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Yes, students must solve the PSEB Class 10 sample papers or previous years' question papers within the time provided as this will help them with time management. This practice will let them know which questions to answer first and those that can be attempted later. This way, students can attempt the questions they definitely know the answer to and score full marks. Thereafter, they can attempt other questions in the time left. 

After donwloading the admit card, if a candidate find some mistake or error in the JEST admit card they must immediately contact the exam authority get it corrected before the exam.

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

Sol.

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

PW Institute of Innovation Noida is a tier 2 college, established as a private institute, offering UG level courses with 3 other campuses. The Institute offers BBA, BVoc, and BTech courses in full-time mode with 100% placement assistance in affiliation with Medhavi Skills University. It offers scholarship facilities and various financial aids along with various opportunities for students to build industry standard business models. 

Indraprastha Institute of Information Technology, Delhi, accepts the JEE Main entrance exam, followed by JAC Delhi counselling for BTech admission. As per the IIIT Delhi JAC Delhi cutoff 2025, the Round 1 closing ranks varied between 7429 and 31355 for the students belonging to the General AI quota. Of all the courses, the most competitive was Computer Science and Economics. For IIIT Delhi branch-wise cutoff 2025 for Round 1, refer to the table below. The cutoff data provided below is for the general AI quota. 

Candidates must visit the official website of JEST and click on “Login, ” and fill in their registered email ID and password to access the download link of JEST admit card. Onlt those who completed the JEST Application Process will be eligible to access the admit card, a mandatory document to carry on the day of exam. 

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

Sol:

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

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

$\begin{array}{l}=12-\frac{3}{}\frac{\frac{4K\pi +54K-108K}{4\pi +54}}{}\\ =12-\frac{3}{}\left[\frac{\frac{4K\pi -54K}{4\pi +54}}{}\right]\\ =12-\frac{3}{}\left[\frac{4K\pi -54K}{.}\right]\\ =12-\frac{6}{}\left[\frac{2\pi -27}{.}\right]\\ =12+\frac{6K}{}\left[\frac{27-2\pi }{.}\right]>0\left[?27-2\pi >0\right]\\ \therefore \frac{d^{2}V}{d{x}^2}>0So,itisminima.\\ Hence,thesumofvolumeismin\end{array}$

According to the official data, BTech placement rate in recent past was between 85% and 98%. The MBA placement rate recorded during Amrita Vishwa Vidyapeetham placements over the past three years was 100%. 

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

Sol:

In 2024, 115 companies were part of Amrita Vishwa Vidyapeetham placement session. Refer to the table below to know more:

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

Sol: 

Amrita Vishwa Vidyapeetham MCA placement report is not released yet.  As per the official website, the overall highest package offered in recent past was 80.4 LPA. 

Self-study can be very effective if done properly. Make your own notes and revise them regularly. Use online resources, videos, and reference books for better understanding. However, if you have doubts, don't hesitate to ask teachers or take extra help from coaching if needed. Regular practice and smart study are the keys to success.

Top recruiters that visited during Amrita Vishwa Vidyapeetham placement in recent past are mentioned below:

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

Sol:

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