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Hi, you can contact me at physicstorv@gmail.com. Please write your status (class etc) and preparation level and expectation from me clearly in the mail.

@@PhysicsTorv i dont have enough money im just asking for ur free guidance and advice under the hope u will and have the time to help me.

yes

In which class are you - XI, XII or XII passed out?

@@PhysicsTorv 12th passout from kerala

@@PhysicsTorv my shift was 28th s1 i panicked and overattempted

@@freemind-b7l ok

you can search in the physicstorv channel by keyword "optics", you will find many tests on ray optics. I just searched and it seems that test 111 to 120 are from ray optics. This might be helpful.

@@PhysicsTorv yes sir thanks a lot!! you were really helpful in my prep and still are. if i can, id love to help you out with your channel. The editing thumbnails and all. I really want to repay my mentor for the impact they had on my prep

@@makimasimp8 thanks for your support. But i guess you need to clear exam and go for higher study. If you really want to help me, maybe, you can share channel on your social media etc for advertisement. All the best.

@ yes sir but after i have secured my college, ill genuinely be more than happy to help you out in any way whatsoever. sir if you ever want to you can contact me, ill mail you on the id you gave too. thanks a lot again sir!! you were/are a big part of my prep and always will be

@@makimasimp8 thanks. send your email id at physicstorv@gmail.com after exam.

thanks for the kind words. you can contact at physicstorv@gmail.com. All the best for upcoming jee main.

@PhysicsTorv yes sir!

Did not get you? Do you mean arithmetic, geometric and other sequences and series?

@@PhysicsTorv Yes sir

@@PhysicsTorv AGP series questions, HP, GP and AP questions from JEE

@@Digitron001Thanks for the clarification. However, unfortunately, i have plan to only provide materials for physics and not math.

@@PhysicsTorv Oh ok, sir can you make 1 video for projectile motion and inclined projectile motion

Ok, I will try.

In my opinion, you must look at the questions of previous shifts. However, you should not jeopardize your revision of formulae and equations (if necessary for you). These questions will give you an idea of topics from where questions are being asked. So, will know which topics from where questions going to be. You, if can, have a look to the concepts, of those topics. You should not spend too much time in solving previous shifts questions as they will not be repeated by just changing the numerical values but will be completely different question from the same topics but little different concepts. So, maybe you can solve a few questions of the related concepts for all topics where questions are being asked. Regarding, getting tensed before examination, it is quite normal. Almost, everyone gets tensed. You need to control your emotion, believe in yourself and having philosophy of giving your 100% should help you during examination.

Remember, one exam can not decide your fate. So, just be bold, and to some extent badass, and face the exam. Nervousness will only damage your chances so avoid them. Of course, you should be badass within the limit so you do not become careless. All the best!

@@PhysicsTorvokay got that

@@PhysicsTorv thank you sir

Thanks for the compliment. This encourages to make more questions for you aspirants, otherwise, i start thinking that my efforts are just futile.

derivation is beyond the scope of school syllabus. Dimensional analysis is very very powerful tool not only to check an equation but to find out the laws of nature and often used by researchers.

@@PhysicsTorv i will keep that in mind from now on thanks

Thanks for your question. I will upload solutions in next one or two days.

@PhysicsTorv ok

thanks

thanks

Classes will be paid one but all materials on youtube will free of cost including answer keys and solutions.

(1) Consider an element of length dl that subtends angle d\theta at the centre of the ring O. Its mass, dm = [ dl / (2 pi * r)] * m (2) There is tension in the ring, say, T . The direction of T is tangentially. Now I need a diagram to explain better but will do w/o that. Extend the line of tensions in backward directions from the tip of dl to meet at a point say C. Join the tips of dl to the centre of the ring. The angle between two tensions at C = 180 - d\theta. (3) Join CO and consider this direction as x-axis. Resolve two tensions along CO. This will be give force along CO = T cos (90 - d\theta/2) + T cos(90 - d\theta/2) = 2 T sin(d\theta/2) = 2 T * (d\theta/2) = T d\theta = T (dl / r) . Here we used the fact that d\theta = small so sin(alpha) = alpha. (4) This force along CO = T dl /r = centripetal force acting on dl . Ans, we know that centripetal force = (dm) omega^2 r. So, (5) T dl /r = (dm) omega^2 r = [dl * m/(2 pi r)] omega^ r . It means T = m omega^2 r / (2 pi). Done!

