Generic selectors
Exact matches only
Search in title
Search in content
Search in posts
Search in pages
Filter by Categories
nmims post
Objective Type Set
Online MCQ Assignment
Question Solution
Solved Question
Uncategorized

# Multiple choice question for engineering

## Set 1

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

Answer: a [Reason:] 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

Answer: b [Reason:] 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

Answer: c [Reason:] 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

Answer: d [Reason:] 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

Answer: a [Reason:] 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

Answer: b [Reason:] 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

Answer: a [Reason:] 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

Answer: b [Reason:] 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

Answer: b [Reason:] 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 = __________________ Answer: a [Reason:] 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

Answer: d [Reason:] 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

Answer: d [Reason:] 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.

## Set 2

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

Answer: c [Reason:] Outer diameter is equal to the sum of inner diameter and 2 times the diameter of 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

Answer: [Reason:] 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.

3. Spring index = _________________ Answer: a [Reason:] 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.

4. Mean diameter of coil is given as 100 mm and diameter of wire is 5 mm. Spring index is________
a) 40
b) 30
c) 25
d) 20

Answer: d [Reason:] Spring index is the ratio of mean diameter of coil to diameter of wire. Spring index = 100mm / 5 mm =20. Spring index does not have units since it is 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

Answer: [Reason:] Spring index is the ratio of 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

Answer: a [Reason:] The ratio of mean diameter of 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

Answer: a [Reason:] 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

Answer: b [Reason:] 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

Answer: a [Reason:] 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

Answer: b [Reason:] 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 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

Answer: c [Reason:] 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

Answer: c [Reason:] 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.

## Set 3

1. Which of the following is Hyperbola equation?
a) y2 + x2/b2 = 1
b) x2= 1ay
c) x2 /a2 – y2/b2 = 1
d) X2 + Y2 = 1

Answer: c [Reason:] The equation x2 + y2 = 1 gives a circle; if the x2 and y2 have same co-efficient then the equation gives circles. The equation x2= 1ay gives a parabola. The equation y2 + x2/b2 = 1 gives an ellipse.

2. Which of the following constructions use hyperbolic curves?
a) Cooling towers
b) Dams
c) Bridges
d) Man-holes

Answer: a [Reason:] Cooling towers, water channels use Hyperbolic curves as their design. Arches, Bridges, sound reflectors, light reflectors etc use parabolic curves. Arches, bridges, dams, monuments, man-holes, glands and stuffing boxes etc use elliptical curves.

3. The lines which touch the hyperbola at infinite distance are ________
a) Axes
b) Tangents at vertex
c) Latus rectum
d) Asymptotes

Answer: d [Reason:] Axis is line passing through the focuses of hyperbola. The line which passes through the focus and perpendicular to major axis is latus rectum. Tangent is the line which touches the curve at only one point.

4. Which of the following is the eccentricity for hyperbola?
a) 1
b) 3/2
c) 2/3
d) 1/2

Answer: b [Reason:] The eccentricity for ellipse is always less than 1. The eccentricity is always 1 for any parabola. The eccentricity is always 0 for a circle. The eccentricity for a hyperbola is always greater than 1.

5. If the asymptotes are perpendicular to each other then the hyperbola is called rectangular hyperbola.
a) True
b) False

Answer: a [Reason:] In ellipse there exist two axes (major and minor) which are perpendicular to each other, whose extremes have tangents parallel them. There exist two conjugate axes for ellipse and 1 for parabola and hyperbola.

6. A straight line parallel to asymptote intersects the hyperbola at only one point.
a) True
b) False

Answer: a [Reason:] A straight line parallel to asymptote intersects the hyperbola at only one point. This says that the part of hyperbola will lay in between the parallel lines through outs its length after intersecting at one point.

7. Steps are given to locate the directrix of hyperbola when axis and foci are given. Arrange the steps.
i. Draw a line joining A with the other Focus F.
ii. Draw the bisector of angle FAF1, cutting the axis at a point B.
iii. Perpendicular to axis at B gives directrix.
iv. From the first focus F1 draw a perpendicular to touch hyperbola at A.
a) i, ii, iii, iv
b) ii, iv, i, iii
c) iii, iv, i, ii
d) iv, i, ii, iii

Answer: d [Reason:] The directrix cut the axis at the point of intersection of angular bisector of lines passing through the foci and any point on hyperbola. Just by knowing this we can find the directrix just by drawing perpendicular at that point to axis.

