A long helical spring having a spring stiffness of 12 kN/m and number of turns 20, breaks into two parts with number of turns 10 each in both the parts. If the two parts are connected in series, then the softness of the resultant spring will be
A. 6 kN/m
B. 12 kN/m
C. 24 kN/m
D. 30 kN/m
Answer: Option B
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Each spring will have a spring stiffness of Nk ...now according to question its is cut into 2 parts then each vill have;
**** 2 x 12 = 24 kN/m,
Now if they are connected in series,
Resultant stiffnes is;
1/R = 1/R1 + 1/ R2.
r1 and r2 are same.
Solving we get R = 12 R= k.