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BARC Topper – Ashish Ranjan Interview – Mechanical Engineering

Here we are sharing the Interview of Ashish Ranjan who qualified BARC interview and selected for BARC OCES, Ashish Ranjan interview was held on 09 June 2017. Ashish Ranjan is from Mechanical Engineering and is currently a BARC employee.

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BARC INTERVIEW

Name: Ashish Ranjan

Date of interview : 9/6/2017

Time: 5:40 pm to 7:00 pm

Branch: Mechanical Engineering

Result: selected as OCES Trainee

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Now the interview experience goes like this:

My interview was last interview of the day by panel given to me (there were 6 panel for Mechanical engineering and I got 3rd panel)

Me: may I come in?

Member 1: yes please sit down

They were in hurry, I didn’t even get chance to wish them good evening (honestly they don’t care either)

M1: (member of interview panel =M from now on) so are you staying at hostel of BARC?

Me:yes sir .

M1:did you properly have lunch?

Me:yes sir.

I wasn’t looking nervous or worried so they didn’t waste much time in introduction and all (they don’t care about that either, if you are terrified then they might spend 10min to calm you down)

They gave me a pencil and an A-4 size page.

M1:ok, Ashish now we want you to write 4-5 subjects you are most comfortable in.

Me: (writing and speaking at the same time as each of the five members were noting down these subjects in a piece of paper of there own) 1.Strength of material

M1:you can use short forms

Me: 2)FM

3)HT

4)MD

5. Basic Thermodynamics

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M1 gave M2 signal to start.

M2:ok, there’s a bar with load of 3P acting on one of it’s side, what will be the stress?

Me:there will be no stress if we neglect all frictions. Body will accelerate with a=3P/m

M2: we add a force of 5P on other side. Now what will there be stress?

Me: (after thinking for few seconds)body will have a stress of 3P/A and acceleration of 2P/m

M2:will the stress be uniform longitudinally or will it vary?

Me: Sir this is a case of dynamic analysis, in btech we have studied till static stress analysis so if I have to go based on that then it will be uniform longitudinally but it might be possible that it varies from maximum at 5P side and minimum at 3P side.

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M1:but will there be stress at all?

Me:yes sir stress will be present.

M1:ok Ashish, if I drop a ball from certain height will there be stress when it freely falls?

Me: are we considering air drag?

M1:No.

Me: Then there will be no stress.

M1:then how is this bar stressed?

Me:sir, if we consider that same ball falling freely and tie it with an elastic band on top and then let it fall then it is both falling and resisting against the band, in this case there will be stress in the ball since it is resisting aginst something.

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M1:ok(indicating M3 to ask his question)

M3: (draws a tank with one hole at bottom face, one at outlet of vertical pipe attached at bottom face and last also at bottom but on vertical face) what are velocities at each of these outlets?

Me:(using bernoulli’s equation derived velocities)

M1:now buckets under each stream at same level, which one will fill first?

Me:first one.

M1:why?

Me:highest velocity

M2:which one will fill second?

Me:other 2 will fill simultaneously…. Ok wait… Stream from third hole is having a projectile type stream

I started using equations of projectile motion.. Filled almost a page…

M1:Is this necessary?

M3:what are you trying to do?

Me:sir, I am trying to find the velocity of stream just before it gets into bucket as that would give us volume flow rate and as we know volume of bucket we can find time taken to fill it.

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M1:ok continue.

Me: further using projectile motion equations in x and y axis.

M4:(after 4-5 seconds) sometimes analysis is not necessary. Use your engineering sense, which one do you think will fill faster.

M2:yes use method of feeling.

Me: (after few seconds, trying to feel) I think second hole will fill first.

M1: sure?

Me: (thinking)

M1:do you think in third hole water has to travel larger distance?

Me:yes sir, I get it.

M1:Is this the only reason for 2nd stream filling bucket before 3rd one?

Me:yes sir.

M1:ok, indicating 5th member to ask question.

M5: have you done tensile test?

Me:sir, I’ve not done it because our college didn’t had the apparatus, but I’ve seen videos on it.

M5:What does the specimen look like?

Me: (drew the shape) the length to diameter ratio is 10:1, lines are marked on specimen at equal interval.

M5:how do we measure stain?

Me:we use stain gauge.

M5: how is it placed?

Me:showed it in drawing

M5:why are lines marked?

Me:after some time elongation becomes large enough, we don’t need stain gauge anymore and we can measure it directly using these lines.

M5: draw stress station curve.

Me:for?

M1:Mild steel

Me: drawing and explaining at the same time.

M5: explain various points.

Me: (explained)

They did not get into dislocation theory and luder’s band.

M2:see Ashish, fracture point has lower stress than what we had even before upper yield point so why doesn’t it fall there itself?

Me: (thinking for 3-5 seconds) sir the stress what we are showing here is the force divided by original area, actually after ultimate stress point necking will take place and actual stress will increase.

M4:what is that stress called?

Me:true stress.

M4:(has a smile on his face) and what is this stress called? the one you have shown on y axis?

Me:Engineering stress.

M3:Now draw the diagram for aluminium.

Me: (drawn)

M1:Where’s the yield point?

Me: specefic yield point is not visible in stress strain curve of aluminium.

M3:how do we know it for engineering purposes?

Me: (even though I knew the answer I knew this is a bit high level question that most Mechanical engineering students are not aware of, so I pretended to think and recall and answered in 2-5 seconds) sir there’s something called proof stress.

M1:what is it?

Me:sir we draw a line with slope equal to modules of elasticity and having intersection at x axis at 0.02% strain. Wherever this line intersect the curve is our yield point.

M1:Good.

