Verified Solution • 2025–26 Session moderate

QJEEAdv08Ph717f: JEE Advanced 2008 Physics - Class 11 JEEAdv08Ph717f Solution: Student I, II and III perform an experiment for measuring the…

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

Least count for length = 0.1 cm

Least count for time = 0.1 s

Student Length of the
pendulum
(cm)		 No. of
oscillations
(n)		 Total time
for(n)
oscillations
(s)		 Time
periods
(s)
I 64.0 8 128.0 16.0
II 64.0 4 64.0 16.0
III 20.0 4 36.0 9.0

If EI, EII and EIII are the percentage errors in g, i.e., (△gg×100)\left(\frac{\triangle g}g\times100\right) for students I, II and III, respectively,then

Multiple Choice Options

  • A EI = 0
  • B EI is minimum
  • C EI = EII
  • D EII is maximum

Step-by-Step Verified Solution

Time period (T) =2πlg= 2\pi \sqrt {{l \over g}}

or tn=2πlg{t \over n} = 2\pi \sqrt {{l \over g}}

∴\therefore g=(4π2)(n2)lt2g = {{(4{\pi ^2})({n^2})l} \over {{t^2}}}

% error in g=Δgg×100g = {{\Delta g} \over g} \times 100

=(Δll×2Δll)×100= \left( {{{\Delta l} \over l} \times {{2\Delta l} \over l}} \right) \times 100

EI=(0.164+2×0.1128)×100=0.3125%{E_I} = \left( {{{0.1} \over {64}} + {{2 \times 0.1} \over {128}}} \right) \times 100 = 0.3125\%

EII=(0.164+2×0.164)×100=0.46875%{E_{II}} = \left( {{{0.1} \over {64}} + {{2 \times 0.1} \over {64}}} \right) \times 100 = 0.46875\%

EIII=(0.120+2×0.136)×100=1.055%{E_{III}} = \left( {{{0.1} \over {20}} + {{2 \times 0.1} \over {36}}} \right) \times 100 = 1.055\%

Hence, EI\mathrm{E_I} is minimum.

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