaring the Time Error: Since kT2k \propto T^{-2}, the error in TT is multiplied by 2, not 1. Many students mistakenly treat it as a linear dependence.
  • Magnitude vs. Direction: The question asks for the percentage error, not the direction. Even if the result is negative, the magnitude is what matters.
Always write down the error propagation formula explicitly to avoid such mistakes.

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