JEE Advance - Physics (2010 - Paper 1 Offline - No. 10)
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?
Error ΔT in measuring T, the time period, is 0.05 seconds
Error ΔT in measuring T, the time period, is 1 second
Percentage error in the determination of g is 5%
Percentage error in the determination of g is 2.5%
Giải thích
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,
$$\Delta T = T \times {{\Delta t} \over t} = 2\,s \times {1 \over {40}} = 0.05\,s$$
The time period of simple pendulum is
$$T = 2\pi \sqrt {{l \over g}} $$
or, $${T^2} = {{4{\pi ^2}l} \over g}$$
or, $$g = {{4{\pi ^2}l} \over {{T^2}}}$$
$$\therefore$$ $${{\Delta g} \over g} = {{2\Delta T} \over T} = 2 \times {1 \over {40}} = {1 \over {20}}$$ ($$\because$$ $${{\Delta T} \over T} = {{\Delta t} \over t}$$)
Percentage error in determination of g is
$${{\Delta g} \over g} \times 100 = {1 \over {20}} \times 100 = 5\% $$