tag:blogger.com,1999:blog-8705083632203739705.post4841030862536574522..comments2021-07-13T13:15:52.215+05:30Comments on PHYSICS Information and Links: GATE 2011 Question Paper Download Answer key pdf file 13 februaryUnknownnoreply@blogger.comBlogger94125tag:blogger.com,1999:blog-8705083632203739705.post-12366211787532862312015-03-01T20:00:54.144+05:302015-03-01T20:00:54.144+05:30GATE 2011 Answer Key<a href="https://physics43.files.wordpress.com/2015/02/gate-2011-answer-keyph-physicskerala.pdf" rel="nofollow">GATE 2011 Answer Key</a>Shttps://www.blogger.com/profile/08625054792778496944noreply@blogger.comtag:blogger.com,1999:blog-8705083632203739705.post-40817201140903291402012-08-19T18:43:41.608+05:302012-08-19T18:43:41.608+05:3042. D50. AHow?42. D<br>50. A<br>How?physicskerala.inhttp://www.physicskerala.in/noreply@blogger.comtag:blogger.com,1999:blog-8705083632203739705.post-33654495587629311902012-02-03T11:21:07.430+05:302012-02-03T11:21:07.430+05:30When \(a \ne 0\), there are two solutions to \(ax^...When \(a \ne 0\), there are two solutions to \(ax^2 + bx + c = 0\) and they are<br><br><br>$$x = {-b \pm \sqrt{b^2-4ac} \over 2a}.$$Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8705083632203739705.post-37687705848054145652011-07-02T17:46:08.493+05:302011-07-02T17:46:08.493+05:30Hi i want to download gate chemistry paper 2011Hi i want to download gate chemistry paper 2011Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-8705083632203739705.post-87193315790225240872011-03-13T19:28:39.195+05:302011-03-13T19:28:39.195+05:30@ CoPRA I am also find a right ans.But when i am u...@ CoPRA <br>I am also find a right ans.<br>But when i am use maxwell forth eq.(Ampeer law) for free space then we get sol. (A).<br>or we are going with mexwell third eq.(farday law) for free space then get ans. (B).<br>Our both sol.'s are not satisfies Mexwell first or second eq.(Gauss or free magnetic Pole) <br>So How can use maxwell eq. solved this qus.Anuj Kumarhttp://www.blogger.com/profile/10650566899948645036noreply@blogger.comtag:blogger.com,1999:blog-8705083632203739705.post-81193280185556713712011-03-12T07:43:15.639+05:302011-03-12T07:43:15.639+05:30To Anuj KumarYou missed the negative sign in Farad...To Anuj Kumar<br>You missed the negative sign in Faraday's law..<br><br>Try to find the E( Electric field) first using the above relation. Don't try to fix the answer arbitrarily. <br><br>The second combination You mentioned above holds good the left hand rule. But Our aim is to find correct answer. not the combination of answersCoPRAhttp://www.blogger.com/profile/14350348626725467837noreply@blogger.comtag:blogger.com,1999:blog-8705083632203739705.post-8814005217357449732011-03-11T03:03:20.016+05:302011-03-11T03:03:20.016+05:30@basudev farday law are Curl E = dB/dt.solved by t...@basudev farday law are <br><br>Curl E = dB/dt.<br><br>solved by this eq.<br>get it problem.<br>because Maxwall have linear Independence solution.<br>So maxwall eq.s have one or more sol.<br>Thanks.Anuj Kumarhttp://www.blogger.com/profile/10650566899948645036noreply@blogger.comtag:blogger.com,1999:blog-8705083632203739705.post-88561541011452425322011-03-09T22:42:49.241+05:302011-03-09T22:42:49.241+05:30@Anuj: I dont follow what you said. By the way, Fa...@Anuj: I dont follow what you said. By the way, Faraday's law merely states del.E = rho/epsilon. Where is the connection between magnetic and electric fields?basudevhttp://www.blogger.com/profile/14917587915550283214noreply@blogger.comtag:blogger.com,1999:blog-8705083632203739705.post-27283138480182295382011-03-09T15:13:58.155+05:302011-03-09T15:13:58.155+05:30@basudev you can say right.but when we take electr...@basudev you can say right.<br>but when we take electric field (B).<br>and use Farday law (Mexwell 2ed eq.)<br>or solved same pattern then we get also this magnetic field.<br>again problem.Anuj Kumarhttp://www.blogger.com/profile/10650566899948645036noreply@blogger.comtag:blogger.com,1999:blog-8705083632203739705.post-65962561598381357042011-03-08T08:47:58.070+05:302011-03-08T08:47:58.070+05:30@Anuj: Instead of sitting and guessing, please kin...@Anuj: Instead of sitting and guessing, please kindly plug the magnetic field into the Maxwell's equations. There is only one unique solution.basudevhttp://www.blogger.com/profile/14917587915550283214noreply@blogger.comtag:blogger.com,1999:blog-8705083632203739705.post-71848954100077780652011-03-07T02:13:25.661+05:302011-03-07T02:13:25.661+05:30About Qus. 53i thing ans(D)Because in oscillator s...About Qus. 53<br>i thing ans(D)<br>Because in oscillator system have orthonormalized states so<br>0th level energy is 1h/2.<br>1st level energy is 3h/2.<br>2ed level energy is 5h/2.