When a helical compression spring is cut into halves, the stiffness of the resulting spring will be

A closely-coiled helical spring is cut into two halves. The stiffness of the resulting spring will be

A

Same

B

Double

C

Half

D

One-fourth

A potato tuber has been cut into two halves. A few drops of iodine solution are placed on the cut surface of one of the halves. What colour change will be noticed?

A

From brown to blue-black

B

From brown to orange-red

C

From blue to pink

D

From pink to blue-green

Two closely-coiled helical springs 'A' and 'B' are equal in all respects but the number of turns of spring 'A' is double that of spring 'B'. The stiffness of spring 'A' will be __________ that of spring 'B'.

A

One-sixteenth

B

One-eighth

C

One-fourth

D

One-half

Two closely coiled helical springs 'A' and 'B' are equal in all respects but the diameter of wire of spring 'A' is double that of spring 'B' The stiffness of spring 'B' will be __________ that of spring 'A'

A

One-sixteenth

B

One-eighth

C

One-fourth

D

One-half

A closely coiled helical spring is acted upon by an axial force. The maximum shear stress developed in the spring is $$\tau $$. The half of the length of the spring if cut off and the remaining spring is acted upon by the same axial force. The maximum shear stress in the spring in new condition will be

A

$$\frac{\tau }{2}$$

B

$$\tau $$

C

$$2\tau $$

D

$$4\tau $$

A coil is cut into two halves, the stiffness of cut coil will be

A

Double

B

Half

C

Same

D

None of these

When an open coiled helical compression spring is subjected to an axial compressive load, the maximum shear stress induced in the wire is (where D = Mean diameter of the spring coil, d = Diameter of the spring wire, K = Wahl's stress factor and W = Axial compressive load on the spring)

A

$$\frac{{{\text{WD}}}}{{\pi {{\text{d}}^3}}} \times {\text{K}}$$

B

$$\frac{{2{\text{WD}}}}{{\pi {{\text{d}}^3}}} \times {\text{K}}$$

C

$$\frac{{4{\text{WD}}}}{{\pi {{\text{d}}^3}}} \times {\text{K}}$$

D

$$\frac{{8{\text{WD}}}}{{\pi {{\text{d}}^3}}} \times {\text{K}}$$

When a closely-coiled helical spring of mean diameter (D) is subjected to an axial load (W), the deflection of the spring ($$\delta $$) is given by (where d = Diameter of spring wire, n = No. of turns of the spring and C = Modulus of rigidity for the spring material)

A

$$\frac{{{\text{W}}{{\text{D}}^3}{\text{n}}}}{{{\text{C}}{{\text{d}}^4}}}$$

B

$$\frac{{2{\text{W}}{{\text{D}}^3}{\text{n}}}}{{{\text{C}}{{\text{d}}^4}}}$$

C

$$\frac{{4{\text{W}}{{\text{D}}^3}{\text{n}}}}{{{\text{C}}{{\text{d}}^4}}}$$

D

$$\frac{{8{\text{W}}{{\text{D}}^3}{\text{n}}}}{{{\text{C}}{{\text{d}}^4}}}$$

When a closely-coiled helical spring of mean diameter (D) is subjected to an axial load (W), the stiffness of the spring is given by

A

$$\frac{{{\text{C}}{{\text{d}}^4}}}{{{{\text{D}}^3}{\text{n}}}}$$

B

$$\frac{{{\text{C}}{{\text{d}}^4}}}{{2{{\text{D}}^3}{\text{n}}}}$$

C

$$\frac{{{\text{C}}{{\text{d}}^4}}}{{4{{\text{D}}^3}{\text{n}}}}$$

D

$$\frac{{{\text{C}}{{\text{d}}^4}}}{{8{{\text{D}}^3}{\text{n}}}}$$

Two closely coiled helical springs 'A' and 'B' are equal in all respects but the number of turns of spring 'A' is half that of spring 'B'. The ratio of deflections in spring 'A' to spring 'B' is

A

$$\frac{1}{8}$$

B

$$\frac{1}{4}$$

C

$$\frac{1}{2}$$

D

2

When a helical compression spring is cut into halves, the stiffness of the resulting spring will be

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