In an experiment to determine the acceleration due to gravity g, the formula used for the time period of a periodic motion is T = 2 π 7 R - r 5 g . The values of R and r are measured to be 60 ± 1   m m   a n d   10 ± 1 m m ,   respectively. In five successive…

Question

In an experiment to determine the acceleration due to gravity g, the formula used for the time period of a periodic motion is T = 2 &#960; 7 R - r 5 g . The values of R and r are measured to be 60 &#177; 1 &#160; m m &#160; a n d &#160; 10 &#177; 1 m m , &#160; respectively. In five successive measurements, the time period is found to be 0.52s, 0.56s, 0.57s, 0.54s and 0.59s. The least count of the watch used for the measurement of time period is 0.01s. Which of the following statements (s) is (are) true?

Aicharya · Accepted Answer

T Absolute error 1 0.52 - &#160; 0.04 2 0.56 00 3 0.57 + &#160; 0.01 4 0.54 - 0.02 5 0.59 + 0.03 T a v g = 0.556 = 0.56 Avg. absolute error 0 . 04 + 0 + 0 . 01 + 0 . 02 + 0 . 03 5 = 0 . 02 Percentage error in time period = &#8710; T T &#215; 100 % = 0 . 02 0 . 56 &#215; 100 % &#8776; 3.57 % Percentage error in radius = &#8710; r r &#215; 100 % = 1 10 &#215; 100 = 10 % Also, g = 28 &#960; R - r 5 T 2 Percentage error in g ,= &#8710; g g = &#8710; R + &#8710; r R - r + 2 &#8710; T T &#8710; g g &#215; 100 % = 1 + 1 50 &#215; 100 % + 2 3.57 % &#8776; 11 %

In an experiment to determine the acceleration due to gravity g, the formula used for the time period of a…

Solution

	T	Absolute error
1	0.52	$- 0.04$
2	0.56	00
3	0.57	$+ 0.01$
4	0.54	$- 0.02$
5	0.59	$+ 0.03$
	$T_{a v g} = 0.556$ $= 0.56$