Questions
A spring of force constant 10 N/m has initial stretch 0.2 m. In changing the stretch to 0.25 m, the increase of PE is about :
a
0. 1 joule
b
0.2 joule
c
0.3 joule
d
0.5 joule
detailed solution
Correct option is A
Increase in PE = =12kx22−kx12=12(10)0.252−0.22=0.1125JTalk to our academic expert!
Similar Questions
A simple pendulum is swinging in a vertical plane. The ratio of its potential energies when it is making angles and with the vertical is,
799 666 8865
support@infinitylearn.com
6th Floor, NCC Building, Durgamma Cheruvu Road, Vittal Rao Nagar, HITEC City, Hyderabad, Telangana 500081.
free study material
JEE
JEE Mock Tests
JEE study guide
JEE Revision Notes
JEE Important Questions
JEE Sample Papers
JEE Previous Year's Papers
NEET
NEET previous year’s papers
NEET important questions
NEET sample papers
NEET revision notes
NEET study guide
NEET mock tests
Popular Books
HC Verma Part 1HC Verma Part 2RS AgarwalRD SharmaLakhmir SinghDK GoelSandeep GargTS GrewalNCERT solutions
Class 12 NCERT SolutionsClass 11 NCERT SolutionsClass 10 NCERT SolutionsClass 9 NCERT SolutionsClass 8 NCERT SolutionsClass 7 NCERT SolutionsClass 6 NCERT Solutions
