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.

In determination of value of acceleration due to gravity (g) by simple

1.	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.
Answer» 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.

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