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In order to get admission at Pearl Academy Jaipur, candidates must meet the eligibility criteria set by the college. For admission candidates need to appear for entrance exam. Here is the Pearl Academy Jaipur detailed eligibility and selection criteria:

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

Sol:

$\begin{array}{l}\underset{x\to 0^{-}}{lim}f\left(x\right)=\frac{x}{\left|x\right|+2x^2}=\underset{h\to 0}{lim}\frac{0-h}{\left|0-h\right|+2{\left(0-h\right)}^2}\\ =\underset{h\to 0}{lim}\frac{-h}{h+2h^2}=\underset{h\to 0}{lim}\frac{-h}{h\left(1+2h\right)}\\ =\underset{h\to 0}{lim}\frac{-1}{1+2h}=\frac{-1}{1+2\left(0\right)}=-1\\ \underset{x\to 0^{+}}{lim}f\left(x\right)=\frac{x}{\left|x\right|+2x^2}=\underset{h\to 0}{lim}\frac{0+h}{\left|0+h\right|+2{\left(0+h\right)}^2}\\ =\underset{h\to 0}{lim}\frac{h}{h+2h^2}=\underset{h\to 0}{lim}\frac{h}{h\left(1+2h\right)}=\frac{1}{1+0}=1\\ As\underset{x\to 0^{-}}{lim}f\left(x\right)\ne \underset{x\to 0^{+}}{lim}f\left(x\right)\\ Hence,f\left(x\right)isdiscontinuousatx=0regardlessthechoiceofk.\end{array}$

This is a short answer type question as classified in NCERT Exemplar

When charge particle moves in a uniform magnetic field the path of particle is circular . so when it moves in circular path its radius Is also constant . so its kinetic energy also constant.

This is a short answer type question as classified in NCERT Exemplar

Average work done by a human heart per beat =0.5J

Total work done during 72 beats= 72 $\times 0.5$ =36J

Power = work done/time

=36/60=0.6W

Kindly go through the solution

Sol:

$\begin{array}{l}\underset{x\to 0^{-}}{lim}f\left(x\right)=\frac{1-coskx}{xsinx}=\underset{h\to 0}{lim}\frac{1-cosk\left(0-h\right)}{\left(0-h\right)sin\left(0-h\right)}\\ =\underset{h\to 0}{lim}\frac{1-cos\left(-kh\right)}{\left(-h\right)sin\left(-h\right)}=\underset{h\to 0}{lim}\frac{1-coskh}{hsinh}\left[\begin{array}{l}?sin\left(-\theta \right)=-sin\theta \\ cos\left(-\theta \right)=cos\theta \end{array}\right]\\ =\underset{h\to 0}{lim}\frac{2si{n}^2\frac{kh}{2}}{hsinh}=\underset{\begin{array}{l}h\to 0\\ kh\to 0\end{array}}{lim}\frac{2sin\frac{kh}{2}}{\frac{kh}{2}}\times \frac{kh}{2}\times \frac{sin\frac{kh}{2}}{\frac{kh}{2}}\times \frac{kh}{2}.\frac{1}{h.\frac{sinh}{h}.h}\\ =2.1.\frac{kh}{2}.1.\frac{kh}{2}.\frac{1}{h^2}.1\left[\begin{array}{l}\underset{h\to 0}{lim}\frac{sinh}{h}=1and\\ \underset{kh\to 0}{lim}\frac{sinkh}{kh}=1\end{array}\right]\\ =\frac{k^2}{2}\\ \underset{x\to 0}{lim}f\left(x\right)=\frac{1}{2}\\ \therefore \underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0}{lim}f\left(x\right)\\ \therefore \frac{k^2}{2}=\frac{1}{2}\Rightarrow k^2=1\Rightarrow k=\pm 1\\ Hence,thevalueofkis\pm 1.\end{array}$

The eligibility criteria for admission to the BTech course at ANA Group of Institutions are that candidates must complete their Class 12 from Science stream (PCM subjects) with a minimum aggregate of 50% (45% for SC/ ST). Additionally, for the admission process, the institute accepts valid scores from CUET UG/ JEE Mains. The institute also gives admissions for management quota seats. 

