Greetings Mechanics Lovers!

Since many people are starting their preparations for entering into most prestigious institute of India,(IITs),lets start a Mechanics Marathon.This Marathon will be topic-oriented,i.e. we will try to post good questions of certain topics for first three weeks.This note will help us to improve our problem solving strategy.

Rules are simple as other contest/marathons.

•I'll post two problems.If anybody solves any one of the two problems,s/he should post next problem within 3 hours.

•There will be max 2 unsolved problems at a time.

•Make sure you know how to solve your own problem before posting it, in case no one else is able to solve it within 24 hours. Then, you must post the solution and you have the right to post a new problem.

•It is NOT compulsory to post original problems. But make sure it has not been posted on brilliant.Also please don't post HC Verma questions or DC Pandey questions.

•There should be only one question. No sub questions allowed.No parts. Only one answer should be there.

•The scope of the problems is Jee-Advance and NSEP.(mechanics only).

**Topics for first three weeks:**

• Vectors

• Kinematics

• Laws of motion

• Work Energy and Power

• Circular Motion

• Center of mass.

Post only the problem and it's solution here. Extremely relevant comments are allowed.You are **allowed** to discuss solutions and problems here.

Format your proof as follows:

```
Solution to Problem (Insert Problem no here)
[Post your solution here]
```

And put your Problem in a new thread following the format :

```
Problem (Insert Problem no here)
[Post your problem here]
```

Remember to reshare this note so it goes to everyone out there. And above all else, **have fun**!

## Comments

Sort by:

TopNewestProblem 5

A chain of length L is placed on a smooth spherical surface of radius R with one of its ends fixed at the top of the sphere . What will be the acceleration of each element of the chain when its upper end is released

It is assumed that the length of chain is less than quarter of the sphere – Prakhar Bindal · 8 months, 1 week ago

Log in to reply

Take a small element dx making angle \(d\theta\) with centre ( at an angle \(\theta\) with horizontal.)

Check for the tangential forces acting on it you will find that only one force act on it tangentially and that is

\( dm g cos\theta \) \( ( dm = \frac{MRd\theta}{L}\) ( where M is assumed as mass of chain which will cancel at end))

[

Edited : Sorry guys , I left one thing { Credit to Deeparaj Bhat }

There is one more tangential force , that is dT , but since tension is an internal force the integration of dT vanish .

]

This is the force on a single element , integrate it to get the total force and then divide it by mass to get acceleration , which you will get as

\( a = \frac{gR}{L} ( 1 - cos(\frac{L}{R})) \)

Calculations I left for you ! – Aniket Sanghi · 8 months, 1 week ago

Log in to reply

@Harsh Shrivastava if you can remove the barrier of topics on posting questions , I will like post my next q that will be original! And will include a combination of topics. – Aniket Sanghi · 8 months, 1 week ago

Log in to reply

Log in to reply

– Aniket Sanghi · 8 months, 1 week ago

Actually my q include rotation , conservation of momentum and all basic chapters , but never mind I will post sth. Related to the topic itselfLog in to reply

Log in to reply

Log in to reply

Log in to reply

– Aniket Sanghi · 8 months, 1 week ago

I will like to see your solution too!Log in to reply

Now, note that tension is an internal force and hence, \(\int dT\) over the chain length vanishes.

We end up with the same answer finally. ;)

I know this is a very small point, but just thought that it's worth mentioning. – Deeparaj Bhat · 8 months, 1 week ago

Log in to reply

And should I delete my solution and you will post yours? – Aniket Sanghi · 8 months, 1 week ago

Log in to reply

Log in to reply

– Aniket Sanghi · 8 months, 1 week ago

Let me update my solution , and you are free to post the next question as I was wrong in one part :(Log in to reply

– Aniket Sanghi · 8 months, 1 week ago

Is the answer a = (gR/L)(1 - cos (L/R))Log in to reply

– Prakhar Bindal · 8 months, 1 week ago

Correct! . Post next question and solutionLog in to reply

Log in to reply

– Prakhar Bindal · 8 months, 1 week ago

Its Not right to say acceleration of the chain As a whole as every element will be having an acceleration of different direction but their magnitude will be the same that i am asking to calculateLog in to reply

Log in to reply

– Prakhar Bindal · 8 months, 1 week ago

Nope of each small element of chain ! . please do some analysis you will realiseLog in to reply

Problem 2:5 equal masses hang as shown in the figure.Find acceleration of each body.Assume pulleys to be frictionless,light and string is inextensible. – Harsh Shrivastava · 8 months, 1 week ago

Log in to reply

By Symmetry Of forces acceleration of left most and right most will be same and acceleration of middle 3 will be same.

