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The overall average package recorded during IME IIT Kanpur MBA placements was INR 18.2 LPA. Refer to the table below to know more:

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

Sol: 

$\begin{array}{l}{\left(1.999\right)}^5={\left(2-0.001\right)}^5\\ Letx=2and\Delta x=-0.001\\ Lety=x^5\\ Differentiatingbothsidesw.r.t.,x,weget\\ \frac{dy}{dx}=5x^4=5{\left(2\right)}^4=80\\ Now\Delta y=\left(\frac{dy}{dx}\right).\Delta x=80.\left(-0.001\right)=-0.080\\ \therefore {\left(1.999\right)}^5=y+\Delta y\\ =x^5-0.080={\left(2\right)}^5-0.080=32-0.080=31.92\\ Hence,approximatevalueof{\left(1.999\right)}^{5}is31.92.\end{array}$

You can share PSEB Class 10 solved paper with teacher, who will help know if the answers are correct and can fetch one full marks.

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

Sol: 

$\begin{array}{l}Asperthegivencondition,\\ \frac{d\theta }{dt}=2\frac{d}{dt}\left(sin\theta \right)\\ \Rightarrow \frac{d\theta }{dt}=2cos\theta \frac{d\theta }{dt}\Rightarrow 1=2cos\theta \\ \therefore cos\theta =\frac{1}{2}\Rightarrow cos\theta =cos\frac{\pi }{3}\Rightarrow \theta =\frac{\pi }{3}\\ Hence,therequiredangleis\frac{\pi }{3}.\end{array}$

Explanation- according to law of conservation of momentum

S0 m_Av+m_b0=m_Av₁+m_Bv₂

So m_A(v-v₁)= m_Bv₂

according to law of conservation of kinetic energy

1/2m_Av²=1/2m_Av₁²+1/2m_Bv₂²

So m_A(v²-v₁²)= m_Bv₂²

From above eqn we can say that v+v₁=v₂ or v=v₂-v₁

So v₁= $\frac{m_A{-m}_B}{m_A{+m}_B}$ v  and v₂= $\frac{2m_A}{m_A{+m}_B}$ v

_$?$ initial=h/m_Av

_$?$ final=h/m_Av₁= $\frac{h\left(m_A{+m}_B\right)}{m_{a\left(m_A{-m}_B\right)v}}$

$d?$ = $?$ _final- $?$ _initial= $\frac{h}{m_{A}v}\{\frac{m_A{+m}_B}{m_A{-m}_B}-1\}$

Backstage Pass Institute of Gaming and Technology fee vary from course to course. Students need to pay a certain amount of fee to confirm their admission. The course-wise fees for top courses are as follows:

• UG Diploma: INR 2.3 lakh
• BSc: INR 6 lakh
• BA: INR 6 lakh

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

Sol: 

$\begin{array}{l}Weknowthat\\ Areaofcircle,A=\pi r^2,wherer=radiusofthecircleandperimeter=2\pi r\\ AsperQuestion,\frac{dA}{dt}={\rm K},whereK=constant\\ \Rightarrow \frac{d}{dt}\left(\pi r^2\right)=K\\ \Rightarrow \pi .2r.\frac{dr}{dt}=K\\ \therefore \frac{dr}{dt}=\frac{{\rm K}}{4\pi r^2}\dots \left(1\right)\\ NowPerimeterc=2\pi r\\ Differentiatingbothsides\text{?}w.r.t.t,weget\\ \Rightarrow \frac{dc}{dt}=\frac{d}{dt}\left(2\pi r\right)\\ \Rightarrow \frac{dc}{dt}=2\pi .\frac{dr}{dt}\\ \Rightarrow \frac{dc}{dt}=2\pi .\frac{K}{2\pi r}=\frac{K}{r}\left[Fromeqn\left(1\right)\right]\\ \Rightarrow \frac{dc}{dt}\propto \frac{1}{r}\\ Hence,theperimeterofthecirclevariesinverselyastheradiusofthecircle.\end{array}$

National Institute of Technology Hamirpur is considered a Tier 2 institute. It is one of the reputed institute and offer quality education to the studnets. It is one of the thirty-one NITs of the country. NITH was established by a collaboration of the Government of India and the Government of Himachal Pradesh.  NIT Hamirpur offers a variety of UG, PG and doctoral courses across streams like Engineering, Architecture,  Sciences, Humanities & Social Sciences, and Management. 

OP Jindal University, Raigarh offers a comprehensive range of courses in various disciplines. The flagship course at OPJU Raigarh is the Bachelor of Technology (BTech) programme which offers technical foundation to Engineering students. The tuition fee for the BTech courses at OPJU ranges from INR 3.15 Lacs to INR 6.7 lakhs. The various BTech specializations at OP Jindal University, Raigarh are mentioned below:

1. BTech Computer Science & Engineering
2. BTech Mechanical Engineering
3. BTech in Metallurgical & Materials Engineering
4. BTech Civil Engineering
5. BTech Mining Engineering
6. BTech Electrical Engineering

Explanation -Given threshold frequency of A is given by v_0A= 5 $*{10}^{14}$ hz

V_OB= 10 $*$ 10¹⁴hz

$?=h{v}_0$

$\frac{{?}_{OA}}{{?}_{OB}}=\frac{5*{10}^{14}}{10*{10}^{14}}1$

${?}_{OA}$ < ${?}_{OB}$

$\left(ii\right)$  for metal A slope=h/e= $\frac{2}{(10-5){10}^{14}}$

$h=\frac{2e}{5*{10}^{14}}=\frac{2*1.6*{10}^{-19}}{5*{10}^{14}}$ = 6.4 $*{10}^{-34}$ js

$\mathrm{f}\mathrm{o}\mathrm{r}\mathrm{}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{a}\mathrm{l}\mathrm{}\mathrm{B}$ slope=h/e= $\frac{2.5}{(15-10){10}^{14}}$ = 8 $*{10}^{-34}$ js

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

Sol.

$\begin{array}{l}Ballofsaltisspherical\\ \therefore Volumeofball,V=\frac{4}{3}\pi r^3,wherer=radiusoftheball.\\ AsperQuestion,\frac{dV}{dt}\propto S,whereS=surfaceareaoftheball\\ \Rightarrow \frac{d}{dt}\left(\frac{4}{3}\pi r^3\right)\propto 4\pi r^2\left[?S=4\pi r^2\right]\\ \Rightarrow \frac{4}{3}\pi .3r^2.\frac{d}{dt}\propto 4\pi r^2\\ \Rightarrow 4\pi r^2.\frac{dr}{dt}={\rm K}.4\pi r^2\left[K=constantofproportionality\right]\\ \Rightarrow \frac{dr}{dt}={\rm K}.\frac{4\pi r^2}{4\pi r^2}\\ \Rightarrow \frac{dr}{dt}={\rm K}.1=K\\ Hence,theradiusoftheballisdecreasingatconstantrate.\end{array}$

During IME IIT Kanpur placements 2023 students were offered various job profiles, some of them are:

• Analytics Consultant

• Data Scientist

• Product Manager

• Project Manager

• Program Manager

• Operation Analyst, etc.

The placement rate for IME IIT Kanpur stood at 100% in 2023. The data tabulated below provides IME IIT Kanpur placement highlights for 2023:

To crack IME IIT Kanpur placement interview students can follow the steps given below:

·       Review your resume

·       Practice common interview questions

·       Showcase your strengths

·       Be confident and articulate

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

Sol:

$\begin{array}{l}Let\text{'}r\text{'}betheradiusofthesphere.\\ \therefore Surfaceareaofthesphere=4\pi r^2\\ Volumeofthesphere=\frac{4}{3}\pi r^3\\ Thesidesoftheparallelopipedarex,2xand\frac{x}{3}\\ \therefore Itssurfacearea=2\left[x\times 2x+2x\times \frac{x}{3}+x\times \frac{x}{3}\right]\\ =2\left[2x^2+\frac{2x^2}{3}+\frac{x^2}{3}\right]=2\left[2x^2+x^2\right]\\ =2\left[3x^2\right]=6x^2\\ Volumeoftheparallelopiped=x\times 2x\times \frac{x}{3}=\frac{2}{3}{x}^3\\ Aspertheconditionsofthequestion,\\ surfaceareaoftheparallelopiped+surfaceareaofthesphere=constant\\ \Rightarrow 6x^2+4\pi r^2=K\left(constant\right)\Rightarrow 4\pi r^2=K-6x^2\\ \therefore r^2=\frac{K-6x^2}{4\pi }\dots \left(i\right)\\ NowletV=Volumeoftheparallelopiped+Volumeofthesphere\\ \Rightarrow V=\frac{2}{3}{x}^3+\frac{4}{3}\pi r^3\\ \Rightarrow V=\frac{2}{3}{x}^3+\frac{4}{3}\pi {\left[\frac{K-6x^2}{4\pi }\right]}^{3/2}\left[Fromeq.\left(i\right)\right]\\ \Rightarrow \text{\&thins}\end{array}$

$\begin{array}{l}Squaringbothsides,weget\\ \Rightarrow 4\pi x^2=9\left(K-6x^2\right)\Rightarrow 4\pi x^2=9K-54x^2\\ \Rightarrow 4\pi x^2+54x^2=9K\\ \Rightarrow K=\frac{4\pi x^2+54x^2}{9}\left(ii\right)\\ \Rightarrow 2x^2\left(2\pi +27\right)=9K\\ \therefore x^2=\frac{9K}{2\left(2\pi +27\right)}=3\\ Nowfromeq.\left(i\right)wehave\\ r^2=\frac{K-6x^2}{4\pi }\\ \Rightarrow r^2=\frac{\frac{4\pi x^2+54x^2}{9}-6x^2}{4\pi }\\ \Rightarrow r^2=\frac{4\pi x^2+54x^2-54x^2}{9\times 4\pi }=\frac{4\pi x^2}{9\times 4\pi }=\frac{x^2}{9}\Rightarrow r=\frac{x}{3}\therefore x=3r\\ Nowwehave\frac{dV}{dx}=2x^2-\frac{3x}{}{\left(K-6x^2\right)}^{1/2}\end{array}$