The period of oscillation of a simple pendulum is T =2 π √( L / g ). Measured value of 'L' is 1.0 m from meter scale having a minimum division of 1 mm and time of one complete oscillation is 1.95 s measured from stopwatch of 0.01 s resolution. The percentage error in the determination of 'g' will be :

Question

The period of oscillation of a simple pendulum is T =2 π √( L / g ). Measured value of 'L' is 1.0 m from meter scale having a minimum division of 1 mm and time of one complete oscillation is 1.95 s measured from stopwatch of 0.01 s resolution. The percentage error in the determination of 'g' will be : (A) 1.13 % (B) 1.03 % (C) 1.33 % (D) 1.30 %

Tardigrade Admin · Accepted Answer

T =2 π √(ℓ/ g ) g =(4 π2 ℓ/ T 2) (Δ g / g )=(Δ ℓ/ℓ)+(2 Δ T / T ) (Δ g / g )=(1 × 10-3/1)+2 × (0.01/1.95) (Δ g / g)=0.0113 or 1.13 %

$1.13%$

$1.03%$

$1.33%$

$1.30%$

$T = 2 π \frac{ℓ}{g}$
$g = \frac{4 π ^{2} ℓ}{T ^{2}}$
$\frac{Δ g}{g} = \frac{Δ ℓ}{ℓ} + \frac{2Δ T}{T}$
$\frac{Δ g}{g} = \frac{1 \times 1 0 ^{- 3}}{1} + 2 \times \frac{0.01}{1.95}$
$\frac{Δ g}{g} = 0.0113$ or $1.13%$

Q. The period of oscillation of a simple pendulum is T=2πgL​​. Measured value of 'L' is 1.0m from meter scale having a minimum division of 1mm and time of one complete oscillation is 1.95 s measured from stopwatch of 0.01s resolution. The percentage error in the determination of 'g' will be :

1.13%

1.03%

1.33%

1.30%

T=2πgℓ​​ g=T24π2ℓ​ gΔg​=ℓΔℓ​+T2ΔT​ gΔg​=11×10−3​+2×1.950.01​ gΔg​=0.0113 or 1.13%

$1.13%$

$1.03%$

$1.33%$

$1.30%$

$T = 2 π \frac{ℓ}{g}$
$g = \frac{4 π ^{2} ℓ}{T ^{2}}$
$\frac{Δ g}{g} = \frac{Δ ℓ}{ℓ} + \frac{2Δ T}{T}$
$\frac{Δ g}{g} = \frac{1 \times 1 0 ^{- 3}}{1} + 2 \times \frac{0.01}{1.95}$
$\frac{Δ g}{g} = 0.0113$ or $1.13%$