The period of oscillation of a simple pendulum is T = 2π\(\sqrt {\frac{{\rm{L}}}{{\rm{g}}}} \). Measured value of 'L' is 1.0 m from meter scale having a minimum division of 1 mm and time of one complete oscillation is 1.95 s measured from stopwatch of 0.01 s resolution. The percentage error in the determination of 'g' will be:

CONCEPT:

To calculate the percentage error we use the formulation as

\(\frac{\Delta A}{A}= \frac{\Delta X}{X}+\frac{\Delta Y}{Y}+... \)       ---- (1)

CALCULATION:

Given: T = 2π\(\sqrt {\frac{{\rm{L}}}{{\rm{g}}}} \)       ----(2)

L = 1.0 m

T = 1.98 s

\(\Delta T = 0.01 s\)

\(\Delta L = 0.001 m \)

Let us simplify the equation (2) we have;

T = 2π\(\sqrt {\frac{{\rm{L}}}{{\rm{g}}}} \)

T² = 4π²\(\times\)\({\frac{{\rm{L}}}{{\rm{g}}}} \)

⇒ \(g=\frac{4\pi^2 L}{T^2}\)

For error we have;

\(\frac{\Delta g}{g}=\frac{\Delta L}{L}+2\frac{\Delta T}{T} \)

Now, On putting all the given values we have;

\(\frac{\Delta g}{g}=\frac{0.001}{1}+\frac{2\times0.01}{1.95} \)

⇒ \(\frac{\Delta g}{g}=0.001+0.01025\)

⇒\(\frac{\Delta g}{g}=0.01125 \)

⇒\(\frac{\Delta g}{g}=1.125 \)%

Hence, option 1) is the correct answer.