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Meddling with Mechanics problems!

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!

Note by Harsh Shrivastava
4 months, 2 weeks ago

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Problem 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 · 4 months, 2 weeks ago

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@Prakhar Bindal Solution to problem 5 :

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 · 4 months, 2 weeks ago

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@Aniket Sanghi @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 · 4 months, 2 weeks ago

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@Harsh Shrivastava Actually my q include rotation , conservation of momentum and all basic chapters , but never mind I will post sth. Related to the topic itself Aniket Sanghi · 4 months, 2 weeks ago

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@Deeparaj Bhat I will like to see your solution too! Aniket Sanghi · 4 months, 2 weeks ago

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@Aniket Sanghi As in your solution, just add a \(dT\) (the change in tension) to the tangential forces. Integrate the expression with the appropiate limits.

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 · 4 months, 2 weeks ago

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@Deeparaj Bhat Thanks! I always like to learn :) , just can you also explain me how inegation of dT vanished

And should I delete my solution and you will post yours? Aniket Sanghi · 4 months, 2 weeks ago

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@Deeparaj Bhat Let me update my solution , and you are free to post the next question as I was wrong in one part :( Aniket Sanghi · 4 months, 2 weeks ago

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@Prakhar Bindal Is the answer a = (gR/L)(1 - cos (L/R)) Aniket Sanghi · 4 months, 2 weeks ago

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@Aniket Sanghi Correct! . Post next question and solution Prakhar Bindal · 4 months, 2 weeks ago

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@Adarsh Kumar 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 calculate Prakhar Bindal · 4 months, 2 weeks ago

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@Adarsh Kumar Nope of each small element of chain ! . please do some analysis you will realise Prakhar Bindal · 4 months, 2 weeks ago

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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 · 4 months, 2 weeks ago

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@Harsh Shrivastava Its an elementary problem on virtual work method

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 · 4 months, 2 weeks ago

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@Prakhar Bindal @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 · 3 months, 3 weeks ago

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

About 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 · 3 months, 3 weeks ago

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@Prakhar Bindal 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... Ishan Dasgupta Samarendra · 3 months, 3 weeks ago

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@Ishan Dasgupta Samarendra @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 · 3 months, 3 weeks ago

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@Harsh Shrivastava 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. Ishan Dasgupta Samarendra · 3 months, 3 weeks ago

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@Harsh Shrivastava I Think i am not the right person to be asked . you must contact the people who have given JEE This year

@Mayank Singh @Deeparaj Bhat @Abhineet Nayyar Prakhar Bindal · 3 months, 3 weeks ago

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@Ishan Dasgupta Samarendra Thanks! . I Am bad at chemistry :P.

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 · 3 months, 3 weeks ago

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@Prakhar Bindal 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. Ishan Dasgupta Samarendra · 3 months, 3 weeks ago

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@Ishan Dasgupta Samarendra Might be. ! Prakhar Bindal · 3 months, 3 weeks ago

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@Prakhar Bindal Fun fact: Himanshu Pandey sir is my chemistry teacher(for inorganic and organic). Harsh Shrivastava · 3 months, 3 weeks ago

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@Harsh Shrivastava Cool! . he must be a master in organic chemistry Prakhar Bindal · 3 months, 3 weeks ago

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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 · 4 months, 2 weeks ago

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@Harsh Shrivastava Let the particles collide at point \(P\), whose position vector is \({ r }_{ 3 }\). Let the time taken be \(t\). By the vector laws of addition, we get

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

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

\(\large t\)= \(\large\frac{({r_1} - {r_2})}{({v_2}-{v_1})}\)
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 · 4 months, 2 weeks ago

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@Swapnil Das Nice solution (+1) :).

