In an experiment to find acceleration due to gravity (g) using simple pendulum, time period of 0.5 s is measured from time of 100 oscillation with a watch of 1 s resolution. If measured value of length is 10 cm known to 1 mm accuracy, The accuracy in the determination of g is found to be x%. The value of x is ___________. Numerical answer type. Work it out, then open the answer.

Answer: 5. g = (4π² l)/(T²), so (Δ g)/(g) = (Δ l)/(l) + 2(Δ t)/(t). (Δ l)/(l) = (0.1 cm)/(10 cm) = 1% Total time t = 100× 0.5 = 50 s, so (Δ t)/(t) = 1/50 = 2% and 2(Δ t)/(t) = 4%. x = 1 + 4 = 5

In an experiment to find acceleration due to gravity (g) using simple… | JEE Main 2022 (28 Jul, Shift 2)

Q78JEE Main 2022 · 28 JUL · SHIFT 2Difficult

Errors in Measurement

In an experiment to find acceleration due to gravity (g) using simple pendulum, time period of $0.5\text{ s}$ is measured from time of 100 oscillation with a watch of $1\text{ s}$ resolution. If measured value of length is $10\text{ cm}$ known to $1\text{ mm}$ accuracy, The accuracy in the determination of $g$ is found to be $x\%$ . The value of $x$ is ___________.

Numerical answer type. Work it out, then open the answer.

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Answer: 5

$g = \dfrac{4\pi^2 l}{T^2}$ , so $\dfrac{\Delta g}{g} = \dfrac{\Delta l}{l} + 2\dfrac{\Delta t}{t}$ .
$\dfrac{\Delta l}{l} = \dfrac{0.1\text{ cm}}{10\text{ cm}} = 1\%$
Total time $t = 100\times 0.5 = 50$ s, so $\dfrac{\Delta t}{t} = \dfrac{1}{50} = 2\%$ and $2\dfrac{\Delta t}{t} = 4\%$ .
$x = 1 + 4 = 5$

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