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In India, there are many universities that provide good placement opportunities. There are many private universities that provide good placement offers to their students. Placement statistics usually showcase things like the highest CTC offers, placement rate, and top recruiters who participated inRead more
In India, there are many universities that provide good placement opportunities. There are many private universities that provide good placement offers to their students.
Placement statistics usually showcase things like the highest CTC offers, placement rate, and top recruiters who participated in the placement drive of the universities.
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Students are mostly attracted to those universities that provide high scholarship placements. As the admission season approaches, universities that provide a high placement percentage with jobs in renowned organizations attract more applicants.
But, there is high competition in the private universities in India that offer the best placements. Here is a list of top universities that provide high placements in top companies.
Which university provides good placements in India?
1. Chandigarh University
This is one of the premier private universities in India, which is located in Chandigarh. It was founded in 2012, but in a short period of time, this university has established a good track record in terms of placement criteria.
Students from this university have been placed in Fortune 500 companies, which has provided a CTC of more than 35 lacs PA. Students were placed in companies like Amazon, Microsoft, IBM, etc.
2. VIT University
Vellore Institute of Technology is one of the best technical universities in India which has good placement records.
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This can be seen from the fact that students in the 20–21 batch received about 4703 job offers. The university is also ranked in 28th position by the NIRF in the overall management and engineering categories.
3. Amity University
Amity University is also one of the top-ranked universities, which is located in Noida, Uttar Pradesh, and was established in 1986. The university has been conducting placements for many years.
The university has also received participation from various famous companies. Of those, 1706 students were offered plum jobs in good sectors of the industry. The highest package was around 30 lacs PA, while the average was about 4.5 lacs.
4. Sastra University
Shanmugha Arts, Science, Technology, and Research Academy University is a deemed university and also one of the top universities in all of Tamil Nadu.
In recent placement drives, 2400 Sastra university students got placed, and many of them got internship offers from GGO-D companies. Many students got good job packages, while the top one was 9.5 LPA.
5. Manipal University
This is one of the top companies that provide high placements and is located in Karnataka, where it was established in 1953. This university also provides UG and PG research courses like B.Tech, MCA, B.arch, and many more.
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In the placement season of 2020, students received a good amount of 22.3 LPA. Top recruiters like Amazon, Bosch, Microsoft, and Samsung
6. Thapar University
Thapar University is an engineering and technology institute, which is ranked as the 29th best engineering institution in the NIRF ranking of 2020.
This university has invited more than 250 recruiters this year, and the B.tech program records the highest number of placements as compared to other programs. The university has a placement record of 90%, in which the highest salary reached UNR 43 LPA.
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These are some of the best universities that provide high placements, but it also depends upon the competition and latest trends of that time.
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You will observe questions like find the position of the final image of the 1.0-cm-tall object in subjects such as Physics and science. You have to understand the question description properly first before attempting to solve the same. We have to solve it as per the Physics formulae that we use forRead more
You will observe questions like find the position of the final image of the 1.0-cm-tall object in subjects such as Physics and science. You have to understand the question description properly first before attempting to solve the same.
We have to solve it as per the Physics formulae that we use for while calculating values of mirror or lens.
(Image Source: https://www.physicsforums.com/threads/double-lens-system-converging-diverging.805553/)
Solved: Find the position of the final image of the 1.0-cm-tall object (Using Formula)
You have to put the values as follows:
Here, we have to find the value of di first of the 1st lens.
Substitution the values of each digit in equation – (i), we get the following figure:
By solving the above equation, we get di (distance) = -20-cm (Because it is located to the left side of 1st lens)
Now, adding the distance between lenses, we have to go through the same formula for the 2nd lens as well.
Now, using the same formula, we can put values as follows:
If you calculate the equation, you will get di= -6.4-cm.
Thus, the final image distance is -6.4-cm located to the left of the 2nd lens.
When we use magnification formula, -V1/ u1 * -V2/ u2
Image Height= M X Height of the Object
It means the image so formed will be erect.
How to Find the position of the final image of the 2.0-cm-tall object.
Reference: A 2.0-cm tall object is positioned 40-cm away from a diverging lens with a focal length of 15 cm. Find the position and size of the image.
If we go by the above formula, we have to calculate it as follows:
We know that Concave Lens = Diverging Lens
Here, Focal Length= 15-Cm
The distance of the object from the lens= 40-cm
Height of the object= 2.0-cm
We have to find the position of the image along with its height of the object.
Going by the lens formula, we know:
By putting the values in above formula, we get the following:
It means the image distance from the lens will be 10.9-cm. The use of negative sign before the distance means it will be formed to the left side or behind the lens.
For further analysis, we can also use magnification formula as here:
Putting the values in their respective positions, we get;
Solving the above sum; we get
It means the height of the image is + 0.54. And the image thus obtained will be erect and virtual.
Thus, the image distance from the lens is 10.90 cm, and the negative sign implies that it is formed behind the lens (on the left side).
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