Question 276: Gravitation

Question

Time period of a simple pendulum on earth's surface is ' T '. It time period becomes ' xT ' when taken to a height ' $2 R$ ' above earth's surface. The value of $x$ will be $(R=$ radius of earth)

Accepted Answer

Step 1: Formula for simple pendulum
On EarthÃ¢ÂÂs surface:
$ T = 2\pi \sqrt{\frac{l}{g}} $
At height $h$:
$ g' = g \left(\frac{R}{R+h}ight)^2 $
So:
$ T' = 2\pi \sqrt{\frac{l}{g'}} = 2\pi \sqrt{ \frac{l}{g \left(	frac{R}{R+h}ight)^2 } } = 2\pi \sqrt{ \frac{l}{g} \cdot \left(\frac{R+h}{R}ight)^2 } $
$ T' = T \cdot \left(\frac{R+h}{R}ight) $
Step 2: Substitute $h = 2R$
$ T' = T \cdot \left(\frac{R+2R}{R}ight) = T \cdot \frac{3R}{R} = 3T $
So
$ x = 3 $
Ã¢ÂÂ Final Answer:
Option D: 3

Answer

2

Answer

4

Answer

1

Answer

3

Gravitation

Time period of a simple pendulum on earth's surface is ' T '. It time period becomes ' xT ' when taken to a height ' 2 R ' above earth's surface. The value of x will be (R= radius of earth)

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