A mass is suspended at the bottom of two springs in series having stiffness 10 N/mm and 5 N/mm. The equivalent spring stiffness of the two Springs is nearly

A mass is suspended at the bottom of two springs in series having stif

1.	A mass is suspended at the bottom of two springs in series having stiffness 10 N/mm and 5 N/mm. The equivalent spring stiffness of the two Springs is nearly
A.	0.3 N/mm
B.	3.3 N/mm
C.	5 N/mm
D.	15 N/mm
Answer» C. 5 N/mm

Discussion

No Comment Found

Related MCQs

Active and inactive, both types of coils support the load although both don’t participate in spring action.
The helix angle is very small about 2⁰. The spring is open coiled spring.
A closely-coiled helical spring is made of 10 mm diameter steel wire, with the coil consisting of 10 turns with a mean diameter 120 mm. The spring carries an axial pull of 200 N. What is the value of shear stress-induced in the spring neglecting the effect of stress concentration and of deflection in the spring, when the modulus of rigidity is 80 kN/mm2?
A machine mounted on a single coil spring has a period of free vibration of T. If the spring is cut into four equal parts and placed in parallel and the machine is mounted of them, then the period of free vibration of the new system will be:
A helical spring has N turns of coil diameter D and a second spring, made of same wire diameter and same material, has N/2 turns of coil of diameter 2D. If the stiffness of the first spring is k, then the stiffness of the second spring will be
A mass is suspended at the bottom of two springs in series having stiffness 10 N/mm and 5 N/mm. The equivalent spring stiffness of the two Springs is nearly
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:
A helical compression spring of stiffness K is cut into two pieces, each having equal number of turns and kept side by side under compression. The equivalent spring stiffness of this new arrangement is equal to:
If two closely coiled helical springs A & B with the mean diameter of spring A is half of that of spring B and having an equal number of active coils and same wire diameter are subjected to a same axial load of W, then the ratio of deflection in spring A to B.
A close helical spring of 100 mm mean diameter is made of 10 mm diameter rod, and has 20 turns. The spring carries an axial load of 200 kN with G = 8.4 × 104 N/mm2. The stiffness of the spring is nearly

A mass is suspended at the bottom of two springs in series having stiffness 10 N/mm and 5 N/mm. The equivalent spring stiffness of the two Springs is nearly

Discussion

No Comment Found

Related MCQs

Reply to Comment