Question

Two co-axial springs are subjected to a force of 1 kN. Spring constant of larger diameter spring is 80 N/mm and that of smaller diameter spring is 120 N/mm. The deformation in the spring combination will be equal to

• A. 5 mm
• B. 15 mm
• C. $\frac{125}{6}mm$
• D. $\frac{135}{7}mm$

Answer 1

The correct option is A 5 mm

Coaxial springs undergo same deformation so they behave as parallel springs. The spring constant of co-axial springs is

$k=k_1+k_2$

$=80+120=200$

$So\Delta =\frac{1000}{200}=5mm$