Verified Solution • 2025–26 Session moderate

QJEEM23Ph71b6: JEE Main 2023 Physics - Class 11 JEEM23Ph71b6 Solution: T is the time period of simple pendulum on the earth's surface. …

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

Multiple Choice Options

  • A 4
  • B 12\frac{1}{2}
  • C 2
  • D 14\frac{1}{4}

Step-by-Step Verified Solution

At surface of earth time period

T=2πℓg\mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{g}}}

At height h=R\mathrm{h}=\mathrm{R}

amp;g′=g(1+hR)2=g4amp;∴ xT=2πℓ(g/4)amp;⇒xT=2×2πℓgamp;⇒xT=2 T⇒x=2\begin{aligned}& \mathrm{g}^{\prime}=\frac{\mathrm{g}}{\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^2}=\frac{\mathrm{g}}{4} \\\\& \therefore \,\mathrm{xT}=2 \pi \sqrt{\frac{\ell}{(\mathrm{g} / 4)}} \\\\& \Rightarrow \mathrm{xT}=2 \times 2 \pi \sqrt{\frac{\ell}{\mathrm{g}}} \\\\& \Rightarrow \mathrm{xT}=2 \mathrm{~T} \Rightarrow \mathrm{x}=2\end{aligned}

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