A block of mass 2Kg is attached to the lower end of a spring of force constant 1960 Nm−1 hanging from rigid support. Initially spring is unstretched, if the block is suddenly released the maximum elongation produced in spring is
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
1cm
b
2cm
c
2.2cm
d
3cm
answer is B.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Loos of P.E= gain of elastic potential energymgx=12kx2 x=2mgk=2×2×9.81960=2100m x=2cm