Question:
The mass and the diameter of a planet are three times the respective values for the Earth. The period of oscillation of a simple pendulum on the Earth is $2 \mathrm{~s}$. The period of oscillation of the same pendulum on the planet would be:
$\frac{\sqrt{3}}{2} s$
$\frac{2}{\sqrt{3}} \mathrm{~s}$
$\frac{3}{2} \mathrm{~s}$
$2 \sqrt{3} \mathrm{~s}$
JEE Main Previous Year Single Correct Question of JEE Main from Physics Oscillations chapter.
JEE Main Previous Year 11 Jan 2019, II
Correct Option: 4
Solution:
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Free
90 Qs. 360 Marks 180 Mins
Concept:
A simple pendulum is one which can be considered to be a point mass suspended from a string or rod of negligible mass. It is a resonant system with a single resonant frequency. For small amplitudes, the period of such a pendulum can be approximated by
Calculation:
Period of motion of a pendulum is given by
----(1)
On the surface of earth, let period of motion is Te and acceleration due to gravity is ge
----(2)
On another planet, let period of motion is TP and gravitational acceleration is gp
----(3)
(∴ Pendulum is same, so l will be same)
From Equations (2) and (3),
----(4)
Now,
and
Given, Mp = 3Me
and Rp = 3Re
----(5)
From Equations. (4) and (5)
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