In a simple pendulum experiment for determination of acceleration due to gravity (.g.) , time taken for 20 oscillations is measures by using a watch of 1 second least count. The mean value of time taken comes out to be 30 s . The length of the pendulum is measured by using a meter scale of least count 1 mm and the value obtained is 55.0 cm . The percentage error in the determination of g is close to

Question

In a simple pendulum experiment for determination of acceleration due to gravity (.g.) , time taken for 20 oscillations is measures by using a watch of 1 second least count. The mean value of time taken comes out to be 30 s . The length of the pendulum is measured by using a meter scale of least count 1 mm and the value obtained is 55.0 cm . The percentage error in the determination of g is close to (A)  text0.2 % (B)  text6.8 % (C)  text3.5 % (D)  text0.7 %

Tardigrade Admin · Accepted Answer

g=(4 π 2 l/T2) (Δ g/g)=(Δ l/l)+(2 Δ T/T) =(0.1/55)+2× ((1/30)) =0.06848 Percentage error ≈6.8 %

$0.2 %$

$6.8 %$

$3.5 %$

$0.7 %$

$g = \frac{4 π ^{2} l}{T ^{2}}$
$\frac{Δ g}{g} = \frac{Δ l}{l} + \frac{2Δ T}{T}$
$= \frac{0.1}{55} + 2 \times (\frac{1}{30})$
$= 0.06848$
Percentage error $\approx 6.8%$

0.2%

6.8%

3.5%

0.7%

g=T24π2l​ gΔg​=lΔl​+T2ΔT​ =550.1​+2×(301​) =0.06848 Percentage error ≈6.8%

$0.2 %$

$6.8 %$

$3.5 %$

$0.7 %$

$g = \frac{4 π ^{2} l}{T ^{2}}$
$\frac{Δ g}{g} = \frac{Δ l}{l} + \frac{2Δ T}{T}$
$= \frac{0.1}{55} + 2 \times (\frac{1}{30})$
$= 0.06848$
Percentage error $\approx 6.8%$