Problem 14 - A Complete Resource Book in Physics for JEE Main

Question

A simple pendulum has a time period \(T_{1}\) when on the earth's surface, and \(T_{2}\) when taken to a height \(R\) above the earth's surface, where \(R\) is radius of earth. The value of \(T_{2} / T_{1}\) is (A) 1 (B) \(\sqrt{2}\) (C) 4 (D) 2

Step-by-Step Solution

Answer

The value of \(\frac{T_2}{T_1}\) is 2. The correct answer is (D) 2.

Step 1: Formula for the time period of a simple pendulum

The formula for the time period of a simple pendulum is given by: \[T = 2\pi \sqrt{\frac{l}{g}}\] where \(T\) is the time period, \(l\) is the length of the pendulum, and \(g\) is the gravitational acceleration.

Step 2: Find gravitational acceleration at Earth's surface and height R

On the Earth's surface, the gravitational acceleration is \(g_1 = g\), where g is approximately 9.81 m/s². At a height R above the Earth's surface, the gravitational acceleration \(g_2\) can be found using the formula: \[g_2 = \frac{g}{(1 + \frac{R}{R_E})^2}\] where \(R_E\) is the radius of the Earth.

Step 3: Calculate the time periods for the two situations

Using the formula for the time period of a simple pendulum, we have: On Earth's surface: \[T_1 = 2\pi \sqrt{\frac{l}{g}}\] At height R above the Earth's surface: \[T_2 = 2\pi \sqrt{\frac{l}{g_2}}\]

Step 4: Find the ratio \(T_2 / T_1\)

Now we need to find the ratio of the time periods: \[\frac{T_2}{T_1} = \frac{2\pi \sqrt{\frac{l}{g_2}}}{2\pi \sqrt{\frac{l}{g}}}\] Simplifying, we get: \[\frac{T_2}{T_1} = \sqrt{\frac{g}{g_2}}\] Now we substitute the formula for \(g_2\) from Step 2: \[\frac{T_2}{T_1} = \sqrt{\frac{g}{\frac{g}{(1 + \frac{R}{R_E})^2}}}\] Further simplification gives: \[\frac{T_2}{T_1} = \sqrt{(1 + \frac{R}{R_E})^2}\] Finally, we find the ratio: \[\frac{T_2}{T_1} = 1 + \frac{R}{R_E}\] Since the height R is equal to the Earth's radius, we have: \[\frac{T_2}{T_1} = 1 + \frac{R_E}{R_E}\] \[\frac{T_2}{T_1} = 1 + 1\] \[\frac{T_2}{T_1} = 2\] The correct answer is (D) 2.