• # question_answer A helical spring has N turns of coil of diameter D, and a second spring, made of same wire diameter and of same material, has N/2 turns of coil of diameter 2D. If the stiffness of the first spring is k, then the stiffness of the second spring will be: A) k/4                               B) k/2C) 2k                                D) 4k

${{n}_{1}}=N,$ ${{D}_{1}}=D,$ ${{d}_{1}}=d\,;\,\,{{d}_{2}}=d,$ ${{G}_{1}}={{G}_{2}},\,{{n}_{2}}=N/2,$ ${{D}_{2}}=2D$ Stiffness, k $=\frac{D{{d}^{4}}}{8{{D}^{3}}n}$ For same G and d, $k\propto \frac{1}{{{D}^{3}}n}$             $\frac{{{k}_{1}}}{{{k}_{2}}}={{\left( \frac{{{D}_{2}}}{{{D}_{1}}} \right)}^{3}}\times \left( \frac{{{n}_{2}}}{{{n}_{1}}} \right)$ $={{\left( \frac{2D}{D} \right)}^{3}}\times \left( \frac{N}{2N} \right)=4$ ${{k}_{2}}=\frac{k}{4}$