The time period of a simple pendulum is given by T =2 π √(ℓ/ g ) The measured value of the length of pendulum is 10 cm known to a 1 mm accuracy. The time for 200 oscillations of the pendulum is found to be 100 second using a clock of 1 s resolution. The percentage accuracy in the determination of 'g' using this pendulum is 'x'. The value of 'x' to the nearest integer is:-

Question

The time period of a simple pendulum is given by T =2 π √(ℓ/ g ) The measured value of the length of pendulum is 10 cm known to a 1 mm accuracy. The time for 200 oscillations of the pendulum is found to be 100 second using a clock of 1 s resolution. The percentage accuracy in the determination of 'g' using this pendulum is 'x'. The value of 'x' to the nearest integer is:- (A) 2 % (B) 3 % (C) 5 % (D) 4 %

Tardigrade Admin · Accepted Answer

g=(4 π2 ℓ/T2) (Δ g/g)=(Δ ℓ/ℓ)+2 (Δ T/T)=(0.1/10)+2(((1/200)/0.5)) (Δ g/g)=(1/100)+(1/50) (Δ g/g) × 100=3 %

$2%$

$3%$

$5%$

$4%$

$g = \frac{4 π ^{2} ℓ}{T ^{2}}$
$\frac{Δ g}{g} = \frac{Δ ℓ}{ℓ} + 2 \frac{Δ T}{T} = \frac{0.1}{10} + 2 (\frac{\frac{1}{200}}{0.5})$
$\frac{Δ g}{g} = \frac{1}{100} + \frac{1}{50}$
$\frac{Δ g}{g} \times 100 = 3%$

2%

3%

5%

4%

g=T24π2ℓ​ gΔg​=ℓΔℓ​+2TΔT​=100.1​+2(0.52001​​) gΔg​=1001​+501​ gΔg​×100=3%

$2%$

$3%$

$5%$

$4%$

$g = \frac{4 π ^{2} ℓ}{T ^{2}}$
$\frac{Δ g}{g} = \frac{Δ ℓ}{ℓ} + 2 \frac{Δ T}{T} = \frac{0.1}{10} + 2 (\frac{\frac{1}{200}}{0.5})$
$\frac{Δ g}{g} = \frac{1}{100} + \frac{1}{50}$
$\frac{Δ g}{g} \times 100 = 3%$