@@PhysicsTorvthanks for this, i can solve it further , but i dont understand the logic of taking element that way

@@Grind21242 If you consider whole ring, then total force = 0 (symmetric and opposite force for diametrically opposite points). Tension is same everywhere but direction is different. So, in order to consider T (in a specified direction), you must take a small element. Again, taking a point is meaningless as it has zero mass. In fact, for a wire, there is nothing like a point. In fact, that point is element which has some mass and T. These T must be in opposite direction to keep wire stretched.

I will come soon. Test 132 solutions is published so its solution will come on its turn.

Thanks. All questions are of JEE standards and aimed for better conceptual understanding.

No, this one is correct. Here, I did not say that mass is constant. The obvious thing is to consider density to be constant unless stated otherwise. So, if you calculate g_0 keeping density constant, you will find g_0 = (4/3) \pi G ho R. So, g_0 is proportional to R, and hence, when R is halved, g_0 also get halved. Thanks for asking.

@@PhysicsTorv that makes sense thanks for answering

@@PhysicsTorv with this ive made 3 mistakes overall also sir do you have any tips to calculate fast ,like calculating random decimals similar to that one in q14 .ive wasted a lot of time on that

@@Grind21242 Well, what would I suggest to you is following: (1) If you think, some question will take too long to solve - either conceptually or silly calculation, mark this question for second round. First finish which can be completed in a quick time. (2) Do not do silly calculation to the infinite precision, unless, all options are too close. Usually, this is not the case. (3) To clarify statement (2), let us use Q(14) as an example. Here, intensity = [(1/2) \epsilon_0 E_0^2] v = [(1/2) (B_0^2/\mu_0)]. Here, you will use second formula. So, I = [B_0^2 / (2 *\mu_0)]*v = [10^(-12) / (2 * 4 * pi* 10^(-7))]*v . Here, check if v = c = 3 x 10^(-8). In this case, v = omega / k = (6 x 10^6/0.02) = 3 x 10^8 =c. But, this may not be the case all the time. EM wave might be travelling in a medium other than vacuum/air and might be considerably smaller. So be vigilant about it. Now, coming back to the calculation, collect all powers of 10 in numerator and others at one place. So, I = 10^(-5) * (3 x 10^8)/(8 * 3.14) =10^3 * (3/25) . Do not try to multiply 8 and 3.14 but just take an approximate value. 8 * 3 = 24 and 8 *0.14 = 1.12 = 1. So, 8 *3.14 = 25. We are underestimating denominator and so our answer would be little higher than the actual one. Now, I = 10^3 * 0.12 = 120 W/m^2. So, the closest option is 119.4 W/m^2. You can also use the trick that pi^=10. If you use, this you will be saved from division. So, I = 3 x 10^3/ (8 * pi) = 3 * pi * 10^3/(80) = 300 * pi/8 =942/8 = 118. So, closest option is (2). I hope, I could be of some use to your fast calculation strategy. Good luck!

@@PhysicsTorv Noted. Thank you so much

Thanks for pointing this out. You are right. Answer of Q.(8) should be (1) and not (2). I have put a notice in the description section for others. Thanks a lot for this.

@@PhysicsTorv not a problem

It is published. Link is here : th-cam.com/video/Q6KtZaE9Bj4/w-d-xo.html

thanks. keep commenting so that i would know that my efforts are helpful to students.

It's paid one.

My own. But based on PYQ.

@@PhysicsTorv got it sir!

thanks

thanks

Ok, i will try to do so.

I realized that many students might have difficulty in friction problems. However, if one know the rules, it is easy.

@@PhysicsTorv thanks would be great

In the link given in the description. I put the link here too: drive.google.com/file/d/1lDPhuqXgd_yborysWaxslAkht1JjsoFe/view?usp=drive_link

thanks

Thanks.

What can I suggest to you is that you go through all the questions of daily test sets and weekly test sets. If your foundation, as you wrote, is little less than desired, you might face difficulty in solving that. However, you will get an idea of questions of JEE level. Do not disheartened. You can appear in January and April in 2025. Most of the students do not do well in January session due to various reasons. So, focus for April session without loosing hope. I will start revision session soon for January session with all formulae and tricks with limited but important questions. Stay tuned. This revision session might be the most helpful to you. Please write back if you have any specific questions.

@PhysicsTorv Got it sir, I will do as you suggested above, Thanks. Will be waiting for the revision series as well, excited. I really appreciate your response, have a gud day sir!

