Search any question & find its solution
Question:
Answered & Verified by Expert
A pendulum is oscillating with frequency 'n' on the surface of earth. If it is taken to
depth $\frac{\mathrm{R}}{2}$ below the surface of earth where $\mathrm{R}$ is radius of earth. New frequency of
oscillations at depth $\frac{\mathrm{R}}{2}$ is
Options:
depth $\frac{\mathrm{R}}{2}$ below the surface of earth where $\mathrm{R}$ is radius of earth. New frequency of
oscillations at depth $\frac{\mathrm{R}}{2}$ is
Solution:
2658 Upvotes
Verified Answer
The correct answer is:
$\frac{n}{\sqrt{2}}$
$\begin{aligned} T=2 \pi \sqrt{\frac{l}{g}} \quad T^{\prime} &=2 \pi \sqrt{\frac{1}{g^{\prime}}} \\ &=2 \pi \sqrt{\frac{1}{g / 2}} \\ T^{\prime} &=\frac{\sqrt{2}}{T} \\ \therefore n^{\prime} &=\frac{n}{\sqrt{2}} \end{aligned}$
$n=1 / T$
$\begin{aligned} g^{\prime} &=g(1-d / R) \\ &=g\left(1-\frac{R / 2}{R}\right) \\ &=g / 2 \end{aligned}$
$n=1 / T$
$\begin{aligned} g^{\prime} &=g(1-d / R) \\ &=g\left(1-\frac{R / 2}{R}\right) \\ &=g / 2 \end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.