Question

The period of oscillation of a simple pendulum is given by T=2×pi×root(l/g),where l is about 100 cm and is known to have 1mm accuracy.The period is qbout 2s.The time of 100 oscillations is measured by a stop watch of least count 0.1s.The percentage of error in g is:

Answer 1

T =2π√( L/g )
t/n = 2π√L/g

t= 4π²Ln²/g

g = 4π²L/t²
here n is no of oscillations

so,
∆g/g = ∆L/L + 2∆t/t

given ,
L = 100 cm
∆L = 1mm = 0.1 cm
t = 200 sec
∆t =0.1 sec

now,

∆g/g = 0.1/100 + 2× 0.1/200

percentage error % of g = ∆g/g ×100
=0.1/100 ×100 + 0.1×100/100
=0.1 + 0.1= 0.2 %

Answer 2

Answer:

5% is the percentage error in g

Explanation:

Below is explanation :

HOPE IT HELPS :)