NEET Solved Paper 2019 Question 41
Question: When a block of mass M is suspended by a long wire of length L, the length of the wire becomes $ (L+l) $ . The elastic potential energy stored in the extended wire is- [NEET 5-5-2019]
Options:
A) $ \frac{1}{2}Mgl $
B) $ \frac{1}{2}MgL $
C) $ Mgl $
D) $ MgL $
Show Answer
Answer:
Correct Answer: A
Solution:
Loss in gravitational P.E. = $ Mgl $ Elastic potential energy $ U=\frac{1}{2}Mgl $