Apply Virtual work method

Call acceleration of extremum's to be a1 and middle ones to be a2

You get

2Ta1 = 6Ta2

a1 = 3a2

Apply Newton's second for two blocks

mg-T =ma1

2T-mg = ma2

Rest is algebra which is left as an exercise for reader

On Solving a1 = 3g/7 and a2 = g/7 – Prakhar Bindal · 8 months, 1 week ago

Log in to reply

@Prakhar Bindal Hello Prakhar. Could you please tell me where I could learn about this Virtual Work method, especially on how to apply it for JEE problems? I had never heard of it before and it seems quite interesting! \[\]Thanks in advance! – Ishan Dasgupta Samarendra · 7 months, 2 weeks ago

Log in to reply

@Ishan Dasgupta Samarendra . Long time no see . how are you?

Hey helloAbout your question actually i was taught this by my teacher in Laws Of Motion in FIITJEE .

I Am giving you some links for learning.

https://books.google.co.in/books?id=imp4CgAAQBAJ&pg=SA6-PA34&lpg=SA6-PA34&dq=virtual+work+method+JEE&source=bl&ots=ka5Dc74yLI&sig=JJ7L4LQfIFhgeWR0dOg_D95gJrw&hl=en&sa=X&ved=0ahUKEwiwjMHy-pnNAhWEk5QKHeykDjwQ6AEIGzAA#v=onepage&q=virtual%20work%20method%20JEE&f=false

https://www.youtube.com/watch?v=lsHXK8nlLBs&feature=youtu.be – Prakhar Bindal · 7 months, 2 weeks ago

Log in to reply

– Ishan Dasgupta Samarendra · 7 months, 2 weeks ago

Thank you very much for the links. I'll go through them right away! \[\]I'm fine, thanks! Just buried under a huge amount of work, and hence sadly I've got no spare time for Brilliant. What about you?\[\] However I do see most of the problems you post on Chemistry; I really like them. Aachchaa, which books do you study from for P, C and M? As in theory as well as practice problems.\[\]Lastly, which center are you at FIITJEE? I seem to recall seeing your name once somewhere...Log in to reply

@Prakhar Bindal @Ishaan Dasgupta Samarendra

How did you guyz managed your time in class 11?

How many hours did you studied?

Please guide me.Thanks! – Harsh Shrivastava · 7 months, 2 weeks ago

Log in to reply

– Ishan Dasgupta Samarendra · 7 months, 1 week ago

I too am not the right person to ask. However one of the major mistakes I made in class 11 was that I never planned by which dates to finish which chapters. One should always have a countdown so that one knows exactly how far behind/ahead one is. For example, say you practise questions for Physics from sources A,B and C. Then you must have days scheduled by when you'll finish A, B and C respectively. That way you'll know whether you have been working the way you had to, or not.Log in to reply

@Mayank Singh @Deeparaj Bhat @Abhineet Nayyar – Prakhar Bindal · 7 months, 2 weeks ago

Log in to reply

You too know the amount of work fiitjee people give :P . so i am left with very less amount of time to complete other books but still i have some

Maths---> I refer Amit Mohan Agarwal's Skills in mathematics Series by Arihant

Physics----> I Have HC Verma for theory and DC Pandey for problems . i too have irodov but i hardly pick it up

Chemistry----> Physical - NCERT For Theory and RC Mukherjee for problems

Organic- Solomon's for theory and Himanshu Pandey for problems

Inorganic---> :P . i have JD Lee but i haven't opened it from last 6 months!

