Consider a planet of radius $r$ having density $\lambda$.A tunnel is dig inside it it at a distance $\frac{r}{2}$ from its centre as shown. An object of mass m is left in the tunnel at the surface at $t=0$

Find the time taken by the object to reach the mid of the tunnel and the normal reaction applied by wall of the tunnel on the object.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

$</code> ... <code>$</code>...<code>."> Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in $</span> ... <span>$ or $</span> ... <span>$ to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestHint, it doesn't matter how far the tunnel is from the center of the Earth.

Log in to reply

@Ronak Agarwal @Azhaghu Roopesh M @Rohit Shah @Krishna Sharma

Log in to reply

How much are you getting in AIITS-8.

@Tanish

Log in to reply

The answer key webpage is showing error so i couldn't match my answers

Log in to reply

I totally messed up with the paper.

Log in to reply

@Abhineet Nayyar @Kushal Patankar

Log in to reply

@Tanishq Varshney

Use dimensional analysis(during exam).

The actual way is to use restoring force and stuff, with it you will get that the motion of the object in the tunnel is S.H.M.

Log in to reply

I want that stuff @Raghav Vaidyanathan

Log in to reply

Hi Raghav

I need your view on this problem

Log in to reply

First of all, thanks for mentioning me @Tanishq Varshney !! Is the answer for time period = $\frac { \pi }{ 2 } \sqrt { \frac { 3 }{ 4\pi \lambda G } }$ ???

Log in to reply

I have same guess.

Log in to reply

Can u provide a hint or some expression to get the answer

Log in to reply

$F=A \omega^2$ here A= √3r/2 and you know what F is. You will get \omega there. And the rest is piece of cake.

Since the motion inside is SHM , we know that force at amplitude isLog in to reply

Tunnel In a Planet

Let's consider that the object starts from the centre and moves towards the surface. Consider a very small displacement $'x'$ from the centre such that the restoring force remains constant. Now, the motion of the body through this $'x'$ is an SHM which i will prove later. According to shell theorem, the Earth around this motion won't apply any force. So, the force is applied by a small sphere with radius $R=\sqrt { \frac { { r }^{ 2 } }{ 2 } +{ x }^{ 2 } }$ which is equal to $\frac { Gm\frac { 4 }{ 3 } \pi \lambda { R }^{ 3 } }{ { R }^{ 2 } }$. But the restoring force will be the vertical component of this force = $\frac { G(\frac { 4 }{ 3 } \pi { R }^{ 3 }\lambda )m }{ { R }^{ 2 } } \frac { x }{ R }$ = $G\frac { 4 }{ 3 } \pi \lambda m(x)$

Clearly this force is directly proportional to $x$ and so, the motion is a simple harmonic one. So, the spring factor in the case is =$G\frac { 4 }{ 3 } \pi \lambda m$ and the time period is given by = $2\pi \sqrt { \frac { Mass\quad factor }{ Spring\quad factor } }$ = $2\pi \sqrt { \frac { m }{ \frac { 4 }{ 3 } \pi \lambda Gm } } =2\pi \sqrt { \frac { 3 }{ 4\pi \lambda G } }$

But since, you have asked about the time to move from extreme to mean, it is equal to $Time period/4$ = $\frac { \pi }{ 2 } \sqrt { \frac { 3 }{ 4\pi \lambda G } }$

And the normal reaction (I guess :P) will be given by the other component of the attractive force = $\frac { Gm\frac { 4 }{ 3 } \pi \lambda { R }^{ 3 } }{ { R }^{ 2 } } (\frac { r/2 }{ R } )\quad =\quad \frac { 2 }{ 3 } Gm\pi \lambda r$

So, how did i do?? Lol:D

Log in to reply

Log in to reply

Log in to reply

Can I have your views over here

Log in to reply