JEE Challenger
More from Units and Measurements
PhyHardAdv 2016 • Paper 2

Error Analysis in Acceleration Due to Gravity Measurement Using Rolling Motion

To determine the acceleration due to gravity gg, the time period formula T=2π7(R−r)5gT = 2\pi \sqrt{\frac{7(R-r)}{5g}} is utilized in an experiment. The recorded values for RR and rr are (60±1) mm(60 \pm 1)\text{ mm} and (10±1) mm(10 \pm 1)\text{ mm}, respectively. Five consecutive time period measurements yield 0.52 s0.52\text{ s}, 0.56 s0.56\text{ s}, 0.57 s0.57\text{ s}, 0.54 s0.54\text{ s}, and 0.59 s0.59\text{ s}. The least count of the measuring stopwatch is 0.01 s0.01\text{ s}. Which of the following statement(s) is/are true?

Options

A

The error in the measurement of rr is 10%10\%

Correct
B

The error in the measurement of TT is 3.57%3.57\%

Correct
C

The error in the measurement of TT is 2%2\%

D

The error in the determined value of gg is 11%11\%

Correct

Topics & Concepts

Units and MeasurementsMeasuring Instruments

Step-by-Step Solution

To determine the correct options, we analyze each statement individually:

1. Error in the Measurement of rr (Statement A)

The given value for rr is (10±1) mm(10 \pm 1)\text{ mm}. The absolute error in rr is Δr=1 mm\Delta r = 1\text{ mm}, and the measured value is r=10 mmr = 10\text{ mm}.

The percentage error in rr is: Percentage error in r=(Δrr)×100%=(1 mm10 mm)×100%=10%\text{Percentage error in } r = \left( \frac{\Delta r}{r} \right) \times 100\% = \left( \frac{1\text{ mm}}{10\text{ mm}} \right) \times 100\% = 10\%

Thus, Option (A) is true.

2. Error in the Measurement of TT (Statements B and C)

The given five consecutive time period measurements are 0.52 s0.52\text{ s}, 0.56 s0.56\text{ s}, 0.57 s0.57\text{ s}, 0.54 s0.54\text{ s}, and 0.59 s0.59\text{ s}.

First, calculate the mean value of the time period (TmeanT_{\text{mean}}): Tmean=0.52+0.56+0.57+0.54+0.595=2.785=0.556 sT_{\text{mean}} = \frac{0.52 + 0.56 + 0.57 + 0.54 + 0.59}{5} = \frac{2.78}{5} = 0.556\text{ s}

Rounding off TmeanT_{\text{mean}} to two decimal places (matching the least count of 0.01 s0.01\text{ s}): Tmean≈0.56 sT_{\text{mean}} \approx 0.56\text{ s}

Next, calculate the absolute errors for each measurement: ∣ΔT1∣=∣0.52−0.556∣=0.036 s|\Delta T_1| = |0.52 - 0.556| = 0.036\text{ s} ∣ΔT2∣=∣0.56−0.556∣=0.004 s|\Delta T_2| = |0.56 - 0.556| = 0.004\text{ s} ∣ΔT3∣=∣0.57−0.556∣=0.014 s|\Delta T_3| = |0.57 - 0.556| = 0.014\text{ s} ∣ΔT4∣=∣0.54−0.556∣=0.016 s|\Delta T_4| = |0.54 - 0.556| = 0.016\text{ s} ∣ΔT5∣=∣0.59−0.556∣=0.034 s|\Delta T_5| = |0.59 - 0.556| = 0.034\text{ s}

The mean absolute error (ΔTmean\Delta T_{\text{mean}}) is: ΔTmean=0.036+0.004+0.014+0.016+0.0345=0.1045=0.0208 s≈0.02 s\Delta T_{\text{mean}} = \frac{0.036 + 0.004 + 0.014 + 0.016 + 0.034}{5} = \frac{0.104}{5} = 0.0208\text{ s} \approx 0.02\text{ s}

The percentage error in TT is: Percentage error in T=(ΔTmeanTmean)×100%=(0.020.56)×100%=20056%=257%≈3.57%\text{Percentage error in } T = \left( \frac{\Delta T_{\text{mean}}}{T_{\text{mean}}} \right) \times 100\% = \left( \frac{0.02}{0.56} \right) \times 100\% = \frac{200}{56}\% = \frac{25}{7}\% \approx 3.57\%

Thus, Option (B) is true and Option (C) is false.

3. Error in the Determined Value of gg (Statement D)

The formula for the time period is: T=2π7(R−r)5gT = 2\pi \sqrt{\frac{7(R-r)}{5g}}

Squaring both sides and solving for gg: T2=28π2(R−r)5g  ⟹  g=28π2(R−r)5T2T^2 = \frac{28\pi^2 (R-r)}{5g} \implies g = \frac{28\pi^2 (R-r)}{5T^2}

The maximum relative error in gg is given by: Δgg=Δ(R−r)R−r+2(ΔTT)\frac{\Delta g}{g} = \frac{\Delta(R-r)}{R-r} + 2\left( \frac{\Delta T}{T} \right)

For the distance terms: R−r=60 mm−10 mm=50 mmR - r = 60\text{ mm} - 10\text{ mm} = 50\text{ mm} Δ(R−r)=ΔR+Δr=1 mm+1 mm=2 mm\Delta(R - r) = \Delta R + \Delta r = 1\text{ mm} + 1\text{ mm} = 2\text{ mm}

Therefore, the fractional error in (R−r)(R-r) is: Δ(R−r)R−r=2 mm50 mm=0.04 or 4%\frac{\Delta(R-r)}{R-r} = \frac{2\text{ mm}}{50\text{ mm}} = 0.04 \text{ or } 4\%

Now, calculating the percentage error in gg: Δgg×100%=4%+2×3.57%=4%+7.14%=11.14%≈11%\frac{\Delta g}{g} \times 100\% = 4\% + 2 \times 3.57\% = 4\% + 7.14\% = 11.14\% \approx 11\%

Thus, Option (D) is true.

Conclusion

The correct statements are A, B, and D.

PreviousNext