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Amrita Vishwa Vidyapeetham has one of the best relaxed and student friendly curriculums among other Private Universities. The First Year hence had minor load with many labs, courses, tutorials, clubs and activities, though in Online Mode, I dont know how much of load will yet be prevalent. However, you must
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The volume of the right circular cone is V= (1/3)*Pi*r^2*h
The new increased height of the cone can be written as (120/100)*h
The new increased based radius of the cone can be written as (110/100)*r
Now the new volume is= (1/3)*Pi*(1.1r)^2*(1.2h) = 1.452*((Pi*r^2*h)/3) which is also equal to 1.452*V
Step-by-step explanation:
Let original radius of cone=R
And original height=H
Hence original volume V=(1/3)πR²H
New radius when reduced by10%=R-(10/100)R =0.9R
New height when increased by 11 1/9 % i.e. 100/9% =H+(100/9)R/100=10/9H
Hence new volume V=(1/3)(0.9R)²(10/9H) =(1/3)(0.81R²)(10/9H=(1/3)(0.90)R²H=0.90V
Reduction in volume =V-0.90V=0.10V
Percentage reduction =(0.10V/V)×100= 10 %
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Yes, you can apply for MA in philosophy in IIT bombay as it offers MA courses now. You need to clear its entrance exam and interview round to secure seat in it.
The minimum eligibility criteria for IIT MA is a bachelor's degree in any discipline(BTech/Bsc/Ba) after 10+2 from
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Banaras Hindu University has yet to release the application form for B.SC Mathematics hons. The admission into the course is based on BHU UET entrance exam. In the month of July 2021(tentative), Banaras Hindu University will issue the BHU UET 2021 application form. The deadline for completing and submitting
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