Question

What is the percentage error in the measurement of time period of simple pendulum if maximum errors in the measurement of "l" and "g" are 2% and 4% respectively....???

Answer 1

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Answer 2

Answer:

The percentage error in the measurement of time period of simple pendulum is 3%.

Explanation:

Given that,

Error of l = 2%

Error of g = 4%

Formula of the Time period is

$T = 2\pi\sqrt{\dfrac{l}{g}}$

The percentage error in the measurement of time period of simple pendulum

$\dfrac{\Delta T}{T}\times100=\dfrac{1}{2}(\dfrac{\Delta l}{l}\times100+\dfrac{\Delta g}{g}\times100)$

$\dfrac{\Delta T}{T}\times100=\dfrac{1}{2}(2\%+4\%)$

$\dfrac{\Delta T}{T}\times100=1\%+2\%$

$\dfrac{\Delta T}{T}\times100=3\%$

Hence, The percentage error in the measurement of time period of simple pendulum is 3%.