Physics And Measurement Question 192
Question: While measuring the acceleration due to gravity by a simple pendulum, a student makes a positive error of 1% in the length of the pendulum and a negative error of 3% in the value of time period. His percentage error in the measurement of g by the relation $ g=4\pi _{{}}^{2}(l/T _{{}}^{2}) $ will be
Options:
A) 2%
B) 4%
C) 7%
D) 10%
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Given that $ \frac{\Delta l}{l}\times $ 100=+1%and $ \frac{\Delta T}{T}\times $ 100=-3%
Percentage error in the measurement of g is
$ [ \frac{4{{\pi }^{2}}l}{T^{2}} ]=100\times \frac{\Delta l}{l}-2\times \frac{\Delta T}{T}\times 100 $ =1%-2[-3%]=+7%
