JEE Advance - Physics (2008 - Paper 1 Offline - No. 1)
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., $$\left(\frac{\triangle g}g\times100\right)$$ for students I, II and III, respectively,then
Explanation
Time period (T) $$ = 2\pi \sqrt {{l \over g}} $$
or $${t \over n} = 2\pi \sqrt {{l \over g}} $$
$$\therefore$$ $$g = {{(4{\pi ^2})({n^2})l} \over {{t^2}}}$$
% error in $$g = {{\Delta g} \over g} \times 100$$
$$ = \left( {{{\Delta l} \over l} \times {{2\Delta l} \over l}} \right) \times 100$$
$${E_I} = \left( {{{0.1} \over {64}} + {{2 \times 0.1} \over {128}}} \right) \times 100 = 0.3125\% $$
$${E_{II}} = \left( {{{0.1} \over {64}} + {{2 \times 0.1} \over {64}}} \right) \times 100 = 0.46875\% $$
$${E_{III}} = \left( {{{0.1} \over {20}} + {{2 \times 0.1} \over {36}}} \right) \times 100 = 1.055\% $$
Hence, $$\mathrm{E_I}$$ is minimum.
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