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The time period of simple pendulum is t=2π.what is the accuracy in determination of g if 10cm length is known to 1mm accuracy and 0.5 s time period is measured from time of 100 oscillations with the watch of 1sec resolution?
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Calculation of Accuracy in Determination of g


Given Data:


  • Length of the pendulum (l) = 10 cm

  • Accuracy in the measurement of length (Δl) = 1 mm

  • Time period of the pendulum (t) = 0.5 s

  • Number of oscillations (n) = 100

  • Resolution of the watch (Δt) = 1 s



Calculation of g:

The time period of a simple pendulum is given by the formula:

t = 2π * sqrt(l/g)

where g is the acceleration due to gravity.

On rearranging the above formula, we get:

g = (4π^2 * l) / t^2

Substituting the given values, we get:

g = (4π^2 * 0.1) / (0.5)^2 = 39.478 m/s^2


Calculation of Accuracy in Determination of g:

The percentage error in the measurement of length is given by:

Δl/l * 100%

Substituting the given values, we get:

Δl/l * 100% = (1/10) * 100% = 10%

Similarly, the percentage error in the measurement of time is given by:

Δt/t * 100% = (1/500) * 100% = 0.2%

The total percentage error in the determination of g is given by:

Δg/g * 100% = (Δl/l + 2Δt/t) * 100%

Substituting the given values, we get:

Δg/g * 100% = (10% + 2*0.2%) * 100% = 10.4%

Therefore, the accuracy in the determination of g is 10.4%.
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