Question 26.3.46: Wire diameter, mean coil diameter and number of turns of a c...

(a)

EXPLANATION

We know that stiffness of a close-coiled helical spring is given by

k=\frac{C d^4}{64 R^3 \times n}

1st case,        k_1=\frac{C d^4}{64\left(\frac{D}{2}\right)^3 \times N} \quad\left(\text { here } R=\frac{D}{2} \text { and } n=N\right)

2nd case,        k_2=\frac{C(2 d)^4}{64\left(\frac{2 D}{2}\right)^3 \times 2 N}

(here dia. of wire = 2d ; Mean coil dia. = 2D and number of turns = 2N)

=\frac{C d^4 \times 16}{64 \times\left(\frac{D}{2}\right)^3 \times 8 \times 2 N}=\frac{C d^4}{64\left(\frac{D}{2}\right) \times N}= \pmb{k _1} \quad \quad \left\lgroup\because \quad k_1=\frac{C d^4}{\left.64\left(\frac{D}{2}\right)^3 \times N\right)}\right\rgroup