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A student performs experiment with simple pendulum and measures time for 10 vibration.if he measures a time for 100 vibrations,the relative error in the measurement of time period will be reduced by

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A students performs an experiment to determine the acceleration due to gravity (g) at a place using a simple pendulum. The length of the pendulum is 60 cm and the total time for 30 oscillations is 100s. What is maximum percentage error for the measurement g ? Given, least count for time $= 0.1 s$ and least count for length $= 0.1 c m$ .

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 :

Q. While measuring

g

using simple pendulum method

%

error in measurement of

ℓ

\pm 2 %

and in measurement time is

\pm 1 %

then relative error in

g

Q. A boy performs an experiment in which he uses a simple pendulum to find the value of 'g' using the formula

g = \frac{4 π^{2} l}{T^{2}}

= Where l = 1 m. Error in measurements of length is

Δ l = 2 c m

, human error in time measurement is 0.15sec and least count of stop watch

Δ T = 0.01 s

, What will be the percentage error in determining value of

g

?

Q. A student measures the time period of

100

oscillations of a simple pendulum four times. That data set

90 s, 91 s, 95 s

92 s

. If the minimum division in the measuring clock is

1 s,

then he reported time should be

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