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The period of oscillation of a simple pe...

The period of oscillation of a simple pendulum is `T = 2 pi sqrt((L)/(g)) .L` is about `10 cm` and is known to `1mm` accuracy . The period of oscillation is about `0.5 s`. The time of 100 oscillation is measured with a wrist watch of `1 s` resolution . What is the accuracy in the determination of `g` ?

A

`2%`

B

`3%`

C

`4%`

D

`5%`

Text Solution

Verified by Experts

The correct Answer is:
D
The accuracy in determination of is found in terms of minimum percentage error in calculation. The percentage error in `g= (Delta g)/(g) xx 100`%. Where `(Delta g)/(g)` is the relative error in determination of g.
`T= 2pi sqrt((L)/(g)) or T^(2) =2pi^(2) sqrt((L)/(g)) or g= (4pi^(2)L)/(T^(2))`,
Now `(Delta g)/(g) = (Delta L)/(L) + 2 xx (Delta T)/(T)`
In terms of percentage, `(Delta g)/(g) xx 100 = (Delta L)/(L) xx 100 + 2 xx (Delta T)/(T) xx 100`
Percentage error in L `100 xx (Delta L)/(L) = 100 xx (0.1)/(10) = 1%`
Percentage error in `T = 100 xx (Delta T)/(T) = 100 xx (1)/(50) = 2%`
Thus percentage error in g `=100 xx (Delta g)/(g) = 1% + 2 xx 2% = 5%`
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