A spring of force constant fr is cut into lengths of ratio 1: 2: 3. They are connected in series and the new force constant is k'. Then they are connected in parallel and force constant is k" . Then k' : k" is :

A
1 : 9
B
1 :11
C
1 :14
D
1 :16

Solution:

Length of the spring segments $= \frac{l}{6}, \frac{l}{3}, \frac{l}{2}$
As we know $k \propto \frac{1}{l}$
So, spring constant for spring segments will be
$k_{1} = 6 k, k_{2} = 3 k, k_{3} = 2 k$
So in parallel combination
$k " = k_{1} + k_{2} + k_{3} = 11 k$
In series combination
$k' = k$ (As it will become original spring)
So, $k' : k " = 1 : 11$

Solution:

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