Yes, you can check your UK 12th result, if your UK 12th hall ticket is lost. You can check the UK 12th result from the school.
Hence, you need not worry, even if your UK 12th hall ticket is lost.

This is a short answer type question as classified in NCERT Exemplar

m =100kg

Height h= 10m

Power= rate of work done =change of energy /time= mgh/t

= $\frac{100\times 9.8\times 10}{20}=5\times 98=490W$

This is a short answer type question as classified in NCERT Exemplar

Total linear momentum of the system of two balls is always conserved. While balls are in contact, there may be deformation which means elastic PE which came from part of KE .But they are in contact while collision so according to above case kinetic energy will not conserved.

This is a short answer type question as classified in NCERT Exemplar

No, work done is not always zero in circular path it is zero only when all the forces are zero.

Bharti Vidya Peeth College Pune offers placement assistance to students. Bharati Vidyapeeth placements 2025 will be updated here once out. The university is visited by top companies such as Accenture, Reliance Industries Limited, Amazon,  and others. According to the BDVU Pune NIRF report 2024, the key highlights of Bharati Vidyapeeth placements 2023 as compared with 2022 are presented below:

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

Sol:

$\begin{array}{l}f\left(x\right)=\frac{2^{x+2}-16}{4^x-16}=\frac{2^2.2^x-16}{{\left(2^x\right)}^2-{\left(4\right)}^2}=\frac{4\left(2^x-4\right)}{\left(2^x-4\right)\left(2^x+4\right)}\\ =\frac{4}{\left(2^x+4\right)}\\ \underset{x\to 2^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}\frac{4}{\left(2^{2-h}+4\right)}=\frac{4}{2^2+4}=\frac{4}{4+4}=\frac{4}{8}=\frac{1}{2}\\ \underset{x\to 2}{lim}f\left(x\right)=k\\ Asthefunctioniscontinuousatx=2\\ \therefore \underset{x\to 2^{-}}{lim}f\left(x\right)=\underset{x\to 2}{lim}f\left(x\right)\\ \therefore k=\frac{1}{2}\\ Hence,thevalueofkis\frac{1}{2}.\end{array}$

This is a short answer type question as classified in NCERT Exemplar

No total mechanical energy of the body falling freely under gravity is not conserved because a small part of its energy is utilised against resistive force of air, which is non conservative force. Gain in KE< loss in PE.

Yes, Bharati Vidyapeeth is ranked by various ranking bodies. Being one of the elite deemed universities of India, Bharati Vidyapeeth has been ranked by several organisations in the past few years. Some of the top BVDU rankings are enlisted below:

Yes, SIT, Pune (Symbiosis Institute of Technology) accepts MHT? CET scores for B.Tech admission. You can also apply with JEE Main or the institute's own entrance exam (SITEEE). Just make sure you've completed 10+2 with Physics and Mathematics and at least 45% marks. Then apply through their admission process based on your entrance score.

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

Sol:

$\begin{array}{l}\underset{x\to 5}{lim}f\left(x\right)=3x-8=\underset{h\to 0}{lim}3\left(5-h\right)-8=15-8=7\\ \underset{x\to 5^{+}}{lim}f\left(x\right)=2k\\ Asthefunctioniscontinuousatx=5\\ \therefore \underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 5^{+}}{lim}f\left(x\right)\\ \therefore 7=2k\Rightarrow k=\frac{7}{2}\\ As\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1}{lim}f\left(x\right)\\ Hence,thevalueofkis\frac{7}{2}.\end{array}$

This is a short answer type question as classified in NCERT Exemplar

Force of gravity acts on the car vertically downward while car is moving along horizontal direction. So angle between them is 90. So work done is zero.

W= fscos90=0