In an experiment to find acceleration due to gravity (g) using simple pendulum, time period of 0.5s is measured… - JEE Main 2022 Physics PYQ

JEE Main 2022 — Physics Mechanics

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

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Held on 28 Jul 2022 · Verified 6 Jul 2026.

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Solution

Time period of oscillation is given by $T=2\pi \sqrt{\frac{l}{g}}$ , here, $l$ is length and $g$ is acceleration due to gravity.

From above relation, we have $g=\frac{1}{4{\pi }^{2}}\frac{{T}^{2}}{l}$ .

Fractional error in $g$ is stated as

$\frac{\Delta g}{g}=\frac{2\Delta T}{T}+\frac{\Delta l}{l}$

$\Rightarrow \frac{\Delta g}{g}\times 100=2\times \frac{1}{100\times 0.5}\times 100+\frac{1\mathrm{mm}}{10\mathrm{cm}}\times 100$

$\Rightarrow \frac{\Delta g}{g}\times 100=\frac{5}{100}\times 100=5%$

Hence, value of $x=5$ .

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Solution

Time period of oscillation is given by $T=2\pi \sqrt{\frac{l}{g}}$ , here, $l$ is length and $g$ is acceleration due to gravity.

From above relation, we have $g=\frac{1}{4{\pi }^{2}}\frac{{T}^{2}}{l}$ .

Fractional error in $g$ is stated as

$\frac{\Delta g}{g}=\frac{2\Delta T}{T}+\frac{\Delta l}{l}$

$\Rightarrow \frac{\Delta g}{g}\times 100=2\times \frac{1}{100\times 0.5}\times 100+\frac{1\mathrm{mm}}{10\mathrm{cm}}\times 100$

$\Rightarrow \frac{\Delta g}{g}\times 100=\frac{5}{100}\times 100=5%$

Hence, value of $x=5$ .

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Solution

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JEE Main Mechanics in other years

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Did you get this right?

Solution

Your note

JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Mechanics