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 the radius of the earth). The value of $ T_2/T_1 $ is
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 the radius of the earth). The value of $ T_2/T_1 $ is
Updated On: Jun 14, 2022
- 1
- $\sqrt 2$
- 4
- 2
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The Correct Option is D
Solution and Explanation
$ T \propto \frac{1}{\sqrt g } i.e \frac{T_2}{T_1} =\sqrt{\frac{g_1}{g_2}}$
where, $ g_1 $ = acceleration due to gravity on earth's surface = g
$g_2 $ = acceleration due to gravity at a height h = R
from earth's surface =g/4
$ \, \, \, \, \bigg[ Using\: g(h) = \frac{g}{\bigg( 1+\frac{h}{R}\bigg)^2} \bigg] $
$\Rightarrow \, \, \, \, \, \, \, \, \, \frac{T_2}{T_1} =\sqrt{\frac{g}{g/4}} =2$
where, $ g_1 $ = acceleration due to gravity on earth's surface = g
$g_2 $ = acceleration due to gravity at a height h = R
from earth's surface =g/4
$ \, \, \, \, \bigg[ Using\: g(h) = \frac{g}{\bigg( 1+\frac{h}{R}\bigg)^2} \bigg] $
$\Rightarrow \, \, \, \, \, \, \, \, \, \frac{T_2}{T_1} =\sqrt{\frac{g}{g/4}} =2$
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Concepts Used:
Gravitation
In mechanics, the universal force of attraction acting between all matter is known as Gravity, also called gravitation, . It is the weakest known force in nature.
Newton’s Law of Gravitation
According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,
- F ∝ (M1M2) . . . . (1)
- (F ∝ 1/r2) . . . . (2)
On combining equations (1) and (2) we get,
F ∝ M1M2/r2
F = G × [M1M2]/r2 . . . . (7)
Or, f(r) = GM1M2/r2
The dimension formula of G is [M-1L3T-2].