JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Elasticity & SHM

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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 = π^{2} ms^{- 2}

]

Answer: 8

Solution

At height

h

above the earth's surface, the acceleration due to gravity is

g^{'} = \frac{g}{( 1 + \frac{h}{R} ) ^{2}}

Here

h = R

(given), so

g^{'} = \frac{g}{( 1 + 1 ) ^{2}} = \frac{g}{4}

The time period of a simple pendulum of length

ℓ = 4 m

at this location is

T = 2 π \frac{ℓ}{g ^{'}} = 2 π \frac{4 ℓ}{g} = 2 π \frac{4 \times 4}{π ^{2}} = 2 π \times \frac{4}{π} = 8 s

JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Integer answer

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