I Am from FIITJEE East Delhi centre (Laxmi Nagar) . Where are you from ? . where did you saw my name? – Prakhar Bindal · 7 months, 2 weeks ago

Log in to reply

– Ishan Dasgupta Samarendra · 7 months, 1 week ago

Oh, I too have many of the books you use! Anyway thanks very much for telling me !\[\] I'm from the South Delhi Center. I must have seen your name in some AIITS result or somewhere else. i really can't remember.Log in to reply

– Prakhar Bindal · 7 months, 1 week ago

Might be. !Log in to reply

– Harsh Shrivastava · 7 months, 1 week ago

Fun fact: Himanshu Pandey sir is my chemistry teacher(for inorganic and organic).Log in to reply

– Prakhar Bindal · 7 months, 1 week ago

Cool! . he must be a master in organic chemistryLog in to reply

Problem 1:Two particles, 1 and 2, move with constant velocities v1 and v2. At the initial moment their radius vectors are equal to r1 and r2. How must these four vectors be interrelated for the particles to collide? – Harsh Shrivastava · 8 months, 1 week ago

Log in to reply

\({ r }_{ 3 }={ { r }_{ 1 } }+{ v }_{ 1 }t\\ { r }_{ 3 }={ r }_{ 2 }+{ v }_{ 2 }t\)

\(\therefore\). \({r_1}- {r_2} = ({v_2} -{v_1})t\)

\(\large t\)= \(\large\frac{({r_1} - {r_2})}{({v_2}-{v_1})}\)

From the system of equations,we get \(\large \frac{({r_1} - {r_2})}{|({r_1}-{r_2})|} = \frac{({v_2} -{v_1})}{(|{v_2}-{v_1}|)}\)

By intuition, we can tell that the particles \(1\) and \(2\) collide when velocity of the second is directed towards the first. In that case, the velocity vectors turn parallel to the position vectors, thus making their unit vectors equal.– Swapnil Das · 8 months, 1 week agoLog in to reply

Don't you think the velocities, in addition of being parallel, must be not be equal to each other in case \(r_{1} ≠ r_{2}\) ? – Aditya Sky · 8 months, 1 week ago

Log in to reply

Log in to reply

– Harsh Shrivastava · 8 months, 1 week ago

Nah.Anyone can participate.Log in to reply

– Kaustubh Miglani · 8 months, 1 week ago

Thanks For The Invite.Is the answer to this question v2 vector-v1vector/mod(v2 vector -v1 vector)=r2vector-r1vector/mod(r2vector-r1 vector)Log in to reply

Problem 10A meteorite of mass m collides with a satellite which was oribiting around a planet of mass \(M\) in a circular orbit of radius R . Due to collision, the meteorite sticks to satellite and satellite is seen to have entered an orbit in which its

minimum distancefrom the planet is \(\frac{R}{2}\). If mass of the satellite is is 10 times that of the meteorite , What was the velocity of the meteorite before collision ? (edit: Assume velocity of meteorite was initially perpendicular to that of the satellite) – Sumanth R Hegde · 2 weeks, 6 days agoLog in to reply

on solving v turns out to be sqrt (58GM/R) – Prakhar Bindal · 2 weeks, 5 days ago

Log in to reply

– Harsh Shrivastava · 2 weeks, 5 days ago

Post the next problem!Log in to reply

– Harsh Shrivastava · 2 weeks, 5 days ago

I might I have done a serious calculation mistake coz I also did the same.Log in to reply

– Sumanth R Hegde · 2 weeks, 5 days ago

Yes :P.......Plz post the next problemLog in to reply

– Harsh Shrivastava · 2 weeks, 5 days ago

Is the answer sqrt(288GM/R)?Log in to reply

– Sumanth R Hegde · 2 weeks, 5 days ago

noLog in to reply

Problem 8:A cylinder of mass \(m\) and radius \(R\) rest on two supports in the same height .One support is stationery while the other slides from the cylinder at velocity \(v\). Determine the force of normal pressure \(N\) by cylinder at stationery support when \(x=R\sqrt{2}\). – Aryan Goyat · 8 months ago

Log in to reply

As the cylinder remains in contact with the stationary support it does circular motion about that point. As the cylinder is in contact with the block so the velocity of block on the normal of point of contact (which passes through center of cylinder) should be equal to that of cylinder.

\(\Rightarrow v_c = \dfrac{v}{\sqrt{2}} \)

As the block does circular motion the forces perpendicular to the point of contact provide the required centripetal force.

