JEE MAIN - Physics (2015 (Offline) - No. 27)
The period of oscillation of a simple pendulum is $$T = 2\pi \sqrt {{L \over g}} $$. Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using wrist watch of 1 s resolution. The accuracy in the determination of g is:
1 %
5 %
2 %
3 %
Explanation
Given $$T = 2\pi \sqrt {{L \over g}} $$
$$ \Rightarrow g = {{4{\pi ^2}L} \over {{T^2}}}$$
$$ \Rightarrow g = {{4{\pi ^2}L{n^2}} \over {{t^2}}}$$
[ as $$T = {t \over n}$$ ]
So, percentage error in $$g$$ =
$${{\Delta g} \over g} \times 100 = {{\Delta L} \over L} \times 100 + 2{{\Delta t} \over t} \times 100$$
= $${{0.1} \over {20.0}} \times 100 + 2 \times {1 \over {90}} \times 100$$
= 2.72 % = 3 %
$$ \Rightarrow g = {{4{\pi ^2}L} \over {{T^2}}}$$
$$ \Rightarrow g = {{4{\pi ^2}L{n^2}} \over {{t^2}}}$$
[ as $$T = {t \over n}$$ ]
So, percentage error in $$g$$ =
$${{\Delta g} \over g} \times 100 = {{\Delta L} \over L} \times 100 + 2{{\Delta t} \over t} \times 100$$
= $${{0.1} \over {20.0}} \times 100 + 2 \times {1 \over {90}} \times 100$$
= 2.72 % = 3 %
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