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Various companies visit IME IIT Kanpur for placements some of them are given below:

The overall average package for IME IIT Kanpur stood at INR 18.2 LPA. Check the below table to know more about the average package at IME IIT Kanpur:

Explanation- F_x= $\frac{1}{4}\frac{q^2}{4?{?}_0{x}^2}$

W= ${?}_d^{?}fdx$ = ${?}_d^{?}\frac{q^{2}dx}{4*4?{?}_0}\frac{1}{x^2}$

= $\frac{q^2}{4*4?{?}_0}\frac{1}{d}$

= $\frac{{(1.6*{10}^{-19})}^2*9*{10}^9}{4*{10}^{-10}}$ J= 3.6eV 

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l}NowputtingthevalueofKineq.\left(i\right)weget\\ 6x^2+4\pi r^2=x^2\left(\pi +6\right)\\ \Rightarrow 6x^2+4\pi r^2=\pi x^2+6x^2\Rightarrow 4\pi r^2=\pi x^2\Rightarrow 4r^2=x^2\\ \therefore 2r=x\\ \therefore x:2r=1:1\\ Nowdifferentiatingeq.\left(ii\right)w.r.t.,x,wehave\\ \frac{d^{2}V}{d{x}^2}=6x-\frac{3}{}\frac{d}{dx}\left[x{\left(K-6x^2\right)}^{1/2}\right]\\ =6x-\frac{3}{}\left[x.\frac{1}{2}\times \left(-12x\right)+{\left(K-6x^2\right)}^{1/2}.1\right]\\ =6x-\frac{3}{}\left[\frac{-6x^2}{}+\right]\\ =6x-\frac{3}{}\left[\frac{-6x^2+K-6x^2}{}\right]=6x+\frac{3}{}\left[\frac{12x^2-K}{}\right]\\ =6+\frac{3}{}\left[\frac{\frac{12K}{\pi +6}-K}{}\right]\left[Putx=\right]\\ =6+\frac{3}{}\left[\frac{12K-\pi K-6K}{}\right]=6+\frac{3}{}\left[\frac{6K-\pi K}{}\right]\\ =6+\frac{3}{\pi }\left[\left(6K-\pi K\right)\right]>0soitisminima.\\ Hence,therequiredratiois1:1whenthecombinedvolumeisminimum.\end{array}$

Minimum

IME IIT Kanpur has an impressive placement record. The placement statistics for 2024 are given in the table below:

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol:

IME IIT Kanpur has announced its 2024 placement report. Have a look at the data tabulated below to know more:

Candidates have to register and submit application form online for Uttarakhand NEET counselling. Registration can be done at the official website, hnbumu.ac.in. As part of registration, candidates ahve to enter personal, academic, contact, domicile, and NEET 2025 details and upload images in right size and format.

Explanation- A= 10^-4m²

So d= 10^-3 and i= 100^-4A

I= 100W/m²

$\mathrm{?}=600\mathrm{n}\mathrm{m}=600*{10}^{-9}$

${\mathrm{?}}_{\mathrm{N}\mathrm{a}}=0.97\mathrm{k}\mathrm{g}/{\mathrm{m}}^3$

$\mathrm{v}\mathrm{o}\mathrm{l}\mathrm{u}\mathrm{m}\mathrm{e}=\mathrm{A}*\mathrm{d}$ 10^-4(10^-3)=10^-7m³

$\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{}23\mathrm{k}\mathrm{g}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{s}\mathrm{o}\mathrm{i}\mathrm{d}\mathrm{u}\mathrm{m}$

So volume Na atoms=23/0.97m³

Volume occupied by one Na atom= $\frac{23}{0.97*6*{10}^{26}}=3.95*{10}^{-26}{\mathrm{m}}^3$

Number of Na atoms in target $\frac{{10}^{-7}}{3.95*{10}^{-26}}=2.53*{10}^{18}$

So energy falling per sec= $\frac{\mathrm{n}\mathrm{h}\mathrm{c}}{\mathrm{?}}=\mathrm{I}\mathrm{A}$

So n= $\frac{\mathrm{I}\mathrm{A}\mathrm{?}}{\mathrm{h}\mathrm{c}}$ = $\frac{100*{10}^{-4}*660*{10}^{-9}}{6.62*{10}^{-34}*3*{10}^8}=3.3*{10}^{16}$

N=P $*\mathrm{n}*\mathrm{N}\mathrm{a}=\mathrm{P}*3.3*{10}^{16}*2.53*{10}^{18}$

I = 100 $*{10}^{-6}={10}^{-4}$ A

I=Ne= $\mathrm{P}*3.3*{10}^{16}*2.53*{10}^{18}$ ( ${10}^{-4}$ A)

P= 7.48 $*{10}^{-21}$ it is less than 1.

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol: 

The number of students placed during IME IIT Kanpur placements 2024 has not been disclosed yet. Once released, updates will be posted here. Meanwhile students can have a look at the placement highlights given below:

Amity University Raipur offers plenty of courses in diverse disciplines. Students often seek out the popular Master of Business Administration (MBA) course for a solid Management career. The various MBA specializations, along with their respective tution fees are mentioned below:

BSEB 12th books are based on NCERT patterns, especially for Science subjects. These books help in building strong basics which are useful for exams like NEET, JEE, and other entrance tests. Subjects like Physics, Chemistry, and Biology in BSEB books explain core concepts clearly. After reading the textbooks, students can move to advanced-level reference books. Regular revision of BSEB books can give students a solid foundation for competitive exams.

The placement package recorded at IME IIT Kanpur placements 2024 is given in the table below:

This is a Long Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Weknowthaty=mx+cwilltouchtheellipse\\ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1if{c}^2=a^2{m}^2+b^2\\ Hereequationofstraightlineisxcos\alpha +ysin\alpha =pandthatofellipseis\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\\ xcos\alpha +ysin\alpha =p\\ \Rightarrow ysin\alpha =-xcos\alpha +p\\ \Rightarrow y=-x\frac{cos\alpha }{sin\alpha }+\frac{p}{sin\alpha }\Rightarrow y=-xcot\alpha +\frac{p}{sin\alpha }\\ Compaingwithy=mx+c,weget\\ m=-cot\alpha andc=\frac{p}{sin\alpha }\\ So,accordingtothecondition,weget{c}^2=a^2{m}^2+b^2\\ \frac{p^2}{si{n}^2\alpha }=a^2{\left(-cot\alpha \right)}^2+b^2\\ \Rightarrow \frac{p^2}{si{n}^2\alpha }=a^2\frac{co{s}^2\alpha }{si{n}^2\alpha }+b^2\Rightarrow p^2=a^{2}co{s}^2\alpha +b^{2}si{n}^2\alpha \\ Hence,a^{2}co{s}^2\alpha +b^{2}si{n}^2\alpha =p^{2}Henceproved.\end{array}$

Yes, BSEB provides 12th class books in both Hindi and English mediums. Students can choose the language based on their medium of instruction or preference. All subjects like Physics, Chemistry, Maths, Biology, Political Science, and History are available in both languages. This helps students understand the concepts better in their chosen language. You can check your school or official websites for both versions.