\(mg \cos {45^{\circ}} - N = \dfrac{mv_c ^2}{R} \)

\(N = \dfrac{mg}{\sqrt{2}} - \dfrac{mv^2}{2R} \) – Neelesh Vij · 8 months ago

Log in to reply

– Aryan Goyat · 8 months ago

go and post the next qLog in to reply

– Neelesh Vij · 8 months ago

How to post image. ? I need image to illustrate the question.Log in to reply

– Aryan Goyat · 8 months ago

when will you post qLog in to reply

– Aryan Goyat · 8 months ago

you just get the link by by copying it from publish place where you post q.Log in to reply

Solution to Problem 8:The answer is \(n=\frac{mg}{\sqrt{2}}-\frac{mv^2}{2R}\). Due to time constraint I can't post the complete solution. But it can be found out by balancing forces. The angle made with horizontal can be found out by basic geometry. The velocity of the cylinder is \(\frac{v}{\sqrt{2}}\). This can be found out by equating components. – Aditya Kumar · 8 months ago

Log in to reply

– Aditya Kumar · 8 months ago

I can't find a good problem now, can anyone else post it?Log in to reply

Problem 4:The trajectory for motion of a particle in \(x-y\) plane can be expressed as given below,

\[e^x+e^y=e^x e^y, x,y \in \mathbb{R}\]

It is given that acceleration of this particle is zero and its initial velocity in \(x\) and \(y\) direction are \(1 \text{m/s}\) and \(2 \text{m/s}\) respectively. Also if its initial coordinates can be expressed in form integers, then find the initial coordinates \((x,y)\) of the particale.

If no such coordinates exist then enter your answer as 2016.

P. S. : Original problem. – Akshay Yadav · 8 months, 1 week ago

Log in to reply

\[1=e^{-x}+e^{-y} \\\implies 0=e^{-x}+2e^{-y} \\ (\text{differentiating wrt time } \frac{dx}{dt}=1\: \frac{dy}{dt}=2) \\\implies -1=e^{-y} \\\text{No real solution}\\\large Q. E. D. \] – Deeparaj Bhat · 8 months, 1 week ago

Log in to reply

– Aditya Sky · 8 months, 1 week ago

Can you please explain third line of you solution ?Log in to reply

– Deeparaj Bhat · 8 months, 1 week ago

I meant that the second line is obtained after differentiating the given equation and using the given facts.Log in to reply

– Akshay Yadav · 8 months, 1 week ago

Correct! post the next question!Log in to reply

Log in to reply

– Akshay Yadav · 8 months, 1 week ago

If that is what you wish, then I will post the next question. Did you liked this problem?Log in to reply

– Harsh Shrivastava · 8 months, 1 week ago

Post the next problem.Log in to reply

– Deeparaj Bhat · 8 months, 1 week ago

It was quite different from the usual ones. And I like such problems ;)Log in to reply

Log in to reply

– Akshay Yadav · 8 months, 1 week ago

Yeah! Write the solution to it too.Log in to reply

Let's start with an easy one!

Problem 3: The velocity of a particle moving in the positive direction of the \(x\) axis varies as \(v=\alpha\sqrt{x}\), where \(\alpha\) is a positive constant. Assuming that at the moment \(t=0\) the particle was located at the point \(x=0\), find the time dependence of the velocity and acceleration of the particle. – Swapnil Das · 8 months, 1 week agoLog in to reply

We know,

\(v=\frac{dx}{dt}=\alpha x^{1/2}\)

\(x^{-1/2}dx=\alpha dt\)

\(\displaystyle \int_{0}^{x}x^{-1/2}dx=\displaystyle \int_{0}^{t} \alpha dt\)

\(2x^{1/2}=\alpha t\)

\(x=\alpha^2 t^2/4\)

By differentiation,

\(v=\alpha^2 t/2\)

\(a=\alpha^2/2\) – Akshay Yadav · 8 months, 1 week ago

Log in to reply

– Swapnil Das · 8 months, 1 week ago

Correct.You perhaps can post the next question.Log in to reply

PROBLEM 11

Moment of inertia of a uniform hexagonal plate about an axis LL' is I . The moment of inertia (about XX') of an equilateral uniform triangular plate of thickness double that of hexagonal plate is P . (specific gravity of triangular plate is 3 times that of hexagonal plate)

