JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Elasticity & SHM

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T is the time period of simple pendulum on the earth's surface. Its time period becomes xT when taken to a height R (equal to earth's radius) above the earth's surface. Then, the value of

x

will be:

(A)4
(B)2✓ Correct
(C) $\frac{1}{2}$
(D) $\frac{1}{4}$

Answer: (B)

Solution

The period of a simple pendulum is

T = 2 π ℓ / g

, so for a fixed length it varies inversely as

g

. At a height

h = R

the acceleration due to gravity falls to

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

. The period there becomes

T^{'} = 2 π \frac{ℓ}{g /4} = 2 \cdot 2 π \frac{ℓ}{g} = 2 T

. The period grows to twice its surface value, so

x = 2

JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Single correct option

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