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Bharati Vidyapeeth Deemed University Pune (BVDU), through 23 constituent colleges, schools of study, and institutes, offers undergraduate, postgraduate, and various other courses to students. The below-mentioned are Bharati Vidyapeeth constituent colleges, schools of study, and the institutes:

Emversity School of Hospitality Faridabad maintains a good network of recruiters who help the institute in placing students. Moreover, as per student reviews, BVoc students have been recruited by the following companies in the past:

• Accord Hospital
• Agilus Lab
• Dr. Lal Path Labs

Note: The above information is sourced from the student reviews on Shiksha. Hence, it is subject to change.

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

Sol:

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

Yes, Bharati Vidyapeeth has collaborated with various organisations, colleges, and industries. The university has signed MoUs with several organisations and companies. The below-mentioned are Bharati Vidyapeeth Institute of Environment Education & Research collaborations:

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

(a) Force is applied on the body to lift it in the upward direction and displacement of the body is also in upward direction. So angle is zero.

Work done = fscos $\theta$ = fs

W= positive

(b) But the gravitational force in downward direction and displacement in upward direction so angle is 180. So work done will be negative. W= fscos 180= -fs

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

Sol:

$\begin{array}{l}\underset{x\to 1^{-}}{lim}f\left(x\right)=\frac{x^2}{2}=\underset{h\to 0}{lim}\frac{{\left(1-h\right)}^2}{2}=\frac{1}{2}\\ \underset{x\to 1}{lim}f\left(x\right)=\frac{x^2}{2}=\frac{{\left(1\right)}^2}{2}=\frac{1}{2}\\ \underset{x\to 1^{+}}{lim}f\left(x\right)=2x^2-3x+\frac{3}{2}=2{\left(1\right)}^2-3\left(1\right)+\frac{3}{2}=2-3+\frac{3}{2}=\frac{1}{2}\\ As\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1}{lim}f\left(x\right)=\frac{1}{2}\\ Hence,f\left(x\right)iscontinuousatx=1.\end{array}$

As per student reviews present on Shiksha, Emversity School of Hospitality Faridabad offers good placements for the BVoc course. According to the reviews, 100% of students get placed during the on-campus placements. Some of the companies that have recruited BVoc students include Accord Hospital, Dr. Lal Path Lab, Agilus Lab, etc.

Note: The above information is sourced from the student reviews on Shiksha. Hence, it is subject to change.

No, you can not get direct admission to Bharati Vidyapeeth. The eligibility criteria for courses are different and are mandatory for students to fulfil to get admission. 

Apart from the 100% scholarship, students scoring 95% or above in Class 12 can get a 50% tuition waiver. Those scoring between 90–95% receive 40%, and 85–90% marks get 25% off. These scholarships apply to eligible students enrolled in BA programmes at Faculty of Liberal Arts, MIT WPU.

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

When the elevator is descending then electric power is required to prevent it from falling freely under gravity.

Also as the weight inside the elevator increases, its speed of descending increases, therefore there should be a limit on the number of passengers in the elevator to prevent the elevator from descending with a large velocity.

So when we increase the weight it fall with more speed . that is why there is limit in elevator.

Yes, ANA Group of Institutions does offer BTech courses in full-time mode. The institute offers five specialisations in BTech courses: Computer Science Engineering with Data Science & AI, Civil Engineering, Electronics and Communication Engineering, Mechanical Engineering, and Electrical Engineering. Moreover, the basic eligibility criteria to get admission into any of the courses are similar for every specialisation i.e., candidates must complete their Class 12 from the PCM stream with a minimum aggregate of 50%.

Bharti Vidya Peeth University Pune popular programmes are BSc,  MBA, and BTech. BTech students are given technical knowledge. Whereas MBA students are taught presentation skills, managerial skills,  and various other skills. BTech and MBA students are placed in top companies such as TCS, Cognizant, HCL, Infosys, IBM,  and others. BTech students can become Software Engineers, Data Analysts, Data Engineers,  or can go for various other job profiles. BTech students can also go for non-technical fields such as Marketing, Sales,  and others.

To qualify for the Dr. Vishwanath Karad Scholarship (100% fee waiver), MIT WPU BA Psychology students must have scored at least 92% in Class 10 and 85% in Class 12. For other BA specialisations, the Class 10 requirement is 90%, with the same Class 12 minimum of 85%.

