A spring is made of a wire of 2 mm diameter having a shear modulus of 80 GPa. The mean coil diameter is 20 mm and the number of active coils is 10. If the mean coil diameter is reduced to 10 mm, the stiffness of the spring is:

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BHEL ET Mechanical Held on May 2019

Option 3 : Increased by 8 times

SSC JE ME Full Test 4

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200 Questions
200 Marks
120 Mins

__Calculation:__

**Stiffness of the spring is given by:**

\(K = \frac{{Gd}}{{8{C^3}N}}\)

where d = Spring wire diameter, D = Mean diameter of the coil, G = Modulus of rigidity, N = Number of active coils

\(C = Spring~index = \frac{D}{d}\)

__Calculation:__

__Given:__

d_{1} = 2 mm, D_{1} = 20 mm

\({C_1} = \frac{{{D_1}}}{{{d_1}}} = \frac{{20}}{2} = 10\)

d_{2} = 2 mm, D_{2} = 10 mm

\({C_2} = \frac{{{D_2}}}{{{d_2}}} = \frac{{10}}{2} = 5\)

As the wire diameter and the number of active coils are constant. Therefore stiffness index varies as,

\(K \propto \frac{1}{{{C^3}}}\)

\(\frac{{{K_1}}}{{{K_2}}} = {\left( {\frac{{{C_2}}}{{{C_1}}}} \right)^3} = {\left( {\frac{5}{{10}}} \right)^3}\)

**∴ K _{2 } = 8K_{1}**