Railways NTPC (Technical Ability) Engineering Mechanics and Strength of Materials Question Bank Strength of Materials

  • question_answer
    A long helical spring having a spring-stiffness of 12 kN/m and number of turns 20, breaks into two equal parts. The resultant spring stiffness will be:

    A) 6 kN/m             

    B) 12 kN/m

    C) 24 kN/m                       

    D) 30 kN/m

    Correct Answer: C

    Solution :

    Spring stiffness, \[k=\frac{G{{d}^{4}}}{8D_{m}^{3}n}\] \[\therefore \,\,\,\,\,\,k\propto \frac{1}{n}\] \[{{k}_{2}}=2{{k}_{1}}=24\,\text{N/M}\]   


You need to login to perform this action.
You will be redirected in 3 sec spinner