Correct Answer - Option 2 :
\(\frac{{64\;W{R^3}n}}{{C{d^4}}}\)
Explanation:
The applied axial load on a spring try to twist it due to which twisting train energy is stored in spring and it is given by:
\(U = \frac{{{T^2}L}}{{2CJ}}\)
Where,
T (Torque) = WR
L is the length of spring = distance covered in one turn × no of turns
L = 2πRn
Assuming closed coil helical spring has circular cross-section, its polar moment of inertia is given by:
\(J = \;\frac{{\pi {d^4}}}{{32}}\)
Now
\(U = \frac{{{{\left( {WR} \right)}^2}\left( {2\pi Rn} \right)}}{{\frac{{2C\pi {d^4}}}{{32}}}}\)
\(U = \frac{{{\rm{\;}}32{{\rm{W}}^2}{{\rm{R}}^3}{\rm{n}}}}{{{\rm{\;C}}{{\rm{d}}^4}}}\)
As per Castigilanos Theorem II,
The partial derivative of strain energy stored in a material w.r.t applied axial load at given point gives the deflection (δ) at that point
\(\delta = \;\frac{{\partial U}}{{\partial W}} = \frac{{\;64{\bf{W}}{{\bf{R}}^3}{\bf{n}}}}{{\;{\bf{C}}{{\bf{d}}^4}}}\;\;\)
Trick:
The deflection has the dimension of length i.e. mm or cm and out all the given options only option ‘2’ has the dimension of length. Hence, it would be the right answer as it is dimensionally correct.
