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 $g = 4 π^{2} (\frac{l}{T^{2}})$ will be

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Solution

The correct option is C
7%

Percentage error in $g = (% e r r o r i n l) + 2 (% e r r o r i n T) = 1 % + 2 (3 %) = 7 %$

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Q. 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 π^{2} (\frac{l}{T^{2}})

will be

Q. While measuring 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 the time period. His percentage error in the measurement of the value of

g

will be:

Q. In an experiment to determine acceleration due to gravity by simple pendulum, a student commits 1 % positive error in the measurement of length and 3% negative error in the measurement of time period. The percentage error in the value of g will be :

A student makes an error of 1% in measuring length of pendulum and negative error of 3% in value of time periods. The percentage error in the measurement of ‘g’ will be

Q. Length

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and Acceleration due to gravity

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are measured with

\pm 6 %

and

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errors respectively. The percentage error in determination of the Time period of a simple pendulum will be,

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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=4π2(lT2) will be

The correct option is C 7% Percentage error in g=(% error in l) + 2(% error in T)=1%+2(3%)=7%

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 $g = 4 π^{2} (\frac{l}{T^{2}})$ will be

The correct option is C
7%

Percentage error in $g = (% e r r o r i n l) + 2 (% e r r o r i n T) = 1 % + 2 (3 %) = 7 %$