JEE Challenger
More from Units and Measurements
PhyMedAdv 2008 • Paper 1

Comparison of Percentage Errors in Acceleration Due to Gravity

Students I, II and III perform an experiment for measuring the acceleration due to gravity (gg) using a simple pendulum. They use different lengths of the pendulum and/or record time for different number of oscillations. The observations are shown in the table.

Least count for length =0.1 cm= 0.1\text{ cm}
Least count for time =0.1 s= 0.1\text{ s}

StudentLength of theNumber ofTotal time forTimependulum (cm)oscillations (n)(n) oscillations (s)period (s)I64.08128.016.0II64.0464.016.0III20.0436.09.0\begin{array}{|c|c|c|c|c|} \hline \text{Student} & \text{Length of the} & \text{Number of} & \text{Total time for} & \text{Time} \\ & \text{pendulum (cm)} & \text{oscillations } (n) & (n) \text{ oscillations (s)} & \text{period (s)} \\ \hline \text{I} & 64.0 & 8 & 128.0 & 16.0 \\ \hline \text{II} & 64.0 & 4 & 64.0 & 16.0 \\ \hline \text{III} & 20.0 & 4 & 36.0 & 9.0 \\ \hline \end{array}

If EI\text{E}_{\text{I}}, EII\text{E}_{\text{II}} and EIII\text{E}_{\text{III}} are the percentage errors in gg, i.e., (Δgg×100)\left(\frac{\Delta g}{g} \times 100\right) for students I, II and III, respectively,

Options

A

EI=0\text{E}_{\text{I}} = 0

B

EI\text{E}_{\text{I}} is minimum

Correct
C

EI=EII\text{E}_{\text{I}} = \text{E}_{\text{II}}

D

EII\text{E}_{\text{II}} is maximum

Topics & Concepts

Units and MeasurementsMeasuring InstrumentsSHM

Step-by-Step Solution

The acceleration due to gravity gg determined from a simple pendulum of length LL is given by:
T=2πLg  ⟹  g=4π2LT2T = 2\pi \sqrt{\frac{L}{g}} \implies g = 4\pi^2 \frac{L}{T^2}

Since the time period TT is determined by measuring the total time tt for nn oscillations, where T=tnT = \frac{t}{n}, we have:
g=4π2n2Lt2g = 4\pi^2 n^2 \frac{L}{t^2}

The fractional error (and thus the percentage error) in the measurement of gg is given by:
Δgg=ΔLL+2Δtt\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta t}{t}

The percentage error is:
E=(Δgg×100)=(ΔLL+2Δtt)×100\text{E} = \left(\frac{\Delta g}{g} \times 100\right) = \left( \frac{\Delta L}{L} + 2\frac{\Delta t}{t} \right) \times 100

Given:

  • Least count in length, ΔL=0.1 cm\Delta L = 0.1\text{ cm}
  • Least count in time, Δt=0.1 s\Delta t = 0.1\text{ s}

Now, we evaluate the error for each student:

  1. For Student I:

    • L=64.0 cmL = 64.0\text{ cm}
    • t=128.0 st = 128.0\text{ s}
      Δgg=0.164.0+2×0.1128.0=0.164.0+0.164.0=0.264.0=1320\frac{\Delta g}{g} = \frac{0.1}{64.0} + 2 \times \frac{0.1}{128.0} = \frac{0.1}{64.0} + \frac{0.1}{64.0} = \frac{0.2}{64.0} = \frac{1}{320}
      EI=100320%≈0.3125%\text{E}_{\text{I}} = \frac{100}{320}\% \approx 0.3125\%

  2. For Student II:

    • L=64.0 cmL = 64.0\text{ cm}
    • t=64.0 st = 64.0\text{ s}
      Δgg=0.164.0+2×0.164.0=0.364.0=3640\frac{\Delta g}{g} = \frac{0.1}{64.0} + 2 \times \frac{0.1}{64.0} = \frac{0.3}{64.0} = \frac{3}{640}
      EII=300640%≈0.46875%\text{E}_{\text{II}} = \frac{300}{640}\% \approx 0.46875\%

  3. For Student III:

    • L=20.0 cmL = 20.0\text{ cm}
    • t=36.0 st = 36.0\text{ s}
      Δgg=0.120.0+2×0.136.0=1200+1180=191800\frac{\Delta g}{g} = \frac{0.1}{20.0} + 2 \times \frac{0.1}{36.0} = \frac{1}{200} + \frac{1}{180} = \frac{19}{1800}
      EIII=1918%≈1.0556%\text{E}_{\text{III}} = \frac{19}{18}\% \approx 1.0556\%

Comparing the values:
EI<EII<EIII\text{E}_{\text{I}} < \text{E}_{\text{II}} < \text{E}_{\text{III}}

Therefore, EI\text{E}_{\text{I}} is minimum.

Hence, the correct option is B.

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