Question

A tunnel is made across the earth passing through its centre. A ball is dropped from a height h in the tunnel. The motion will be periodic with time period:

• A. $2\pi \sqrt{\frac{R}{g}}+\sqrt{\frac{2h}{g}}$
• B. $2\pi \sqrt{\frac{R}{g}}+4\sqrt{\frac{h}{g}}$
• C. $2\pi \sqrt{\frac{R}{g}}+4\sqrt{\frac{2h}{g}}$
• D. $2\pi \sqrt{\frac{R}{g}}+\sqrt{\frac{h}{g}}$

Answer 1

The correct option is C $2\pi \sqrt{\frac{R}{g}}+4\sqrt{\frac{2h}{g}}$

The ball is dropped from a point A at a height $h$ above the surface. Thus during the entire motion, let the time taken by the ball to move move from $AtoB$ be $t_1$, from $BtoC$ and $CtoB$ be $t_2$, from $CtoD$ be $t_3$, from $DtoC$ be $t_4$, and from $BtoA$ be $t_5$,

Let $T$ be the time period of the complete motion.

Now for the motion $AtoB$; $h=0+\frac{1}{2}g(t_1{)}^2$

$⟹t_1=\sqrt{\frac{2h}{g}}$

Similarly , $t_3=t_4=t_5=t_1=\sqrt{\frac{2h}{g}}$

As the time period of the particle executing SHM inside the tunnel is given by $t=2\pi \sqrt{\frac{R_e}{g}}$

Thus the time period of the motion of the ball inside the tunnel i.e $BtoC$ and $CtoB$, $t_2=2\pi \sqrt{\frac{R}{g}}$

Now the total time period of the motion $T=t_1+t_2+t_3+t_4+t_5$

$⟹T=2\pi \sqrt{\frac{R}{g}}+4\sqrt{\frac{2h}{g}}$