Correct Answer - Option 1 :
\(\frac{1}{8}\)
Concept:
For closed coil helical spring,
\(Deflection\;under\;load,\;δ = \frac{{8W{D^3}n}}{{G{d^4}}}\)
Where, W = Load, D = Mean diameter of coil spring, n = Number of turns, G = Modulus of elasticity and d = Wire diameter of coil
From the above formula, it is clear that keeping all other parameters same,
Deflection, δ ∝ D3
Calculation:
Given:
\(\frac{D_A}{D_B} = \frac{1}{2}\),
∵ Deflection, δ ∝ D3
∴ \(\frac{{{δ _A}}}{{{δ _B}}} = (\frac{{{D_A}}}{{{D_B}}})^3\)
\(\frac{{{δ _A}}}{{{δ _B}}} = (\frac{1}{2})^3\)
\(\frac{{{δ _A}}}{{{δ _B}}} = \frac{1}{8}\)
Hence the ratio of deflection in spring A to spring B will be \(\frac{1}{8}\)