Use app×
QUIZARD
QUIZARD
JEE MAIN 2026 Crash Course
NEET 2026 Crash Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
172 views
in Physics by (91.2k points)
closed by
A force of `(2500 +- 5)`N is applied over an area of `(0.32 +- 0.02)m^2` Calculate the pressure exerted over the area.

1 Answer

0 votes
by (91.8k points)
selected by
 
Best answer
Here, `F =(2500 +- 5)N`
`A = (0.32 +- 0.02)m^2, P= ?`
`P = (F)/(A) = (2500)/(0.32) = 7812.5N//m^2`
Now, `(DeltaP)/(P) = +-((DeltaF)/(F) +(DeltaA)/(A)) = +- ((5)/(2500)+(0.02)/(0.32)) = +- (0.002 + 0.0625) = +- 0.0645`
`DeltaP =+- 0.0645 P = +- 0.0645xx7812.5 = +- 503.9N//m^2`

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...