Gravitation

JEE Main level, single correct, medium

52 of 223

Question 52 of 223

A simple pendulum doing small oscillations at a place R\mathrm{R} height above earth surface has time period of T1=4 sT_1=4 \mathrm{~s}. T2T_2 would be it's time period if it is brought to a point which is at a height 2R2 \mathrm{R} from earth surface. Choose the correct relation [R=[R= radius of earth ]] :

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Answer: (D) 3 T1=2 T23 \mathrm{~T}_1=2 \mathrm{~T}_2

T1=2πℓGM(2R)2 T2=2πℓGM(3R)2∴T1 T2=23\begin{aligned} & \mathrm{T}_1=2 \pi \sqrt{\frac{\ell}{\mathrm{GM}}(2 \mathrm{R})^2} \\ & \mathrm{~T}_2=2 \pi \sqrt{\frac{\ell}{\mathrm{GM}}(3 \mathrm{R})^2} \\ & \therefore \frac{\mathrm{T}_1}{\mathrm{~T}_2}=\frac{2}{3}\end{aligned}

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