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JEE Main 2022 (28 Jul, Shift 2) · Simple Pendulum

Simple Pendulum question from JEE Main 2022 (28 Jul, Shift 2)

Q56JEE Main 2022 · 28 JUL · SHIFT 2Medium

Simple Pendulum

Assume there are two identical simple pendulum clocks. Clock - 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 RE=6400 kmR_E = 6400\text{ km} and gg on earth 10 m/s210\text{ m/s}^{2})

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Answer: (c)

  1. T∝1gT \propto \frac{1}{\sqrt{g}} and g∝1(R+h)2g \propto \frac{1}{(R+h)^2}, so T∝(R+h)T \propto (R+h).
  2. T2T1=64=R+hR⇒h=0.5R\frac{T_2}{T_1} = \frac{6}{4} = \frac{R+h}{R} \Rightarrow h = 0.5R.
  3. h=0.5×6400=3200 kmh = 0.5 \times 6400 = 3200\text{ km}.
  4. So the answer is (c) 3200 km.
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