Q108 / 150 · MHT CET 2026 (16 April Shift 2) · Physics
The time taken by simple pendulum for one oscillation is T on earth's surface. Its time period becomes xT when taken to a height R (equal to earth's radius) above the earth's surface. The value of x is
- A$\dfrac{1}{4}$
- B$\dfrac{1}{2}$
- Correct answerC$2$
- D$4$
Solution
The time period of a simple pendulum on the surface of the earth is given by $T = 2\pi \sqrt{\dfrac{L}{g}}$.
The acceleration due to gravity at a height $h$ above the earth's surface is $g' = \dfrac{g}{\left(1 + \dfrac{h}{R}\right)^2}$.
Given $h = R$, we have $g' = \dfrac{g}{\left(1 + \dfrac{R}{R}\right)^2} = \dfrac{g}{4}$.
The new time period $T'$ at height $R$ is $T' = 2\pi \sqrt{\dfrac{L}{g'}} = 2\pi \sqrt{\dfrac{L}{\dfrac{g}{4}}} = 2 \times 2\pi \sqrt{\dfrac{L}{g}} = 2T$.
Since $T' = xT$, we get $x = 2$.
Answer: $2$
The acceleration due to gravity at a height $h$ above the earth's surface is $g' = \dfrac{g}{\left(1 + \dfrac{h}{R}\right)^2}$.
Given $h = R$, we have $g' = \dfrac{g}{\left(1 + \dfrac{R}{R}\right)^2} = \dfrac{g}{4}$.
The new time period $T'$ at height $R$ is $T' = 2\pi \sqrt{\dfrac{L}{g'}} = 2\pi \sqrt{\dfrac{L}{\dfrac{g}{4}}} = 2 \times 2\pi \sqrt{\dfrac{L}{g}} = 2T$.
Since $T' = xT$, we get $x = 2$.
Answer: $2$
Hence, the correct answer is option: 'C'.