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Question

A simple spring has length l and force constant k, is cut into two springs of length l1 and l1=nl2 (n= an integer). The force constant of spring of length l1 is:
  1. k(1+n)
  2. kn(1+n)
  3. kn+1
  4. k

A
kn(1+n)
B
kn+1
C
k(1+n)
D
k
Solution
Verified by Toppr

If l is cut into l1 and l2
l1+l2=l
nl2=l1

l2=ln+1;l1=nln+1
As we know the relationship of elasticity K=EAl
K1=EAl1=(n+1)EAnl=(n+1)Kn

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