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

1. Steps are given to determine the centre of curvature at a given point on a conic. Arrange the steps. Let P be the given point on the conic and F is the focus.

Join P with F.

Draw a line NR perpendicular to PN and cutting PF or PF-extended at R.

Draw a line RO perpendicular to PR and cutting PN-extended at O which is centre of curvature.

At P, draw a normal PN, cutting the axis at N.

a) i, iv, ii, iii

b) iv, i, iii, ii

c) iii, i, iv, ii

d) ii, iv, i, iii

View Answer

Explanation: The centre O of the circle of curvature lies on the normal to the curve at P. This centre is called center of curvature at P. So for that we first found normal and accordingly the curve we found center of curvature.

2. Steps are given to determine the centre of curvature at a given point on an Ellipse. Arrange the steps. Let P be the given point on the conic and F and F1 are the foci.

i. Produce F1G to H so that GH = VF. Join H with F.

ii. Then O is the required centre of curvature.

iii. Draw a line GO parallel to HF and intersecting the axis at O.

iv. Draw a line F1G inclines to the axis and equal to VF1.

a) i, iv, ii, iii

b) iv, i, iii, ii

c) iii, i, iv, ii

d) ii, iv, i, iii

View Answer

Explanation: First we just took the arbitrary line passing through one of the foci and then extended up to the length from that focus to opposite vertex and then extended further up to length of distance between vertex and respective focus. Drawing parallel lines on to the focus gave us O.

3. Steps are given to determine the centre of curvature at a given point on an Ellipse. Arrange the steps. Let P be the given point on the conic and F is one of the focus.

i. Join A with C.

ii. Then O1 and O2 are the centres of curvature when the point P is at A and C respectively.

iii. Draw a rectangle AOCE in which AO = ½ major axis and CO = ½ minor axis.

iv. Through E, draw a line perpendicular to AC and cutting the major axis at O1 and the minor axis O2.

a) i, iv, ii, iii

b) iv, i, iii, ii

c) iii, i, iv, ii

d) ii, iv, i, iii

View Answer

Explanation: First we just took the arbitrary line passing through one of the foci and then extended up to the length from that focus to opposite vertex and then extended further up to length of distance between vertex and respective focus. Drawing parallel lines on to the focus gave us O.

4. Steps are given to determine the centre of curvature at a given point on a hyperbola. Arrange the steps. Let P be the given point on the conic, V is vertex and F and F1 are the foci.

i. Draw a line GO parallel to HF and cutting the axis at O.

ii. Draw a line F1G inclined to the axis and equal to FV1.

iii. Then O is the centre of curvature at the vertex V.

iv. On F1G, mark a point H such that HG = VF. Join H with F.

a) i, iv, ii, iii

b) iv, i, iii, ii

c) iii, i, iv, ii

d) ii, iv, i, iii

View Answer

Explanation: First we just took the arbitrary line passing through one of the foci and then extended up to the length from that focus to opposite vertex and then extended further up to length of distance between vertex and respective focus. Drawing parallel lines on to the focus gave us O.

5. Steps are given to draw the evolute of a cycloid. Arrange the steps.

i. Mark a point P on the cycloid and draw the normal PN to it.

ii. Similarly, mark a number of points on the cycloid and determine centres of curvature at these points.

iii. The curve drawn through these centres is the evolute of the cycloid. It is an equal cycloid.

iv. Produce PN to Op so that NOp = PN. Op is the centre of curvature at the point P.

a) i, iv, ii, iii

b) iv, i, iii, ii

c) iii, i, iv, ii

d) ii, iv, i, iii

View Answer

Explanation: Evolute is generally the locus of center of curvature from point on any curve. So for center of curvature we first need to draw normals at the point on curve and then center of curvature and then similarly other center of curvatures and joining the whole gives us the evolute.

6. Steps are given to draw the evolute of a hypocycloid. Arrange the steps.

i. Draw the diameter PQ of the rolling circle. Join Q with O, the centre of the directing circle.

ii. Mark a number of points on the hypocycloid and similarly, obtain centres of curvature at these points. The curve drawn through these centres is the evolute of the hypocycloid.

iii. Produce PN to cut OQ- produced at Op, which is the centre of curvature at the point P.

iv. Mark a point P on the hypocycloid and draw the normal PN to it.

a) i, iv, ii, iii

b) iv, i, iii, ii

c) iii, i, iv, ii

d) ii, iv, i, iii

View Answer

Explanation: Evolute is generally the locus of center of curvature from point on any curve. So for that we first found the center of curvature of a point and then similarly other joining the whole gives us the evolute.

7. The evolute of the involute of a circle is the circle itself.

a) True

b) False

View Answer

Explanation: In the involute of a circle, the normal NM at any point N is tangent to the circle at the point of contact M. M is the centre of curvature at the point N. Hence, the evolute of the involute is the circle itself.

8. The difference between two consecutive crest/root of a screw is called __________

a) Helix

b) Mean diameter

c) Pitch

d) Revolution

View Answer

Explanation: Mean diameter is the average of maximum diameter and the minimum diameter which is caused by the crest and root of screws, bolts etc. revolution is the one complete turn of helix around its own axis.

9. Pitch of the given bolt is 10 mm. The bolt completed the ½ revolution in forward direction. How much the bolt advances through axis?

a) 10 mm

b) 5 mm

c) 2.5 mm

d) 20 mm

View Answer

Explanation: The axial advance of the point during one complete revolution is called the pitch of the helix. So here pitch is 10 mm and the point start upwards from the base of the cylinder, in ½ revolutions, the point will move up to a distance of 5mm from base.

10. Helix angle can be expressed as tanӨ = __________________

View Answer

Explanation: The helix is seen as a straight line and is the hypotenuse of a right-angled triangle having base equal to the circumference of the circle and the vertical side equal to the pitch of the helix. The angle Ɵ which it makes with the base, is called the helix angle.

11. Number of revolutions are 10 and pitch is 2mm. Find the length of spring.

a) 10

b) 40

c) 30

d) 20

View Answer

Explanation: Here there is mention the type of edges of spring so there would be no additional length. Length of the bolt = pitch x number of revolutions, L = 2 mm x 10, L = 20 mm.

12. Length of spring is 5cm and pitch measured is 4mm. Find the number of revolutions.

a) 20

b) 12.5

c) 13

d) 12

View Answer

Explanation: Here there is mention the type of edges of spring so there would be no additional length. Length of the bolt = pitch x number of revolutions, 5cm = 50 mm =4 x (r), r = 50/4 =12.5 mm.

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