Question 295: Units , Dimensions and Error Analysis

Question

A student uses a simple pendulum of exactly 1m length to determine g, the acceleration due to gravity. He
uses a stop watch with the least count of 1 sec for this and records 40 seconds for 20 oscillations. For this
observation, which of the following statement(s) is (are) true?

Accepted Answer

Relative error in measurement of time,
$${{\Delta t} \over t} = {{1\,s} \over {40\,s}} = {1 \over {40}}$$
Time period, $$T = {{40\,s} \over {20}} = 2\,s$$
Error in measurement of time period,
$$\Delta T = T 	imes {{\Delta t} \over t} = 2\,s 	imes {1 \over {40}} = 0.05\,s$$
The time period of simple pendulum is
$$T = 2\pi \sqrt {{l \over g}} $$
or, $${T^2} = {{4{\pi ^2}l} \over g}$$
or, $$g = {{4{\pi ^2}l} \over {{T^2}}}$$
$$	herefore$$ $${{\Delta g} \over g} = {{2\Delta T} \over T} = 2 	imes {1 \over {40}} = {1 \over {20}}$$ ($$\because$$ $${{\Delta T} \over T} = {{\Delta t} \over t}$$)
Percentage error in determination of g is
$${{\Delta g} \over g} 	imes 100 = {1 \over {20}} 	imes 100 = 5\% $$

Answer

Error ΔT in measuring T, the time period, is 0.05 seconds

Answer

Error ΔT in measuring T, the time period, is 1 second

Answer

Percentage error in the determination of g is 5%

Answer

Percentage error in the determination of g is 2.5%

Units , Dimensions and Error Analysis

A student uses a simple pendulum of exactly 1m length to determine g, the acceleration due to gravity. He uses a stop watch with the least count of 1 sec for this and records 40 seconds for 20 oscillations. For this observation, which of the following statement(s) is (are) true?

Choose the correct answer: