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Yes, admissions are ongoing for UG and PG courses at Backstage Pass Institute of Gaming and Technology. Students can apply directly on the college's official website and fill the application form. However, they can also contact at the college's campus to enquire more about the course and the college.

A part of TMUIC values is inclusivity and Diversity, so students from all over the world apply here. This university offers the pathway programme to Toronto Metropolitan University. 

While the data for Toronto Metropolitan University International College is published, according to some unofficial sources, this university currently hosts 2,000 international students from 140+ countries.

National Institute of Technology Hamirpur offers UG course across various streams such as Engineering, Architecture, and others. The fee strcture of UG programmes has been discussed below: 

Yes, direct admission is valid to some specific courses at Backstage Pass Institute of Gaming and Technology. Students can apply for direct admission to PG and Diploma courses at the college. However, admission to UG courses is based on college-level entrance exam score.

National Institute of Technology Hamirpur admission 2026 is about to open. The instiute offers admisison based on merit and entrance exams. The institute accepts national entrance exams such as CAT, MAT, XAT, JEE Main, and others. Candidates willing to get admission to several courses at the National Institute of Technology Hamirpur must apply online on the official website of the university. 

BBA (Hons) is an undergraduate programme and the duration of this programme is 4 years including an internship program within 7.5 months. Students will be able to acquire numerous skills including Marketing, HR, Operations Management, Economics, Accounting, Finance along with hands-on training of analytical tools. Students must submit the following documents along with the application during the admission process.

• Academic transcripts
• Valid copy of passport
• English Proficiency Scores

Explanation- velocity of a freely falling body is v= $\sqrt[]{2gh}$

And $?=\frac{h}{mv}=\frac{h}{m\sqrt[]{2gh}}$

$?=h$ ^-1

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

Sol:

$\begin{array}{l}Letareaofthefirstsquare{A}_1=x^2\\ andareaofthesecondsquare{A}_2=y^2\\ Now{A}_1=x^{2}and{A}_2=y^2={\left(x-x^2\right)}^2\\ Differentiatingboth{A}_{1}and{A}_{2}w.r.t.t,weget\\ \frac{d{A}_1}{dt}=2x.\frac{dx}{dt}and\frac{d{A}_2}{dt}=2\left(x-x^2\right).\left(1-2x\right).\frac{dx}{dt}\\ \therefore \frac{d{A}_2}{d{A}_1}=\frac{\frac{d{A}_2}{dt}}{\frac{d{A}_1}{dt}}=\frac{2\left(x-x^2\right).\left(1-2x\right).\frac{dx}{dt}}{2x.\frac{dx}{dt}}\\ =\frac{x\left(1-x\right)\left(1-2x\right)}{x}=\left(1-x\right)\left(1-2x\right)\\ =1-2x-x+2x^2=2x^2-3x+1\\ Hence,therateofchangeofareaofthesecondsquarewithrespecttofirstis2x^2-3x+1.\end{array}$

Backstage Pass Institute of Gaming and Technology applications are accepted in online mode. The basic steps to consider during Backstage Pass Institute of Gaming and Technology application process are as follows:

1. Visit the official website at backstagepass.co.in.
2. Tap on the "Application Form" appears on the Home Page.
3. Fill the form with details around personal and academic background.
4. Complete the form and press the "submit" button.

BITS Pilani Dubai offers various programs under five majors. These are as follows:

• Business and Management Studies
• IT and Software
• Architecture and Planning
• Engineering
• Science

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

Sol.

$\begin{array}{l}Letxbethelengthofthecube\\ \therefore VolumeofthecubeV=x^3\dots \left(i\right)\\ Giventhat\frac{dV}{dt}=K\\ Differentiatingeq\left(i\right)w.r.t.t,weget\\ \frac{dV}{dt}=3x^2.\frac{dx}{dt}=K\left(constant\right)\\ \therefore \frac{dx}{dt}=\frac{K}{3x^2}\\ Nowsurface\text{?}\text{?}areaofthecube,S=6x^2\\ Differentiatingbothsidesw.r.t.t,weget\\ \frac{dS}{dt}=6.2.x.\frac{dx}{dt}=12x.\frac{K}{3x^2}\\ \Rightarrow \frac{dS}{dt}=\frac{4K}{x}\Rightarrow \frac{dS}{dt}\propto \frac{1}{x}\left(4K=constant\right)\\ Hence,surface\text{?}\text{?}areaofthecubevariesinverselyasthelengthoftheside.\end{array}$

National Institute of Technology Hamirpur has affordable fee structure. The fee structure of the programme ranges between INR 1.41 lakh- INR 9.61 lakh. The fee structure consists of various components, namely tuition, security deposit, and hostel fee, etc. Among these components, some are to be paid only once during NIT Hamirpur admission, while the remaining are charged according to the fee guidelines. 

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

Sol.

$\begin{array}{l}GiventhatL=200{\left(10-t\right)}^2\\ whereLrepresentsthenumberoflitresofwaterinthepool.\\ Differentiatingbothsidesw.r.t.,t,weget\\ \frac{dL}{dt}=200\times 2\left(10-t\right)\left(-1\right)=-400\left(10-t\right)\\ Buttherateatwhichthewaterisrunningout\\ =-\frac{dL}{dt}=400\left(10-t\right)\dots \left(i\right)\\ Rateatwhichthewaterisrunningafter5seconds\\ =400\left(10-5\right)=2000L/s\left(finalrate\right)\\ Forinitialrateputt=0\\ =400\left(10-0\right)=4000L/s\\ Theaveragerateatwhichthewaterisrunningout\\ =\frac{Initialrate+Finalrate}{2}=\frac{4000+2000}{2}=\frac{6000}{2}=3000L/s\\ Hence,therequiredrate=3000L/s.\end{array}$

The key highlights recorded during the IME IIT Kanpur MBA placements are provided in the table below:

 MBA graduates from IME IIT Kanpur get placed into various sectors such as:

• General Management

• Management Consultant

• Investment Banking

• Business Consultant

• Project Management, etc. 

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

Sol: 

$\begin{array}{l}Internalradiusr=3cm\\ ExternalradiusR=r+\Delta r=3.0005cm\\ \therefore \Delta r=3.0005-3=0.0005cm\\ Lety=r^3\Rightarrow y+\Delta y={\left(r+\Delta r\right)}^3=R^3={\left(3.0005\right)}^3\dots \left(i\right)\\ Differentiatingbothsidesw.r.t.,r,weget\\ \frac{dy}{dx}=3r^2\\ \therefore \Delta y=\left(\frac{dy}{dr}\right).\Delta r=3r^2\times 0.0005\\ =3\times {\left(3\right)}^2\times 0.0005=27\times 0.0005=0.0135\\ \therefore {\left(3.0005\right)}^3=y+\Delta y\left[Fromeqn\left(i\right)\right]\\ ={\left(3\right)}^3+0.0135=27+0.0135=27.0135\\ Volumeoftheshell=\frac{4}{3}\pi \left[R^3-r^3\right]\\ =\frac{4}{3}\pi \left[27.0135-27\right]=\frac{4}{3}\pi \times 0.0135\\ =4\pi \times 0.0045=4\times 3.14\times 0.0045=0.018\pi c{m}^3\\ Hence,approximateVolumeofthemetalintheshellis0.018\pi c{m}^3.\end{array}$