# Engineering Drawing Questions and Answers – Construction of Helical Springs

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This set of Engineering Drawing Multiple Choice Questions & Answers (MCQs) focuses on “Construction of Helical Springs”.

1. A Spring is made of wire whose cross-section is a square of 15 mm side. Inner diameter of spring is 60 mm then outer diameter will be _________
a) 45
b) 75
c) 90
d) 80

Explanation: Outer diameter is equal to the sum of the inner diameter and 2 times the diameter of a wire. Here the cross section of wire is square so diameter can be considered as 15 mm. outer diameter = 60 +2 x 15 = 90 mm.

2. A Spring is made of wire whose cross-section is an equilateral triangle of 8 mm side. Inner diameter of spring is 40 mm then outer diameter will be _________
a) 57.88 mm
b) 54.88 mm
c) 60 mm
d) 56 mm

Explanation: Outer diameter = inner diameter + (2 x diameter of wire), Here wire has cross section of equilateral triangle of side 8 mm so it covers a length of 8.94 (square root (82 + 42 )) mm from inner to outer end of spring. Outer diameter = 40 + 2 x 8.94 = 57.88 mm.
Explanation: Spring index is the ratio of mean diameter of coil to diameter of wire. Pitch to circumference ration is helix angle. In mechanical components usually have some standard in sizes and shapes etc. for which they should maintain some ratio among particular things to indicate some of various sized similar components.
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4. Mean diameter of the coil is given as 100 mm and diameter of the wire is 5 mm. Spring index is________
a) 40
b) 30
c) 25
d) 20

Explanation: Spring index is the ratio of the mean diameter of a coil to diameter of the wire. Spring index = 100mm / 5 mm =20. Spring index does not have units since it is a ration of similar units.

5. Spring index is given as 12.5 and diameter of wire given is 5 mm. Mean diameter of coil is _______
a) 60 mm
b) 62.5 mm
c) 6 cm
d) 56.2 mm

Explanation: Spring index is the ratio of the mean diameter of coil to diameter of wire. 12.5 = mean diameter of coil / 5mm, mean diameter of coil = 12.5 x 5 mm = 62.5 mm. We need to use same units while substituting in formulae.

6. Spring index is given as 15 and mean diameter of coil is 90 mm. Diameter of wire is __________
a) 6 mm
b) 5 mm
c) 7 mm
d) 8 mm

Explanation: The ratio of a mean diameter of the coil to diameter of wire gives spring index. 90 mm / diameter of wire = 15, diameter of wire = 90 mm / 15 = 6 mm. This spring index sometimes gives the strength to spring and used in calculating stress through it.

7. Mean diameter of coil is 170 mm and spring index is 17. Diameter of wire is _________
a) 1 cm
b) 5 mm
c) 153 mm
d) 1.5 cm

Explanation: The ratio of mean diameter of coil to diameter of wire gives spring index. 170 mm / diameter of wire = 17, diameter of wire = 170 mm / 17 = 10 mm= 1 cm.

8. Diameter of wire is 7.5 mm and spring index is 15. Outer diameter of the coil is ___________
a) 112.5 mm
b) 120 mm
c) 1.2 cm
d) 20 cm

Explanation: The ratio of mean diameter of coil to diameter of wire gives spring index. Mean diameter is average of outer and inner diameter of coil in other words outer diameter = diameter of wire +mean diameter or inner diameter = mean diameter – diameter of wire. Mean diameter = 7.5 x 15 = 112.5 mm. Outer diameter = 112.5 mm + 7.5 mm = 120 mm.

9. Mean diameter of coil is 100 mm and inner diameter is 95 mm, spring index is __________
a) 10
b) 5
c) 12
d) 15

Explanation: Spring index is the ratio of mean diameter of coil to diameter of wire. Outer diameter = inner diameter + 2 x diameter of wire. So here diameter of wire is 10 mm. Spring index = 100mm/10mm = 10.

10. Outer diameter is 95 mm and inner diameter is 88 mm. Mean diameter is ________
a) 90 mm
b) 91.5 mm
c) 95.1 mm
d) 88 mm

Explanation: Mean diameter is average of outer and inner diameter of spring. Difference between the outer and inner diameter gives diameter of wire and ratio of mean diameter to the diameter of wire gives spring index.

11. Inner diameter of the coil is 70 mm and diameter of wire is 8 mm, spring index is ________
a) 9.25
b) 8.75
c) 9.75
d) 7.8

Explanation: Outer diameter = inner diameter + 2 x diameter of wire, outer diameter = 70 +2 x 8 = 86 mm. Spring index = mean diameter of coil / diameter of wire. Spring index = ((86+70)/2)/8 = 9.75.

12. Spring index is 10 and diameter of wire is 10 mm. Outer diameter of coil is __________
a) 100 mm
b) 90 mm
c) 110 mm
d) 120 mm

Explanation: Spring index = mean diameter of coil / diameter of wire, 10 = mean diameter /10 mm, mean diameter = 10 x 10 mm = 100 mm. Outer diameter = mean diameter + diameter of wire, Outer diameter = 100 mm+ 10 mm = 110 mm.

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