• # question_answer A length of 10 mm diameter steel wire is coiled to a close coiled helical spring having 8 coils of 75 mm mean diameter, and the spring has a stiffness k. If the same length of wire is coiled to 10 coils of 60 mm mean diameter, then the spring stiffness will be: A) k                                  B) 1.25 kC) 1.56 k              D) 1.95 k

Correct Answer: C

Solution :

d=10 mm, ${{n}_{1}}=8,$ ${{D}_{ml}}=75\,mm$ ${{n}_{2}}=10,$ ${{D}_{m2}}=60\,mm$ Spring stiffness, $k=\frac{G{{d}^{4}}}{8D_{m}^{3}n}$ $\therefore \,\,k\propto \frac{1}{D_{m}^{3}n}$ $\frac{{{k}_{2}}}{{{k}_{1}}}=\frac{D_{m1}^{3}{{n}_{1}}}{D_{m2}^{3}{{n}_{2}}}={{\left( \frac{75}{60} \right)}^{3}}\times \frac{8}{10}$ ${{k}_{2}}=1.5625\,\,k$

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