Physics And Measurement Question 164
Question: The period of oscillation of a simple pendulum is given by $ T=2\pi \sqrt{\frac{l}{g}} $ where l is about 100 cm and is known to have 1mm accuracy. The period is about 2s. The time of 100 oscillations is measured by a stop watch of least count 0.1 s. The percentage error in g is
Options:
A) 0.1%
B) 1%
C) 0.2%
D) 0.8%
Show Answer
Answer:
Correct Answer: C
Solution:
$ T=2\pi \sqrt{l/g} $
$ \Rightarrow T^{2}=4{{\pi }^{2}}l/g $
$ \Rightarrow g=\frac{4{{\pi }^{2}}l}{T^{2}} $
Here % error in l = $ \frac{1mm}{100cm}\times 100=\frac{0.1}{100}\times 100=0.1% $
and % error in T = $ \frac{0.1}{2\times 100}\times 100=0.05% $
% error in g = % error in l + 2(% error in T)
$ =0.1+2\times 0.05 $ = 0.2 %
BETA
