Gravitation Question 5
Question 5 - 25 January - Shift 1
Assume that the earth is a solid sphere of uniform density and a tunnel is dug along its diameter throughout the earth. It is found that when a particle is released in this tunnel, it executes a simple harmonic motion. The mass of the particle is $100 g$. The time period of the motion of the particle will be (approximately) (take $g=10 ms^{-2}$,radius of earth $=6400 km$ )
(1) 24 hours
(2) 1 hour 24 minutes
(3) 1 hour 40 minutes
(4) 12 hours
Show Answer
Answer: (2)
Solution:
Formula: Non conducting solid sphere
Let at some time particle is at a distance $x$ from centre of Earth, then at that position field
$E=\frac{GM}{R^{3}} x$
$\therefore$ Acceleration of particle
$\overrightarrow{{}a}=-\frac{GM}{R^{3}} \overrightarrow{{}x}$
$\Rightarrow \omega=\sqrt{\frac{GM}{R^{3}}}=\sqrt{\frac{g}{R}}$
Now T $=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{R}{g}}$
$\Rightarrow T=2 \times 3.14 \times \sqrt{\frac{6400 \times 10^{3}}{10}}$
$=2 \times 3.14 \times 800 sec \approx 1$ hour 24 minutes
