JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Gravitation

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Assume there are two identical simple pendulum Clocks-1 is placed on the earth and Clock-2 is placed on a space station located at a height h above the earth surface. Clock-1 and Clock-2 operate at time periods 4 s and 6 s respectively. Then the value of h is (consider radius of earth

R_{E} = 6400 km

and g on earth

10 m / s^{2}

)

(A)1200 km
(B)1600 km
(C)3200 km✓ Correct
(D)4800 km

Answer: (C)

Solution

For a simple pendulum,

T \propto 1/ g

and

g \propto 1/ (R + h)^{2}

. Therefore

\frac{T _{1}}{T _{2}} = \frac{g ^{'}}{g} = \frac{R ^{2}}{( R + h ) ^{2}} = \frac{R}{R + h}

. Substituting

T_{1} = 4

T_{2} = 6

R = 6400

km:

\frac{4}{6} = \frac{6400}{6400 + h}

. Solving:

6400 + h = 9600

, so

h = 3200

km.

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Single correct option

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