In an experiment to find acceleration due to gravity (g) using simple pendulum, time period of 0.5 s is measured from time of 100 oscillation with a watch of 1 s resolution. If measured value of length is 10 cm known to 1 mm accuracy. The accuracy in the determination of g is found to be x %. The value of x is

Question

In an experiment to find acceleration due to gravity (g) using simple pendulum, time period of 0.5 s is measured from time of 100 oscillation with a watch of 1 s resolution. If measured value of length is 10 cm known to 1 mm accuracy. The accuracy in the determination of g is found to be x %. The value of x is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

T =2 π √(ℓ/ g ) g =(1/4 π2) ( T 2/ℓ) (Δ g / g )=(2 Δ T / T )+(Δ ℓ/ℓ) (Δ g / g )=2 ⋅ (1/100 × 0.5)+(1 mm /10 cm ) (Δ g / g )=(5/100)

T=2πgℓ​​ g=4π21​ℓT2​ gΔg​=T2ΔT​+ℓΔℓ​ gΔg​=2⋅100×0.51​+10cm1mm​ gΔg​=1005​

$T = 2 π \frac{ℓ}{g}$
$g = \frac{1}{4 π ^{2}} \frac{T ^{2}}{ℓ}$
$\frac{Δ g}{g} = \frac{2Δ T}{T} + \frac{Δ ℓ}{ℓ}$
$\frac{Δ g}{g} = 2 \cdot \frac{1}{100 \times 0.5} + \frac{1 mm}{10 c m}$
$\frac{Δ g}{g} = \frac{5}{100}$