In a simple pendulum experiment for determination of acceleration due to gravity (g), time taken for 20 oscillations is measured by using a watch of 1 second least count. The mean value of time taken comes out to be 30 s. The length of pendulum is measured by using a meter scale of least count 1 mm and the value obtained is 55.0 cm. The percentage error in the determination of g is close to :

Question

In a simple pendulum experiment for determination of acceleration due to gravity (g), time taken for 20 oscillations is measured by using a watch of 1 second least count. The mean value of time taken comes out to be 30 s. The length of pendulum is measured by using a meter scale of least count 1 mm and the value obtained is 55.0 cm. The percentage error in the determination of g is close to : (A) 0.7 % (B) 0.2 % (C) 3.5 % (D) 6.8 %

Tardigrade Admin · Accepted Answer

T = (30 sec/20) Δ T = (1/20) sec L = 55 cm Δ L = 1 mm = 0.1 cm g = (4π2L/T2) percentage error in g is (Δ g/g) ×100 % = ((Δ L/L) + (2Δ T/T)) 100 % = ((0.1 /55) + (2((1/20))/(30)20)) 100 % ≃ 6.8 %

0.7 %

0.2 %

3.5 %

6.8 %

T=2030sec​ΔT=201​sec L=55cmΔL=1mm=0.1cm g=T24π2L​ percentage error in g is gΔg​×100%=(LΔL​+T2ΔT​)100% =(550.1​+2030​2(201​)​)100%≃6.8%

$T = \frac{30 s e c}{20} Δ T = \frac{1}{20} sec$
$L = 55 c m Δ L = 1 mm = 0.1 c m$
$g = \frac{4 π ^{2} L}{T ^{2}}$
percentage error in g is
$\frac{Δ g}{g} \times 100% = (\frac{Δ L}{L} + \frac{2Δ T}{T}) 100%$
$= (\frac{0.1}{55} + \frac{2 ( \frac{1}{20} )}{\frac{30}{20}}) 100% ≃ 6.8%$