1
Two springs A and B having spring constant KA and KB (KA = 2KB) are stretched by applying force of equal magnitude. If energy stored in spring A is EA, then energy stored in B will be
- 2EA
- EA4
- EA2
- 4EA
Answer
(1) 2EA
Solution
Same force F: extension x = FK
E = 12Kx2 = F22K, i.e. E 1K
EBEA = KAKB = 2
EB = 2EA
Answer: (1)
AIPMT 2001