At a certain place, the acceleration due to gravity is determined using a simple pendulum. The length of the pendulum is 25.0cm measured with an instrument with least count 0.1cm and a stopwatch with least count of 1s measures the time taken by the pendulum for 40 oscillations to be 50s . The accuracy in the measurement of g is

Question

At a certain place, the acceleration due to gravity is determined using a simple pendulum. The length of the pendulum is 25.0cm measured with an instrument with least count 0.1cm and a stopwatch with least count of 1s measures the time taken by the pendulum for 40 oscillations to be 50s . The accuracy in the measurement of g is (A) 4.40 % (B) 3.40 % (C) 2.40 % (D) 5.40 %

Tardigrade Admin · Accepted Answer

g=(4 (π )2 l/T2)(Îg/g)=(2 ÎT/T)+(ÎL/L);=2((1/50))+(0 . 1/25 . 0)=4.4 %

$4.40%$

$3.40%$

$2.40%$

$5.40%$

$g = \frac{4 ( π ) ^{2} l}{T ^{2}} \frac{Δ g}{g} = \frac{2Δ T}{T} + \frac{Δ L}{L}; = 2 (\frac{1}{50}) + \frac{0.1}{25.0} = 4.4%$

4.40%

3.40%

2.40%

5.40%

g=T24(π)2l​gΔg​=T2ΔT​+LΔL​;=2(501​)+25.00.1​=4.4%

$4.40%$

$3.40%$

$2.40%$

$5.40%$

$g = \frac{4 ( π ) ^{2} l}{T ^{2}} \frac{Δ g}{g} = \frac{2Δ T}{T} + \frac{Δ L}{L}; = 2 (\frac{1}{50}) + \frac{0.1}{25.0} = 4.4%$