An experiment is performed to obtain the value of acceleration due to gravity g by using a simple pendulum of length L. In this experiment time for 100 oscillations is measured by using a watch of 1 second least count and the value is 90.0 seconds. The length L is measured by using a meter scale of least count 1 mm and the value is 20.0 cm. The error in the determination of g would be :

Question

An experiment is performed to obtain the value of acceleration due to gravity g by using a simple pendulum of length L. In this experiment time for 100 oscillations is measured by using a watch of 1 second least count and the value is 90.0 seconds. The length L is measured by using a meter scale of least count 1 mm and the value is 20.0 cm. The error in the determination of g would be : (A) 1.7 % (B) 2.7 % (C) 4.4 % (D) 2.27 %

Tardigrade Admin · Accepted Answer

T2=(4π2ℓ/g) g=4π2 (ℓ/T2) (Δ g/g)×100=((Δℓ/ℓ)× 100)+2((Δ T/T)× 100) =((0.1/20)× 100)+2((0.01/.9)× 100) =0.5+2× (10/9)=0.5+2.2=2.7 %

1.7 %

2.7 %

4.4 %

2.27 %

T2=g4π2ℓ​ g=4π2T2ℓ​ gΔg​×100=(ℓΔℓ​×100)+2(TΔT​×100) =(200.1​×100)+2(.90.01​×100) =0.5+2×910​=0.5+2.2=2.7%

$T^{2} = \frac{4 π ^{2} ℓ}{g}$
$g = 4 π^{2} \frac{ℓ}{T ^{2}}$
$\frac{Δ g}{g} \times 100 = (\frac{Δ ℓ}{ℓ} \times 100) + 2 (\frac{Δ T}{T} \times 100)$
$= (\frac{0.1}{20} \times 100) + 2 (\frac{0.01}{.9} \times 100)$
$= 0.5 + 2 \times \frac{10}{9} = 0.5 + 2.2 = 2.7%$