<br><br>every state have unity probability.<br>So simply add every state energy.<br>Get total energy.<br>Plz tell me.<br>How can get ans.(B)<br>thanksAnuj Kumarhttp://www.blogger.com/profile/10650566899948645036noreply@blogger.comtag:blogger.com,1999:blog-8705083632203739705.post-72896961677237709132011-03-06T14:09:13.005+05:302011-03-06T14:09:13.005+05:30About the Qus. no. 54 & 55We are give ans (A) ...About the Qus. no. 54 & 55<br>We are give ans (A) of Qus.(54)<br>Then direction of properation <br>is -(i+j)/2^1/2.<br>OR Poynting Vector Ans (D).<br>Poynting Vector & Direction of properation is same OR Hold left hand rule.<br>It's combnation is correct. <br><br>But, When <br>we are take Ans (B) of Qus.54. Then direction of properation <br>is (i-j)/2^1/2.<br>OR Poynting Vector Ans (A).<br>Poynting Vector & Direction of properation is same OR Hold left hand rule.<br><br>it's combnation is also correct.<br>so plz tell me.<br><br>What is this?Anuj Kumarhttp://www.blogger.com/profile/10650566899948645036noreply@blogger.comtag:blogger.com,1999:blog-8705083632203739705.post-12259710757209201412011-03-03T23:06:24.229+05:302011-03-03T23:06:24.229+05:30basudev,You are right. I just missed them. The ans...basudev,<br>You are right. I just missed them. The answers for 16 and 17 is A. 28 ofcourse I was not sure. Thanks anyway.Remyahttp://www.blogger.com/profile/02737088415712728399noreply@blogger.comtag:blogger.com,1999:blog-8705083632203739705.post-9235201887231836052011-03-02T11:41:57.717+05:302011-03-02T11:41:57.717+05:30@remyai believeanswer of question 4 is DCarnot eng...@remya<br>i believe<br>answer of question 4 is D<br>Carnot engine is an ideal theoretical engine.<br>its cycle is ideal and reversible,<br>look its TS diagramSREEJITH Phttp://www.blogger.com/profile/08625054792778496944noreply@blogger.comtag:blogger.com,1999:blog-8705083632203739705.post-79557131445155877772011-03-02T09:30:10.452+05:302011-03-02T09:30:10.452+05:30@Remya: I am afraid the problem you mentioned in G...@Remya: I am afraid the problem you mentioned in Griffiths is not quite equivalent to the problem in the paper. There is an issue of handedness involved. The system of coordinates has to behave like a right handed system by convention. The answer given in the key is correct. <br><br>So goes for both 16 and 28 (which I explicitly mentioned how to do in the discussion above).basudevhttp://www.blogger.com/profile/14917587915550283214noreply@blogger.comtag:blogger.com,1999:blog-8705083632203739705.post-85246016315874611282011-03-01T14:47:44.724+05:302011-03-01T14:47:44.724+05:30One error in the answer key is ques.4. I strongly ...One error in the answer key is ques.4. I strongly believe that the ans is C.Remyahttp://www.blogger.com/profile/02737088415712728399noreply@blogger.comtag:blogger.com,1999:blog-8705083632203739705.post-30199604826719648422011-03-01T12:48:49.413+05:302011-03-01T12:48:49.413+05:30The answer key has some errors. I wud like to corr...The answer key has some errors. I wud like to correct few which i am confident. 16. C ; 17. C (Refer D.J.Griffith page no. 226, example 5.8); 28. B(I am not sure).Remyahttp://www.blogger.com/profile/02737088415712728399noreply@blogger.comtag:blogger.com,1999:blog-8705083632203739705.post-38611355495238445572011-02-26T14:47:41.162+05:302011-02-26T14:47:41.162+05:30Check this out:http://en.wikipedia.org/wiki/Beta_d...Check this out:<br><br>http://en.wikipedia.org/wiki/Beta_decay<br><br>Neutrinos in beta decay<br>Historically, the study of beta decay provided the first physical evidence of the neutrino. In 1911 Lise Meitner and Otto Hahn performed an experiment that showed that the energies of electrons emitted by beta decay had a continuous rather than discrete spectrum. This was in apparent contradiction to the law of conservation of energy, as it appeared that energy was lost in the beta decay process. A second problem was that the spin of the Nitrogen-14 atom was 1, in contradiction to the Rutherford prediction of ½.basudevhttp://www.blogger.com/profile/14917587915550283214noreply@blogger.comtag:blogger.com,1999:blog-8705083632203739705.post-11728858430420331482011-02-26T14:38:35.225+05:302011-02-26T14:38:35.225+05:30Good to see that the answers have finally been sum...Good to see that the answers have finally been summarized. I find the answer to 29 as (C) particularly interesting since the generalised potential has dissipation and is not a conservative potential. I missed this point. Having said this, I am not sure if the person posting the answers understood why the answers were going to be such. <br><br>For example, the answer to 16, being a sitter problem for high school students, has not been provided. Clearly, the person putting the answer up does not understand it. <br><br>Coming to 27, which as I mentioned in my solution, has to be transformed into a volume integral to become a 3 second problem, has been wrongly evaluated. The answer should be (B).