A mass of M kg is suspended by a weight less string the horizontal force that requires to displace it until the string making an angle 45° with the initial vertical direction 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
Mg(2−1)
b
Mg(2+1)
c
Mg2
d
Mg2
answer is A.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
work done by horizontal force is W= F×ℓ sin45° Work done by horizontal force =Fℓ2---------(1) gain in PE of pendulum = Mgh PE=Mg(ℓ−ℓ cos45°) PE=Mgℓ(1-12) PE =Mgℓ (2−12)-----------(2) substitute equation (1) in (2) ∴ Fℓ2 = Mgℓ(2−12) ⇒ F = Mg (2−1)