Find I/P – Prakhar Bindal · 2 weeks, 4 days ago

Log in to reply

– Prakhar Bindal · 2 weeks, 4 days ago

https://s28.postimg.org/vhjxmwknx/Untitled.pngLog in to reply

– Prakhar Bindal · 2 weeks, 4 days ago

http://tinypic.com/view.php?pic=2ludvn5&s=9#.WGoB81N96M8Log in to reply

visit this page for image – Prakhar Bindal · 2 weeks, 4 days ago

Log in to reply

– Aryan Goyat · 2 weeks, 3 days ago

ans is 5Log in to reply

– Harsh Shrivastava · 2 weeks, 3 days ago

Post the next problem , a tough one!Log in to reply

– Prakhar Bindal · 2 weeks, 3 days ago

This was very similar to a problem which appeared in our AITS 1Log in to reply

– Prakhar Bindal · 2 weeks, 3 days ago

wasn't this one tough? :PLog in to reply

– Harsh Shrivastava · 2 weeks, 3 days ago

It was tough indeed, though I have only tried it once.Log in to reply

– Prakhar Bindal · 2 weeks, 3 days ago

Yup correct!Log in to reply

Is this dead? Shall I participate – Harry Jones · 3 weeks, 2 days ago

Log in to reply

– Harsh Shrivastava · 3 weeks, 2 days ago

If you want you may post problems here and I will participate here.So post the next problem !Log in to reply

– Harry Jones · 3 weeks, 2 days ago

Can u derive the equation of trajectory of each particle in the Irodov question no 1.12?Log in to reply

Is the contest dead? – Harsh Shrivastava · 7 months, 3 weeks ago

Log in to reply

– Swapnil Das · 7 months, 2 weeks ago

Oh, sad. But no worries! I'll be coming up with an Electromagnetism contest on Monday, be sure to take part!Log in to reply

– Prakhar Bindal · 7 months, 3 weeks ago

Yeah :P . No more activity as we all are quite busy with our class 12th!Log in to reply

Problem 9A metal rod is of length \(1m\) is clamped at points \(5cm \) and \(15 cm \) from either ends. Find the minimum frequency of natural longitudinal oscillations of rod.

Given :

Young's modulus of rod \( = 1.6 \times 10^{11} Pa\)

Mass per unit volume \( \rho = 2500 kgm^{-3} \)

Log in to reply

firstnode is at the hinge 5cm from it.(This also takes care of a node being at the other point that is hinged) . hence max wavelength is 20cm. Thus min frequency is \(40,000 Hz\) – Sumanth R Hegde · 2 weeks, 6 days agoLog in to reply

Post the next problem please! – Harsh Shrivastava · 2 weeks, 6 days ago

Log in to reply

– Neelesh Vij · 2 weeks, 6 days ago

correct!Log in to reply

Log in to reply

– Neelesh Vij · 3 weeks, 1 day ago

noLog in to reply

– Aryan Goyat · 8 months ago

picLog in to reply

– Neelesh Vij · 8 months ago

Rod is clamped at a distance 5cm from one end and 15 cm from other end. Its easily understandable i guess.Log in to reply

– Aryan Goyat · 8 months ago

yep it is but i thought you were not able to do as guided by me.Log in to reply

– Neelesh Vij · 8 months ago

yes i didn't understand the method u said to post the pic.Log in to reply

– Aryan Goyat · 8 months ago

be on slackLog in to reply

– Neelesh Vij · 8 months ago

I need to go to coaching right now. so talk to you laterLog in to reply

– Aryan Goyat · 8 months ago

ok bye today i have coaching too from 2-30 so better talk at 9-00Log in to reply

Problem 7:\[\] A lift of total mass M kg is raised by cables from rest to rest through a height h.The greatest tension which the cables can safely bear is n Mg Newtons(n>1).Find the shortest interval of time in which the ascent can be made,in terms of n,h,g. – Adarsh Kumar · 8 months ago

Log in to reply

– Aryan Goyat · 8 months ago

now this can have multiple ans depending on if cables are also attached at bottom for retardation or not i assume that the cable bring only acceleration and retardation is brought up by g only in that case i hope t(min)=root(2h/{n(n-1)g})(considering at a ht h v=0)Log in to reply

– Adarsh Kumar · 8 months ago

The latter case is correct!Well done!Post the next question please!Log in to reply

Hey guyz,please delete unnecessary comments as the note is getting slow.Thanks. – Harsh Shrivastava · 8 months ago

Log in to reply

Problem 6 : [Original]

Take a bomb of mass m and give it a velocity \( v = 10m/sec \) at an angle \( \theta = 53° \)with horizontal. The bomb bursts into two identical parts when it's velocity made an angle +45° with horizontal. Just after the breakdown, One of the part's velocity was found to be \( 6 m/sec \) in horizontal direction. (i. e. It retained it's horizontal component of velocity ) . Find the final displacement of one part with respect to other after they come to rest.