<br><br>For (39), the existence of antineutrinos certainly weren't inferred from parity violation experiments. Look up any text book. Nowhere does it say that. It cannot be forward backward asymmetry of electrons!<br><br>Coming to 42, could anybody explain why there should be 150 combinations? I can't get that figure by any stretch of imagination! Dont tell me you put it up simply because someone suggested the answer. <br><br>By the way, for 56, it is actually straight out of barrons GRE book. Type the sentence asked on google and the answer would show up straightaway. It is indeed (B).basudevhttp://www.blogger.com/profile/14917587915550283214noreply@blogger.comtag:blogger.com,1999:blog-8705083632203739705.post-37142685621241081942011-02-25T08:13:24.004+05:302011-02-25T08:13:24.004+05:30@creepy: (37) is indeed (A). I corrected my answer...@creepy: (37) is indeed (A). I corrected my answer in the discussion above.basudevhttp://www.blogger.com/profile/14917587915550283214noreply@blogger.comtag:blogger.com,1999:blog-8705083632203739705.post-74604231384814823002011-02-25T07:39:18.501+05:302011-02-25T07:39:18.501+05:30To basudev..... Yes man U r ryt... I caught on the...To basudev.....<br> Yes man U r ryt... I caught on the wright trap.. Q. 48 the ans id D. and I had a Nice discussion with U and some other Guys... Thanks to the blogspot owner and Friends whoever posted their view.CoPRAhttp://www.blogger.com/profile/14350348626725467837noreply@blogger.comtag:blogger.com,1999:blog-8705083632203739705.post-77725510417611591432011-02-24T22:09:20.551+05:302011-02-24T22:09:20.551+05:30By the way, about 28, (y+1)dy=-xdx, so that y^2/2 ...By the way, about 28, (y+1)dy=-xdx, so that y^2/2 + y = -x^2 +c.... rearrange terms, and a circle with center at x=0 and y=-1/2 shows up with a variable radius.basudevhttp://www.blogger.com/profile/14917587915550283214noreply@blogger.comtag:blogger.com,1999:blog-8705083632203739705.post-87837797654518331812011-02-24T22:01:09.391+05:302011-02-24T22:01:09.391+05:30@dreamz: My friend, you are putting the repulsive ...@dreamz: My friend, you are putting the repulsive interaction between the +q and 2q charges as +ve. Thing is, this force is going to be directed in the same direction as the ones for the negative charges since this positive charge is located in the opposite side of the negative charges.<br><br>@shishir: Thanks for your wishes. About JEST,unfortunately I could not appear for it due to a freak registration related incident. I can't be of much help on that. Sorry about it.basudevhttp://www.blogger.com/profile/14917587915550283214noreply@blogger.comtag:blogger.com,1999:blog-8705083632203739705.post-83054077368168497842011-02-21T13:41:48.056+05:302011-02-21T13:41:48.056+05:30@copra i dont understand the meaning of -'then...@copra<br> i dont understand the meaning of -<br>'then according to the ORDER only we should differentiate.. ' <br>and 'its not diverging and has to be differentiated... ' <br>this is hilarious. <br><br>whether we care about residue or not, if you want to classify the singular points, we should stick to Lorentz series. there are many ways of getting the ans here i wil list 2. <br><br>simplest way is look at the <b>numerator</b> of the function it has a <b>simple zero</b> at z=pi. so overall the function has singularity at z=pi of order one. <br>simplest example is ' (sin(z))/z ', it doesnt have singular point. the num and denominator has zero at z=0.<br> please expand it and see, no principal part for lorentz series. <br> <br>when we consider lorentz series<br>expand <br>z*Sin(z) about 'pi' (because the denominator is already of that form)<br>you can see that it terminates at the first principal part. <br><br>so ans for <b>48.) D </b><br><br>@basudev . u did a great job of collecting the ans and active in discussion<br><br>as u said we have to place the image charges in those positions.<br>so by couloub's law, the total force acting on q is,(it obeys superposition) <br><br>k(-q^2 + (2q)^2 + -((2q)^2)/4) = k(q^2)/2<br>so the ans is <b> 16.) D </b><br><br>similarly, last yrs ans-key has lot of mistakesfroZendreamzhttp://www.blogger.com/profile/06794017580229692345noreply@blogger.comtag:blogger.com,1999:blog-8705083632203739705.post-17587715059553468232011-02-20T23:54:21.847+05:302011-02-20T23:54:21.847+05:3047,Dwhat is the answer of qns no.36??47,D<br>what is the answer of qns no.36??Rupakhttp://www.blogger.com/profile/10341823716732055181noreply@blogger.com