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 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

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