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hi aspirant,
F = G * M * m / r*r
where F represents the force in Newtons, M and m represent the two masses in kilograms, and r represents the separation in meters. G represents the gravitational constant, which has a value of 6.674⋅10−11N(m/kg)26.674⋅10−11N(m/kg)2 .
Hope this help you
Hello aspirant
Top universities to study BSC in physics ( with fees for first year)
Loyola college, chennai,
St. Xavier's college, Mumbai ,
Fees 7187
Christ University, Bangalore
35,00
Madras Christian college,
Chandigarh University
60,000
Stella college
22,895
Ramjas college, new Delhi
14,610
LPU
53,200
Hello,
In 2019 the All India cut off rank for admission to Computer Science Engineering in DTU for general category was 5896 in round 1 and the cut off in round 4 was 8852 under Delhi region. You need to score minimum 99 percentile and score minimum 180-200 marks in
As you have mentioned that you have secured 6648th rank in KCET. Firstly, know that it is a very good rank and you may get the branch of your desire in any top college in Karnataka. However, talking about placements there are many colleges whose placement is good. Also
Hi Aspirant,
As per your rank as per previous year cutoff, the government/public colleges you will get are:-
University Visvesvaraya College of Engineering
Government Sri Krishnarajendra Silver Jubilee Technological Institute
PES College of Engineering, Karnataka
Dr. Ambedkar Institute Of Technology
MCE - Malnad College of Engineering
University BDT College of
Hello
Please refer to the links below
https://www.careers360.com/university/maulana-azad-national-institute-of-technology-bhopal
https://www.careers360.com/university/school-of-planning-and-architecture-delhi
Hope it helps!
Hi Ojaswini,
As per your rank and as per previous year cutoff, the colleges you will get in chemical engineering are:-
MVJ College of Engineering, Bangalore
Bheemanna Khandre Institute of Technology, Bhalki
BMS College of Engineering - [BMSCE], Bangalore last year cutoff rank for Chemical engineering is 8761. So, you
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