Details And Assumptions :

The coefficient of restitution for the ball ground collision is \( e = 0.4 \)

The ground is smooth.

Take \( g = 10m/s^2 \)

Log in to reply

Log in to reply

@Aniket Sanghi i request to find a mistake in this. – Aryan Goyat · 8 months ago

this gives ans as 16.71 as suggested byLog in to reply

Log in to reply

@Adarsh Kumar i would request to find a mistake in this????? – Aryan Goyat · 8 months ago

this gives ans 18.1527 as mentioned byLog in to reply

@aryan goyat for posting this! – Adarsh Kumar · 8 months ago

Thanx!Log in to reply

– Aniket Sanghi · 8 months ago

Go! Post the next question for now @AdarshKumarLog in to reply

– Adarsh Kumar · 8 months ago

Ohk,gimme a minute I just woke up :P.Log in to reply

– Aryan Goyat · 8 months ago

welcomeLog in to reply

@Aniket Sanghi – Harsh Shrivastava · 8 months ago

Log in to reply

Log in to reply

@aryan goyat @Aniket Sanghi – Prakhar Bindal · 8 months ago

Log in to reply

– Aryan Goyat · 8 months ago

be on slackLog in to reply

– Aniket Sanghi · 8 months ago

Nope , my answer is differentLog in to reply

– Prakhar Bindal · 8 months ago

bro come on slackLog in to reply

Log in to reply

– Aryan Goyat · 8 months, 1 week ago

ya and only one correct ans 'bohot nainsafe hai'Log in to reply

Log in to reply

– Aryan Goyat · 8 months, 1 week ago

ok will be there nowLog in to reply

Log in to reply

– Aniket Sanghi · 8 months, 1 week ago

NopeLog in to reply

– Mayank Chaturvedi · 8 months, 1 week ago

Can you figure out mistake in process?- What i did was conserving momentum along x and y axis. Both masses had 6m/s velocity in x direction , one having 12m/s up. Using these, i solved for time( using GP sum) and then difference of distance.Log in to reply

– Adarsh Kumar · 8 months, 1 week ago

I am getting 18.15.Log in to reply

– Mayank Chaturvedi · 8 months, 1 week ago

did you solve in same way as i did?Log in to reply

– Adarsh Kumar · 8 months, 1 week ago

I got the same numbers as you, but after that I used difference of ranges by using infinite gp.Log in to reply

– Mayank Chaturvedi · 8 months, 1 week ago

Yes the same wayLog in to reply

– Adarsh Kumar · 8 months, 1 week ago

Ohk,so you first found out difference in ranges after 1st collision with the ground and then added it to difference in range before 1st collision?Log in to reply

– Mayank Chaturvedi · 8 months, 1 week ago

Indirectly the same. Instead i solved for the time before and after the first collision, then multiplying 6(horizontal speed), differenced them.Log in to reply

– Adarsh Kumar · 8 months, 1 week ago

Yup!That is almost the same way,I think :P !Log in to reply

@Prakhar Bindal we really need your help! – Adarsh Kumar · 8 months, 1 week ago

Log in to reply

Log in to reply

@Prakhar Bindal i request to find a mistake in it?? – Aryan Goyat · 8 months ago

this gives ans as 14.782 as suggested byLog in to reply

@Aniket Sanghi @Deeparaj Bhat please post the next question! – Adarsh Kumar · 8 months, 1 week ago

Log in to reply

I am inviting few people to join the contest, sorry for mass taging.

@Akshay Yadav @Aniket Sanghi @Adarsh Kumar @Chinmay Sangawadekar @Vaibhav Prasad @Kalash Verma @Hrishik Mukherjee. @Nihar Mahajan @Prakhar Bindal @Swapnil Das @Rajdeep Dhingra @Shreya R @milind prabhu @Kaustubh Miglani @Vighnesh shenoy – Harsh Shrivastava · 8 months, 1 week ago

Log in to reply

– Swapnil Das · 8 months, 1 week ago

Thanks for inviting :)Log in to reply

– Harsh Shrivastava · 8 months, 1 week ago

Reshare the note so that it may reach out to others.Thanks!Log in to reply