Question

Two tunnels are dug from one side of the earth's surface to the other side, one along a diameter and the other along a chord, now two partcles are dropped from one end to reach the other end of tunnels. Both the particles oscillate simple harmonically along the tunnels. Let $T_1$ and $T_2$ be the time periods and $V_1$ and $V_2$ be the maximum speeds of the particles in these two tunnels. Then

• A. $T_1=T_2$
• B. $T_1>T_2$
• C. $V_1=V_2$
• D. $V_1>V_2$

Answer 1

Answer: (A), (D)

$V_1>V_2$

$\text{Time Period}:$
The time period of oscillations in tunnel dug at any position inside the earth is given by$$T=2\pi \sqrt{\frac{R}{g}}$$Where,
$R$ is the radius of the earth and $g$ is the acceleration due to gravity.

Thus, $T$ will remain the same everywhere. So, $T_1=T_2$

$\text{Maximum velocity}:$

Maximum velocity of the particle for the tunnel along the diameter is given by, $$V_1=\sqrt{\frac{GM}{R}}$$Maximum velocity of the particle for the tunnel along chord at distance $r$ from the centre is given by, $$V_2=\sqrt{\frac{GM}{R}\left(\frac{R^2-r^2}{R^2}\right)}$$So, $V_1>V_2$,

Hence, options (a) and (d) are the correct answers.