In determination of value of acceleration due to gravity (g) by simple pendulum, the time period is measured by a topwatch (L.C. = 0.5 second) and length of the thread is measured with metre scale (L.C. 0.1 cm) and diametre of bob is measured with vernier callipers (L.C. = 0.1 cm). The following observations are recorded :
Length of the thread `=105.3 cm`
Diameter of the bob `=2.45 cm`
Time period `=2.07 s`
Number of oscillations `=10`
Calculate the value of g, estimate error and write the result in proper significant figures.

Explanation

The time period of simple pendulum is given by
`T=2pi sqrt(L/g)`
Here, `L=l+r=105.3+1.225=106.525 cong 106.5 cm`
or `g=(4pi^(2)L)/T^(2)`
`=(4xx9.87xx106.5)/((2.07)^(2))=982 cm//s^(2)`
The error in g is given by
`(Deltag)/(g)=(DeltaL)/L+2 (Delta T)/(T)=(Delta(l+r))/((l+r))+2(Delta T)/(T)`
`=((0.1+0.005))/((105.3+1.225))+2xx0.5/(2.07xx10)`
`=0.00098+0.048 cong 0.048 cong 5%`
`:. Deltag=0.048xxg=0.048xx982=47 cm//s^(2)`
Thus, the value of g is `(982 pm 47) cm//s^(2)`.
Note : Let us see the result if we increase the number of oscillations from 10 to 20.
`(Delta g)/g=0.00098xx(2xx0.5)/(2.07xx20) cong 0.024 cong 2%`
Thus, the relative error in g falls down from `5 %` to `2 %` by increasing number of oscillations from 10 to 20.