Alokananda's Sand Tower

Classical Mechanics Level 5

Directly below her $10\text{ m}$ tall tower, Alokananda has piled up a heap of sand $1\text{ m}$ high.

When she drops a ball of density $\rho$ from the top of the tower, the ball does reach the bottom of the sand (whose density, by the way, is $\sigma$ ) pit, but by then, it is completely exhausted of its kinetic energy.

Modelling the heap of sand as a pool of fluid and assuming that the Archimedes' principle is the only reason for the resistance provided by the sand, what is the value of $\dfrac{\sigma}{\rho}$ ?

Assume that the ball has a positive radius negligible compared to the height of fall. Further, also assume that the ball immerses completely and no energy is lost in the air-sand transition.

Inspired by Alokananda Sarkar .

The answer is 10.

3 solutions

A Former Brilliant Member
Jul 17, 2016

We can directly apply the work energy theorem in this case.

Let $m$ be the mass of the ball.

Let's figure out the work done on this ball during it's trip down.

$\text{Work done by gravity }= mg\times 10\mathrm{m} \\ \text{Work done by buoyant force } = -\frac{m}{\rho} \sigma g \times 1 \mathrm{m}$

As change in kinetic energy is zero the total work done must also be zero.

So, we get,

$mg\times 10\mathrm{m} - \frac{m}{\rho} \sigma g \times 1 \mathrm{m} = 0 \\ \implies \frac{\sigma}{\rho} = \boxed{10}$

Can't we do like this when the ball just reaches above the sand,its velocity is √(2g×10). On penetrating through the sand,net acceleration isp{(p- sigma)/p}g So, 2×net acceleration ×1= 2g×10( as velocity when it reaches bottom of sand is zero). By this i am getting ans 11. Can someone tell me what is the mistake in this?

Sargam yadav - 4 years, 11 months ago

When it just above the pile, it's velocity is $\sqrt{2g \times 9\mathrm{m}}$ .

Otherwise, it's right.

A Former Brilliant Member - 4 years, 11 months ago

Thanks deeparaj bhat..☺

Sargam yadav - 4 years, 11 months ago

Too easy for level 5.

Subhra Patra - 4 years, 4 months ago

Lol why is this even a level 5 problem when it's a basic application of fluid mechanics and work?

Abhijeet Vats - 4 years, 11 months ago

Devansh Shringi
Jul 17, 2016

There is a problem in question Since change in kinetic energy is net work done by all forces . So net work done is equal to 0 . Therefore (density of ball)* g* 10+(density of sand) g (-1)=0 Therefore (density of sand /density of ball )=10 The guy who formed the question forgot that gravity will be doing work even when the ball is falling inside sand and took 9 instead of 10 so i ask the authorities yo correct the answer .

Ok thanks i will do it now U guys r really active good job

devansh shringi - 4 years, 11 months ago

Thanks. I've updated the answer from 9 to 10. Those who previously answered 10 will be marked correct.

In future, if you spot any errors with a problem, you can “report” it by selecting "report problem" in the “line line line” menu in the top right corner. This will notify the problem creator who can fix the issues.

Brilliant Mathematics Staff - 4 years, 11 months ago

Abhijeet Vats
Jul 17, 2016

Is this a flawed question?

Does something seem wrong?

Agnishom Chattopadhyay - 4 years, 11 months ago