JEE WhatsApp Group
Sign in
Please select an account to continue using cracku.in
↓ →
JEE WhatsApp Group
Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
A simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is 4 m, then the time period of small oscillations will be _________ s. [take $$g = \pi^2$$ m s$$^{-2}$$]
Correct Answer: 8
At height $$h = R$$ from surface, distance from center = $$2R$$.
Effective $$g' = \frac{g}{(1 + h/R)^2} = \frac{g}{4}$$.
$$T = 2\pi\sqrt{\frac{l}{g'}} = 2\pi\sqrt{\frac{4}{g/4}} = 2\pi\sqrt{\frac{16}{g}} = 2\pi \times \frac{4}{\sqrt{g}}$$
With $$g = \pi^2$$: $$T = 2\pi \times \frac{4}{\pi} = 8$$ s.
Therefore, the answer is $$\boxed{8}$$.
Click on the Email ☝️ to Watch the Video Solution
Create a FREE account and get:
Educational materials for JEE preparation
Ask our AI anything
AI can make mistakes. Please verify important information.
AI can make mistakes. Please verify important information.
