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

1. Mathematical equation for Involute is ___________

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Explanation: x= a cos3 Ɵ is equation for hypocycloid, x= (a+ b) cosƟ – a cos ( (a+b)/aƟ) is equation for epicycloid, y = a (1-cosƟ) is equation for cycloid and x = r cosƟ + r Ɵ sinƟ is equation for Involute.

2. Steps are given to draw involute of given circle. Arrange the steps f C is the centre of circle and P be the end of the thread (starting point).

i. Draw a line PQ, tangent to the circle and equal to the circumference of the circle.

ii. Draw the involute through the points P1, P2, P3 ……..etc.

iii. Divide PQ and the circle into 12 equal parts.

iv. Draw tangents at points 1, 2, 3 etc. and mark on them points P1, P2, P3 etc. such that 1P1 =P1l, 2P2 = P2l, 3P3= P3l etc.

a) ii, i, iv, iii

b) iii, i , iv, ii

c) i, iii, iv, ii

d) iv, iii, i, ii

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Explanation: Involute is curve which is formed by thread which is yet complete a single wound around a circular object so thus the thread having length equal to circumference of circular object. And the involute curve follows only the thread is kept straight while wounding.

3. Steps are given to draw tangent and normal to the involute of a circle (center is C) at a point N on it. Arrange the steps.

i. With CN as diameter describe a semi-circle cutting the circle at M.

ii. Draw a line joining C and N.

iii. Draw a line perpendicular to NM and passing through N which is tangent.

iv. Draw a line through N and M. This line is normal.

a) ii, i, iv, iii

b) iii, i , iv, ii

c) i, iii, iv, ii

d) iv, iii, i, ii

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Explanation: The normal to an involute of a circle is tangent to that circle. So simply by finding the appreciable tangent of circle passing through the point given on involute gives the normal and then by drawing perpendicular we can find the tangent to involute.

4. Steps given are to draw an involute of a given square ABCD. Arrange the steps.

i. With B as centre and radius BP1 (BA+ AD) draw an arc to cut the line CB-produced at P2.

ii. The curve thus obtained is the involute of the square.

iii. With centre A and radius AD, draw an arc to cut the line BA-produced at a point P1.

iv. Similarly, with centres C and D and radii CP2 and DP3 respectively, draw arcs to cut DC-produced at P3 and AD-produced at P4.

a) ii, i, iv, iii

b) iii, i , iv, ii

c) i, iii, iv, ii

d) iv, iii, i, ii

View Answer

Explanation: It is easy to draw involutes to polygons. First we have to point the initial point and then extending the sides. Then cutting the extended lines with cumulative radiuses of length of sides gives the points on involute and then joining them gives involute.

5. Steps given are to draw an involute of a given triangle ABC. Arrange the steps.

i. With C as centre and radius C1 draw arc cutting AC-extended at 2.

ii. With A as center and radius A2 draw an arc cutting BA- extended at 3 completing involute.

iii. B as centre with radius AB draw an arc cutting the BC- extended at 1.

iv. Draw the given triangle with corners A, B, C.

a) ii, i, iv, iii

b) iii, i , iv, ii

c) i, iii, iv, ii

d) iv, iii, i, ii

View Answer

Explanation: It will take few simple steps to draw involute for a triangle since it has only 3 sides. First we have to point the initial point and then extending the sides. Then cutting the extended lines with cumulative radiuses of length of sides gives the points on involute and then joining them gives involute.

6. Steps given are to draw an involute of a given pentagon ABCDE. Arrange the steps.

i. B as centre and radius AB, draw an arc cutting BC –extended at 1.

ii. The curve thus obtained is the involute of the pentagon.

iii. C as centre and radius C1, draw an arc cutting CD extended at 2.

iv. Similarly, D, E, A as centres and radius D2, E3, A4, draw arcs cutting DE, EA, AB at 3, 4, 5 respectively.

a) ii, i, iv, iii

b) iii, i , iv, ii

c) i, iii, iv, ii

d) iv, iii, i, ii

View Answer

Explanation: It is easy to draw involutes to polygons. First we have to point the initial point and then extending the sides. Then cutting the extended lines with cumulative radiuses of length of sides gives the points on involute and then joining them gives involute.

7. For inferior trochoid or inferior epitrochoid the curve touches the directing line or directing circle.

a) True

b) False

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Explanation: Since in the inferior trochoids the generating point is inside the generating circle the path will be at a distance from directing line or circle even if the generating circle is inside or outside the directing circle.

8. ‘Hypo’ as prefix to cycloids give that the generating circle is inside the directing circle.

a) True

b) False

View Answer

Explanation: ‘Hypo’ represents the generating circle is inside the directing circle. ‘Epi’ represents the directing path is circle. Trochoid represents the generating point is not on the circumference of generating circle.

**Sanfoundry Global Education & Learning Series – Engineering Drawing.**

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