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MCQEasyJEE 2023Variation with Altitude & Depth

JEE Physics 2023 Question with Solution

T is the time period of a simple pendulum on the Earth's surface. Its time period becomes x Tx \, T when taken to a height RR (equal to Earth's radius) above the Earth's surface. Then, the value of xx will be:

  • A

    44

  • B

    22

  • C

    12\frac{1}{2}

  • D

    14\frac{1}{4}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Time period on Earth's surface is TT. The pendulum is taken to height h=Rh = R above the Earth's surface.

Find: The value of xx such that the new time period is xTxT.

For a simple pendulum,

T∝1gT \propto \frac{1}{\sqrt{g}}

At height hh above Earth,

g′=g(1+hR)2g' = \frac{g}{\left(1+\frac{h}{R}\right)^2}

With h=Rh = R,

g′=g(1+1)2=g4g' = \frac{g}{(1+1)^2} = \frac{g}{4}

Hence the new time period T′T' is

T′=Tgg′T' = T \sqrt{\frac{g}{g'}}

Substituting g′=g4g' = \frac{g}{4},

T′=Tgg/4=T4=2TT' = T \sqrt{\frac{g}{g/4}} = T\sqrt{4} = 2T

Therefore, x=2x = 2. The correct option is B.

Using the pendulum formula explicitly

Given: On the surface,

T=2πℓgT = 2\pi\sqrt{\frac{\ell}{g}}

At height h=Rh = R, acceleration due to gravity becomes

g′=g(1+hR)2=g4g' = \frac{g}{\left(1+\frac{h}{R}\right)^2} = \frac{g}{4}

Find: The factor xx in T′=xTT' = xT.

At the new height,

T′=2πℓg′T' = 2\pi\sqrt{\frac{\ell}{g'}}

So,

T′=2πℓg/4=2π4ℓg=2⋅2πℓg=2TT' = 2\pi\sqrt{\frac{\ell}{g/4}} = 2\pi\sqrt{\frac{4\ell}{g}} = 2\cdot 2\pi\sqrt{\frac{\ell}{g}} = 2T

Thus,

xT=2T⇒x=2xT = 2T \Rightarrow x = 2

Therefore, the value of xx is 22.

Common mistakes

  • Using g′=g2g' = \frac{g}{2} at height RR is incorrect because gravity varies as the inverse square of distance from Earth's center. Use g′=g(1+h/R)2g' = \frac{g}{(1+h/R)^2} instead.

  • Assuming time period is directly proportional to g\sqrt{g} is wrong. For a simple pendulum, T∝1gT \propto \frac{1}{\sqrt{g}}, so the time period increases when gravity decreases.

  • Taking the distance from Earth's center as RR instead of R+hR+h is incorrect. At height h=Rh = R, the distance becomes 2R2R, which gives g′=g4g' = \frac{g}{4}.

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