A simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is 4 m, then the time period of small oscillations will be __________ s. [take g = π² m s⁻² ] Numerical answer type. Work it out, then open the answer.

Answer: 8. At h = R: g' = g(R/2R)² = g/4 = (π²)/(4) T = 2π√(L/g') = 2π√((4 × 4)/(π²)) T = 2π × 4/π = 8 s So T = 8 s.

A simple pendulum is placed at a place where its distance from the… | JEE Main 2024 (30 Jan, Shift 2)

Q39JEE Main 2024 · 30 JAN · SHIFT 2Medium

Acceleration Due to Gravity

A simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is $4\text{ m}$ , then the time period of small oscillations will be __________ s. [take $g = \pi^2\text{ m s}^{-2}$ ]

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

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Answer: 8

At $h = R$ : $g' = g\left(\frac{R}{2R}\right)^2 = \frac{g}{4} = \frac{\pi^2}{4}$
$T = 2\pi\sqrt{\frac{L}{g'}} = 2\pi\sqrt{\frac{4 \times 4}{\pi^2}}$
$T = 2\pi \times \frac{4}{\pi} = 8$ s
So $T = 8$ s.

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