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BARC Interview Tip:

See there are two ways of impressing interviewers, either give all the answers you know instantly in which they will keep on raising difficulty level and ultimately you will reach a point where whatever you have studied is covered and now they will ask questions in portion you have not studied and you are forced to think and use your imagination. Or play in area you are comfortable in, just pretend to be uncomfortable, show that you are thinking and the answers are not recorded in your head. Because how far you can think beyond what you have studied is all they want to see. For you 1st law of thermodynamics might be easy and entropy might be difficult, for them everything is easy like A, B, C, D… to you. They just want to see if you are able to think beyond what you’ve studied. Obviously they also check whether you know your basic subjects well but that part everyone is good at, that is necessary condition but not sufficient. Now I’m not saying that do this pretending thing at every question because then they will see right through it. But whenever you get a chance add it. But for doing this you have to remain calm. BARC was the only interview I had as my GATE 2017 rank was 1825 and I had already dropped an year after btech so pressure was high but I learned to control it. I knew if I have to make it happen tension and worry will not help me and that a calm and composed mind can do anything it wants to.

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M4: two vertical bars, one with area A and another with area 2A,same material ,fixed at top to a rigid support and at bottom to a rigid bar which remains parallel to horizontal as we apply Force in vertical direction. Which bar will fail first?

Me:(using compatibility equation to find force in both members)

M4: what are you trying to do?

Me: sir, I’m trying to find out stress in each of the bars the i’ll divide it with area to find stress, which ever will have higher stress will fail first.

M4:ok, continue.

Me: (found out that F2=2F1)sir, stresses will be same (I was little surprised myself)

M1: but how is that possible? Area of cross sections are different.

Me:Sir, forces will be different but area of cross sections are also different so stresses are same. We can see if in another way too, as lower rigid bar remains parallel, strain in both bars are same and since material is both members are same modules of elasticity are same thus stresses will be same in both.

M1: there is a pipe with uniform heat flux through out its length, water is passing through it, inlet temperature is 25C outlet temperature is 160C, pressure is 1atm..draw the temperature profile of water?

Me:(drew a straight line with positive slope from inlet to outlet)

M1:are you sure?

Me: ( thinking for few seconds) ok sir there’s a mistake.

M1:use the eraser.

Me:(drawing again, drew a curve-positive slope-decreasing slips) sir as the water passes is temperature will rise so heat flux will decrease thus rate of heat transfer will decrease…..

M1:heat flux is constant.

Me: ok, that slipped my mind, then slope will also be constant i.e. We will get a straight line… (thinking for few seconds..)

M1:see, water is entering,what do you think is leaving ?

Me:ok, it will become stream till it reaches outlet.

M1:draw the curve .

Me:(drew straight line, positive slope then at 100C 0 slope till some distance then again positive slope till 160C)

M1:which portion will have maximum length?

Me: q.x=mc(dT) and q.x= m(LH). If mass flow rate is m, heat flux per unit length is q an x is unknown for each of the tgree portions. We can find x factor r each portion.

M1: use the values and find out which lenghth is largest.

Me: c of water in s 4.18kj/kgK latent heat of evaporation is 2300 Kj/kgK and Cp of steam( trying to recall for few seconds) sir, I dont remember.

M1:ok, compare the first two.

Me: x is proportional to c and LH. So second portion will be much longer than first portion.

M1: what is triple point?

Me: it is the point at which all three phases, solid, liquid and gases exist in equilibrium.

M1:what is critical point temperature for water.

Me: 273.16C

M1: are you talking about critical point?

Me: oh sorry sir, did you ask critical point?

M1:yes, what is it?

Me: ok, critical point is something completely different than triple point, it is the point at or beyond which liquid will dirrectly convert to gaseous phase whithout consuming any latent heat.

M1: ok what is critical point pressure and temperature for water?

Me: (trying to recal for 2-3 seconds) sir I cannot remember.

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M1:Ok(indicating M4 to ask next question)

M4:(drew a structure with conical part on left and attached to it a cylindrical part, like a nail, left side temp. Is 400k,left side it is 300k) draw the temperature profile.

Me:(drew curve with increasing slope in left portion and staight line on right portion.

M1:draw the curve outside too. Surrounding’s temperature is 300 and 400 not the surface temp.

Me: ok sir then there will be formation of thermal boundary layer.

M2:draw it.

Me: (drew it) sir natural connection will be taking place.

M2:No no draw the temperature profile.

Me:(drew it, decreasing slope outside)

M2:what do you think will be the thickness of this thermal boundary layer?

Me: it will be small, we can actually feel it if we are physically present near the surface.

M2:how thick? Is it in mm or cm or m?

Me: few cms sir.

M1: if in this stucture(nail one),we drill a hole throughout how will the curve change.

Me:it will be like this only.

M1: no no, there will be some chabges.

Me:sir, the hole is if constant cross section so in left part area is area will change but it is still similar function of x as before.

M2:you are not listening what he is getting to say, listen carefully (by now more than 75 minutes had passed and my enery was completely drained out since I already waited for 6 hours before interview, and on top of that I was the only one, the interviewers didn’t offer anything to eat or drink for all the time I was in interview room)

M1: See Ashish, the graph will be similar but do you think slope will change?

Me:Yes sir, slopes will definitely change.

M1:will or increase or decrease?

Me: sir, q=kA(dt/dx)

M1:what is that law called?

Me: ( thinking for 5-6 seconds, I was confused between Fourier’s law and Newton’s law of cooling, taking your time to give correct answer in one shot is always better than speaking quickly and being wrong) Fourier’s law.

M1: so, increase our decrease?

Me: slope is inversely related to area of cross section. As in drilling hole, area decreases so slope will increase.

M1: Ok Ashish you can go now.

Me:Thank you sir(going out )

M4: please leave the pencil.

Me: (with a smile) sorry sir.

M4:Thank you Ashish, all the best.

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