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 · 4 months, 2 weeks ago

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@Deeparaj Bhat Nah.Anyone can participate. Harsh Shrivastava · 4 months, 2 weeks ago

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@Harsh Shrivastava Thanks For The Invite.Is the answer to this question v2 vector-v1vector/mod(v2 vector -v1 vector)=r2vector-r1vector/mod(r2vector-r1 vector) Kaustubh Miglani · 4 months, 2 weeks ago

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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 · 4 months, 2 weeks ago

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@Aryan Goyat Solution to problem 8:

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 · 4 months, 1 week ago

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@Neelesh Vij go and post the next q Aryan Goyat · 4 months, 1 week ago

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@Aryan Goyat How to post image. ? I need image to illustrate the question. Neelesh Vij · 4 months, 1 week ago

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@Neelesh Vij when will you post q Aryan Goyat · 4 months, 1 week ago

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@Neelesh Vij you just get the link by by copying it from publish place where you post q. Aryan Goyat · 4 months, 1 week ago

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@Aryan Goyat 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 · 4 months, 1 week ago

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@Aditya Kumar I can't find a good problem now, can anyone else post it? Aditya Kumar · 4 months, 1 week ago

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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 · 4 months, 2 weeks ago

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@Akshay Yadav Given equation:

\[1=e^{-x}+e^{-y} \\\implies 0=e^{-x}+2e^{-y} \\ (\text{differentiating wrt time } \frac{dx}{dt}=1\: \frac{dy}{dt}=2) \\\implies -1=e^{-y} \\\text{No real solution}\\\large Q. E. D. \] Deeparaj Bhat · 4 months, 2 weeks ago

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@Deeparaj Bhat Can you please explain third line of you solution ? Aditya Sky · 4 months, 2 weeks ago

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@Aditya Sky I meant that the second line is obtained after differentiating the given equation and using the given facts. Deeparaj Bhat · 4 months, 2 weeks ago

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@Deeparaj Bhat Correct! post the next question! Akshay Yadav · 4 months, 2 weeks ago

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@Deeparaj Bhat If that is what you wish, then I will post the next question. Did you liked this problem? Akshay Yadav · 4 months, 2 weeks ago

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@Akshay Yadav Post the next problem. Harsh Shrivastava · 4 months, 2 weeks ago

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@Akshay Yadav It was quite different from the usual ones. And I like such problems ;) Deeparaj Bhat · 4 months, 2 weeks ago

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@Harsh Shrivastava Yeah! Write the solution to it too. Akshay Yadav · 4 months, 2 weeks ago

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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 · 4 months, 2 weeks ago

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@Swapnil Das \(v=\alpha^2t/2\) and \(a=\alpha^2/2\).

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 · 4 months, 2 weeks ago

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@Akshay Yadav Correct.You perhaps can post the next question. Swapnil Das · 4 months, 2 weeks ago

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Is the contest dead? Harsh Shrivastava · 4 months ago

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@Harsh Shrivastava Oh, sad. But no worries! I'll be coming up with an Electromagnetism contest on Monday, be sure to take part! Swapnil Das · 3 months, 4 weeks ago

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@Harsh Shrivastava Yeah :P . No more activity as we all are quite busy with our class 12th! Prakhar Bindal · 4 months ago

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Problem 9

A 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 \( \rho = 2500 kgm^{-3} \)

Neelesh Vij · 4 months, 1 week ago

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@Neelesh Vij pic Aryan Goyat · 4 months, 1 week ago

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@Aryan Goyat Rod is clamped at a distance 5cm from one end and 15 cm from other end. Its easily understandable i guess. Neelesh Vij · 4 months, 1 week ago

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@Neelesh Vij yep it is but i thought you were not able to do as guided by me. Aryan Goyat · 4 months, 1 week ago

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@Aryan Goyat yes i didn't understand the method u said to post the pic. Neelesh Vij · 4 months, 1 week ago

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@Neelesh Vij be on slack Aryan Goyat · 4 months, 1 week ago

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@Aryan Goyat I need to go to coaching right now. so talk to you later Neelesh Vij · 4 months, 1 week ago

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@Neelesh Vij ok bye today i have coaching too from 2-30 so better talk at 9-00 Aryan Goyat · 4 months, 1 week ago

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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 · 4 months, 2 weeks ago

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@Adarsh Kumar 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) Aryan Goyat · 4 months, 2 weeks ago

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@Aryan Goyat The latter case is correct!Well done!Post the next question please! Adarsh Kumar · 4 months, 2 weeks ago