8. Steps are given to locate asymptotes of hyperbola if its axis and focus are given. Arrange the steps.
i. Draw a perpendicular AB to axis at vertex.
ii. OG and OE are required asymptotes.
iii. With O midpoint of axis (centre) taking radius as OF (F is focus) draw arcs cutting AB at E, G.
iv. Join O, G and O, E.
a) i, iii, iv, ii
b) ii, iv, i, iii
c) iii, iv, i, ii
d) iv, i, ii, iii

Answer: b [Reason:] Asymptotes pass through centre is main point and then the asymptotes cut the directrix and perpendiculars at focus are known and simple. Next comes is where the asymptotes cuts the perpendiculars, it is at distance of centre to vertex and centre to focus respectively.

9. The asymptotes of any hyperbola intersects at __________
a) On the directrix
b) On the axis
c) At focus
d) Centre

Answer: d [Reason:] The asymptotes intersect at centre that is midpoint of axis even for conjugate axis it is valid. Along with the hyperbola asymptotes are also symmetric about both axes so they should meet at centre only.

## Set 4

1. Mathematical equation for Involute is ___________ Answer: b [Reason:] 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

Answer: c [Reason:] 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

Answer: a [Reason:] 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

Answer: b [Reason:] 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

Answer: d [Reason:] 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

Answer: c [Reason:] 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

Answer: b [Reason:] 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

Answer: a [Reason:] ‘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.

## Set 5

1. Which of the following is incorrect about Ellipse?
a) Eccentricity is less than 1
b) Mathematical equation is x2 = 4ay
c) Length of latus rectum is 4a
d) The distance from focus to vertex is equal to perpendicular distance from vertex to directrix

Answer: a [Reason:] The eccentricity is equal to one. That is the ratio of perpendicular distance from point on curve to directrix is equal to distance from point to focus. The eccentricity is less than 1 for ellipse, greater than one for hyperbola, zero for circle.

2. Which of the following constructions use parabolic curves?
a) Cooling towers
b) Water channels
c) Light reflectors
d) Man-holes

Answer: a [Reason:] Arches, Bridges, sound reflectors, light reflectors etc use parabolic curves. Cooling towers, water channels use Hyperbolic curves as their design. Arches, bridges, dams, monuments, man-holes, glands and stuffing boxes etc use elliptical curves.

3. The length of the latus rectum of the parabola y2 =ax is ______
a) 4a
b) a
c) a/4
d) 2a

Answer: b [Reason:] Latus rectum is the line perpendicular to axis and passing through focus ends touching parabola. Length of latus rectum of y2 =4ax, x2 =4ay is 4a; y2 =2ax, x2 =2ay is 2a; y2 =ax, x2 =ay is a.

4. Which of the following is not a parabola equation?
a) x2 = 4ay
b) y2 – 8ax = 0
c) x2 = by
d) x2 = 4ay2

Answer: d [Reason:] The remaining represents different forms of parabola just by adjusting them we can get general notation of parabola but x2 = 4ay2 gives equation for hyperbola. And x2 + 4ay2 =1 gives equation for ellipse.

5. The parabola x2 = ay is symmetric about x-axis.
a) True
b) False

Answer: b [Reason:] From the given parabolic equation x2 = ay we can easily say if we give y values to that equation we get two values for x so the given parabola is symmetric about y-axis. If the equation is y2 = ax then it is symmetric about x-axis.

6. Steps are given to find the axis of a parabola. Arrange the steps.
i. Draw a perpendicular GH to EF which cuts parabola.
ii. Draw AB and CD parallel chords to given parabola at some distance apart from each other.
iii. The perpendicular bisector of GH gives axis of that parabola.
iv. Draw a line EF joining the midpoints lo AB and CD.
a) i, ii, iii, iv
b) ii, iv, i, iii
c) iii, iv, i, ii
d) iv, i, ii, iii

Answer: b [Reason:] First we drawn the parallel chords and then line joining the midpoints of the previous lines which is parallel to axis so we drawn the perpendicular to this line and then perpendicular bisector gives the axis of parabola.

7. Steps are given to find focus for a parabola. Arrange the steps.
i. Draw a perpendicular bisector EF to BP, Intersecting the axis at a point F.
ii. Then F is the focus of parabola.
iii. Mark any point P on the parabola and draw a perpendicular PA to the axis.
iv. Mark a point B on the on the axis such that BV = VA (V is vertex of parabola). Join B and P.
a) i, ii, iii, iv
b) ii, iv, i, iii
c) iii, iv, i, ii
d) iv, i, ii, iii

Answer: c [Reason:] Initially we took a parabola with axis took any point on it drawn perpendicular to axis. And from the point perpendicular meets the axis another point is taken such that the vertex is equidistant from before point and later point. Then from that one to point on parabola a line is drawn and perpendicular bisector for that line meets the axis at focus.

8. Which of the following is not belonged to ellipse?
a) Latus rectum
b) Directrix
c) Major axis
d) Axis