Concave mirror forms virtual and erected image when the object is placed in between pole and F (focal point). Here, magnification m = +4 (+ because person would like to see erected image of his face: common sense). but, m = -v/u, so that v = -m*u = -4u. Using mirror formula 1/f = 1/v + 1/u. => 1/f = -1/4u + 1/u => 1/f = 3/4u => u = 3f/4 => u = 3*(-20)/4 = -15 cm. Hope, it explains. If you consider, m = -4 instead of +4 (you want to see your face upside down), then you will find u = -25 cm.

(i) Tensions are internal forces and hence sum of the work done by all internal forces = 0. Otherwise, world would have been very different than what we see them today. (ii) Here, tensions are different in left and right strings, say, T1 in left and T2 in right. T1 does two works (a) moving m1 upwards by say s distance (b) rotating pulley by an angle, say, (-\theta). Negative sign, because m2 comes downward. Similarly, T2 does two work: (a) moving m2 downward : W = -T2s (negative because T2 and movement of m2 are in opposite direction). (iii) If you do calculations, you will find acceleration of m1 and m2 = 4 m/s^2, T1 = 14 N and T2 = 18 N, distance travelled by individual masses = 8 m in 2 seconds. So, work done by T1 and T2 in moving m1 and m2 are W1 = 14* 8 = 112 J; W2= -18 * 8 = -144 J. (iv) Work done in rotating pulley by both T1 and T2 : W1R = torque * theta = -T1*R * theta = -T1*R*(0.5*alpha*t^2) = -T1*R*(0.5*(a/R)*4) =-2T1*a = -2 *14*4=-112 J. (v) Similar calculation will yield, work done in rotating pulley by T2= W2R = 2 * 18*4=144 J (vi) So, total work done by T1+T2 = (W1+W1R) + (W2+W2R) = (112-112) + (-144 + 144 ) = 0. I hope it helps you. You can note that work done by either tension, in this case, is zero, separately.

It will come tonight at 20:00.

Questions on Bohr's atomic model for JEE Main with answer keys. Does it answer your question?

At the beginning, concise description of Bohr's model is also presented.

@@PhysicsTorv okay thank you!

th-cam.com/video/j27Hek_tLx8/w-d-xo.html Have you seen this?

@PhysicsTorv no sir I'll check ryt now

@@AnubhavV-tr7jw th-cam.com/play/PLT6rLbGyfT-YMFu5C1gNi8xGpdkeVyY5T.html You should have a look of the playlist for all relevant videos th-cam.com/play/PLT6rLbGyfT-Y__8Dvln17y6ispCNwN0d2.html

All are correct except 4th one which should be arctan(1/3). In short video I typed E = 3i instead of E =sqrt{3} i. So, the correct answer would be for later one 30 degree. Please look at the regular video. You can also find answer keys if you search in the playlist.

@PhysicsTorv sir where I can access all test

@@Vaibhav-o6g th-cam.com/video/hSpL7frsCN0/w-d-xo.html

Search Physics Torv on youtube. Click on logo. It will take you to Physics Torv channel. Click on the playlist button. Search JEE (Main) Daily Test Series. All tests total 98 are there. You can also look for other playlist such as weekly test series.

Thanks for your question. I have not made video for mechanics chapter. Earlier I was making video with random questions, then some of you suggested to make video chapter-wise. Set 1 to Set 27 consists random questions. They might have one question from mechanics. You can check them out. Chapter wise video for mechanics will come later. After modern physics I have plan to make videos of optics, then, I will make video of mechanics.

​@@PhysicsTorvok sir thanks a lot

thanks .

Not really. However jee 2025 syllabus do not mention explicitly radioactivity but they mention "nuclear fission and fusion". Without radioactivity one can not study fission. So, I do not think that radioactivity is deleted from the syllabus.

I just checked 2024 year questions to find that no questions from radioactivity was asked. So, you are right. Radioactivity has been removed from 2025 JEE syllabus. It is a shame but this is how it is. Thanks a lot for pointing this to me.

th-cam.com/video/DB55Z81vlQY/w-d-xo.htmlsi=TKQ8V1SR6x-C6Kw6 You should visit Physics Torv youtube channel and look for various playlist. There is a plethora of material for JEE. In the playlist you can also access of Answer Keys.

You can also find pdf of daily Question set in the description of regular video.

Thank you sir ​@@PhysicsTorv

These are not pyq questions but inspired by pyq. Most of them are very different from pyq, however of the levels of jee main. Some of these are also of the level of jee advanced.