Problem 26 - A Complete Resource Book in Physics for JEE Main

Question

If the length of a simple pendulum is equal to the radius \(R\) of the earth, its time period will be (A) \(2 \pi \sqrt{R / g}\) (B) \(2 \pi \sqrt{R / 2 g}\) (C) \(2 \pi \sqrt{2 R / g}\) (D) \(\pi \sqrt{R / 2 g}\)

Step-by-Step Solution

Answer

The correct answer for the time period of a simple pendulum with length equal to the radius of the Earth is option (C) \(2 \pi \sqrt{2 R / g}\).

Step 1: Recall the time period formula for a simple pendulum

The time period \(T\) of a simple pendulum is given by the formula: \[ T = 2 \pi \sqrt{\frac{L}{g}} \] where \(L\) is the length of the pendulum and \(g\) is the acceleration due to gravity. In this exercise, the length of the pendulum is equal to the radius of Earth \(R\).

Step 2: Derive the altitude-dependent gravity

At the surface of the Earth, the acceleration due to gravity is \(g = 9.81 m/s^2\). However, gravity weakens as you go away from the Earth's surface. The altitude-dependent gravity can be expressed as: \[ g_a = \frac{GM}{(R+a)^2} \] where \(g_a\) is the acceleration due to gravity at an altitude \(a\), \(G\) is the gravitational constant, \(M\) is the mass of the Earth, and \(a\) is the altitude.

Step 3: Calculate the gravity at an altitude equal to the Earth's radius

In this case, we need to find the gravity at an altitude \(a = R\). So, the gravity at this altitude is: \[ g_a = \frac{GM}{(R+R)^2} = \frac{GM}{(2R)^2} \]

Step 4: Substitute the values in the time period formula

Now, we can substitute the length \(L=R\) and the gravity \(g=g_a\) in the time period formula: \[ T = 2\pi\sqrt{\frac{R}{\frac{GM}{(2R)^2}}} \]

Step 5: Simplify the formula to find the correct option

After substituting the values and simplifying the equation, we get the formula for the time period of the pendulum: \[ T = 2 \pi \sqrt{\frac{2R^3}{GM}} \] This formula corresponds to the option (C) because it has the same form, but we have to divide both numerator and denominator by \(R\): \[ T = 2 \pi \sqrt{\frac{2 R}{g}} \] Therefore, the correct answer is option (C) \(2 \pi \sqrt{2 R / g}\).