In order to determine acceleration due to gravity 'g' , a simple pendulum experiment is performed, the time taken for 20 oscillations is measured by using a watch with least count 1 second. The length of pendulum is measured by meter scale with least count 1 mm. After the experiment, the mean value of time taken comes out to be 30s and the length of pendulum obtained is 55cm . The percentage error in the determination of 'g' is approximately equal to:

Question

In order to determine acceleration due to gravity 'g' , a simple pendulum experiment is performed, the time taken for 20 oscillations is measured by using a watch with least count 1 second. The length of pendulum is measured by meter scale with least count 1 mm. After the experiment, the mean value of time taken comes out to be 30s and the length of pendulum obtained is 55cm . The percentage error in the determination of 'g' is approximately equal to: (A) 0.2 % (B) 0.7 % (C) 6.8 % (D) 3.5 %

Tardigrade Admin · Accepted Answer

g=(4 π 2 ℓ /T2) (Δ g/g)=(Δ ℓ /ℓ )+2(Δ T/T) (Δ g/g)=(0 . 1/55)+2(1/30) (Δ g/g)× 100=(.(10/55)+(20/3).)× 100=6.8 %

$0.2%$

$0.7%$

$6.8%$

$3.5%$

$g = \frac{4 π ^{2} ℓ}{T ^{2}}$
$\frac{Δ g}{g} = \frac{Δ ℓ}{ℓ} + 2 \frac{Δ T}{T}$ $\frac{Δ g}{g} = \frac{0.1}{55} + 2 \frac{1}{30}$
$\frac{Δ g}{g} \times 100 = (\frac{10}{55} + \frac{20}{3}) \times 100 = 6.8%$

0.2%

0.7%

6.8%

3.5%

g=T24π2ℓ​ gΔg​=ℓΔℓ​+2TΔT​ gΔg​=550.1​+2301​ gΔg​×100=(5510​+320​)×100=6.8%

$0.2%$

$0.7%$

$6.8%$

$3.5%$

$g = \frac{4 π ^{2} ℓ}{T ^{2}}$
$\frac{Δ g}{g} = \frac{Δ ℓ}{ℓ} + 2 \frac{Δ T}{T}$ $\frac{Δ g}{g} = \frac{0.1}{55} + 2 \frac{1}{30}$
$\frac{Δ g}{g} \times 100 = (\frac{10}{55} + \frac{20}{3}) \times 100 = 6.8%$