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

Sol:

$\begin{array}{l}\underset{x\to 0^{-}}{lim}f\left(x\right)=\frac{e^{1/x}}{1+e^{1/x}}\\ =\underset{h\to 0}{lim}\frac{e^{\frac{1}{0-h}}}{1+e^{\frac{1}{0-h}}}=\underset{h\to 0}{lim}\frac{e^{-1/h}}{1+e^{-1/h}}\\ =\underset{h\to 0}{lim}\frac{1}{e^{1/h}\left(1-e^{-1/h}\right)}=\underset{h\to 0}{lim}\frac{1}{e^{1/h}-1}=\underset{h\to 0}{lim}\frac{1}{e^{1/0}-1}\\ =\underset{h\to 0}{lim}\frac{1}{e^{\infty }-1}=\underset{h\to 0}{lim}\frac{1}{0-1}=-1\left[?e^{\infty }=0\right]\\ \underset{x\to 0^{+}}{lim}f\left(x\right)=\frac{e^{1/x}}{1+e^{1/x}}=\underset{h\to 0}{lim}\frac{e^{\frac{1}{0+h}}}{1+e^{\frac{1}{0+h}}}=\underset{h\to 0}{lim}\frac{e^{1/h}}{1+e^{1/h}}\\ =\underset{h\to 0}{lim}\frac{1}{e^{-1/h}\left(1+e^{1/h}\right)}=\underset{h\to 0}{lim}\frac{1}{e^{-1/h}+1}=\underset{h\to 0}{lim}\frac{1}{e^{-1/0}+1}\\ =\underset{h\to 0}{lim}\frac{1}{e^{-\infty }+1}=\underset{h\to 0}{lim}\frac{1}{0+1}=1\left[?e^{-\infty }=0\right]\\ \underset{x\to 0}{lim}f\left(x\right)=0\\ As\underset{x\to 0^{-}}{lim}f\left(x\right)\ne \underset{x\to 0^{+}}{lim}f\left(x\right)\ne \underset{x\to 0}{lim}f\left(x\right)\\ Hence,f\left(x\right)isdiscontinuousatx=0.\end{array}$

No, only a passport is accepted for IELTS registration. Aadhaar, PAN, or other Indian ID cards are not valid for booking the IELTS exam.

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

As the block M is at rest and frictional force =Mgsin $\theta$

The force of friction acting between the blocks and incline opposes the tendency of sliding of the block . since block not in motion therefore no work is done. Hence no dissipation of energy.

Yes, admissions at Bharati Vidyapeeth are easy for hardworking students. Candidates must know their Bharati Vidyapeeth Pune eligibility criteria and apply accordingly. Follow the Bharati University Pune application process mentioned below:

Step 1: Visit the official website of Bharati Vidyapeeth.

Step 2: Apply for the accepted entrance exam.

Step 3: Appear with valid scores.

Step 4: Take part in the admission process.

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

Sol:

$\begin{array}{l}\underset{x\to a^{-}}{lim}f\left(x\right)=\left|x-a\right|sin\frac{1}{x-a}\\ =\underset{h\to 0}{lim}\left|a-h-a\right|sin\frac{1}{\left(a-h-a\right)}=\underset{h\to 0}{lim}h.sin\frac{1}{-h}\\ =\underset{h\to 0}{lim}-h.sin\frac{1}{h}\left[?sin\left(-\theta \right)=-sin\theta \right]\\ =0\times \left[anumberoscillatebetween-1and1\right]=0\\ \underset{x\to a^{+}}{lim}f\left(x\right)=\left|x-a\right|sin\frac{1}{x-a}=\underset{h\to 0}{lim}\left|a+h-a\right|.sin\frac{1}{\left(a+h-a\right)}=\underset{h\to 0}{lim}h.sin\frac{1}{h}\\ =0\times \left[anumberoscillatebetween-1and1\right]=0\\ \underset{x\to a}{lim}f\left(x\right)=0\\ As\underset{x\to a^{-}}{lim}f\left(x\right)=\underset{x\to a^{+}}{lim}f\left(x\right)=\underset{x\to a}{lim}f\left(x\right)=0\\ Hence,f\left(x\right)iscontinuousatx=a.\end{array}$