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Hey guyz,please delete unnecessary comments as the note is getting slow.Thanks. Harsh Shrivastava · 4 months, 2 weeks ago

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

Aniket Sanghi · 4 months, 2 weeks ago

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@Aniket Sanghi

Aryan Goyat · 4 months, 2 weeks ago

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@Aryan Goyat this gives ans as 16.71 as suggested by @Aniket Sanghi i request to find a mistake in this. Aryan Goyat · 4 months, 2 weeks ago

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@Aniket Sanghi

Aryan Goyat · 4 months, 2 weeks ago

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@Aryan Goyat this gives ans 18.1527 as mentioned by @Adarsh Kumar i would request to find a mistake in this????? Aryan Goyat · 4 months, 2 weeks ago

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@Aryan Goyat Thanx!@aryan goyat for posting this! Adarsh Kumar · 4 months, 2 weeks ago

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@Adarsh Kumar Go! Post the next question for now @AdarshKumar Aniket Sanghi · 4 months, 2 weeks ago

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@Aniket Sanghi Ohk,gimme a minute I just woke up :P. Adarsh Kumar · 4 months, 2 weeks ago

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@Adarsh Kumar welcome Aryan Goyat · 4 months, 2 weeks ago

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@Aniket Sanghi You may post the solution and the next problem.

@Aniket Sanghi Harsh Shrivastava · 4 months, 2 weeks ago

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@Aryan Goyat Got at same 14.78379 m .

@aryan goyat @Aniket Sanghi Prakhar Bindal · 4 months, 2 weeks ago

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@Prakhar Bindal be on slack Aryan Goyat · 4 months, 2 weeks ago

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@Aryan Goyat Nope , my answer is different Aniket Sanghi · 4 months, 2 weeks ago

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@Aniket Sanghi bro come on slack Prakhar Bindal · 4 months, 2 weeks ago

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@Adarsh Kumar ya and only one correct ans 'bohot nainsafe hai' Aryan Goyat · 4 months, 2 weeks ago

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@Adarsh Kumar ok will be there now Aryan Goyat · 4 months, 2 weeks ago

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@Mayank Chaturvedi Nope Aniket Sanghi · 4 months, 2 weeks ago

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@Aniket Sanghi 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. Mayank Chaturvedi · 4 months, 2 weeks ago

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@Mayank Chaturvedi I am getting 18.15. Adarsh Kumar · 4 months, 2 weeks ago

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@Adarsh Kumar did you solve in same way as i did? Mayank Chaturvedi · 4 months, 2 weeks ago

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@Mayank Chaturvedi I got the same numbers as you, but after that I used difference of ranges by using infinite gp. Adarsh Kumar · 4 months, 2 weeks ago

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@Adarsh Kumar Yes the same way Mayank Chaturvedi · 4 months, 2 weeks ago

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@Mayank Chaturvedi 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? Adarsh Kumar · 4 months, 2 weeks ago

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@Adarsh Kumar Indirectly the same. Instead i solved for the time before and after the first collision, then multiplying 6(horizontal speed), differenced them. Mayank Chaturvedi · 4 months, 2 weeks ago

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@Mayank Chaturvedi Yup!That is almost the same way,I think :P ! Adarsh Kumar · 4 months, 2 weeks ago

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@Adarsh Kumar @Prakhar Bindal we really need your help! Adarsh Kumar · 4 months, 2 weeks ago

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@Aniket Sanghi

Aryan Goyat · 4 months, 2 weeks ago

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@Aryan Goyat this gives ans as 14.782 as suggested by @Prakhar Bindal i request to find a mistake in it?? Aryan Goyat · 4 months, 2 weeks ago

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@Aniket Sanghi @Deeparaj Bhat please post the next question! Adarsh Kumar · 4 months, 2 weeks ago

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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 · 4 months, 2 weeks ago

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@Harsh Shrivastava Thanks for inviting :) Swapnil Das · 4 months, 2 weeks ago

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@Swapnil Das Reshare the note so that it may reach out to others.Thanks! Harsh Shrivastava · 4 months, 2 weeks ago

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