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Multiple choice question for engineering

Set 1

1. When a solid is placed such that axis is inclined with the H.P and parallel to the V.P. Its projections are drawn in __________ stages.
a) 1
b) 4
c) 2
d) 3

View Answer

Answer: c [Reason:] In the initial stage, the axis is kept perpendicular to the H.P and parallel to V.P and projections are drawn and then turning the axis to given angle of rotation with H.P and then again projections are based on previous vertices and edges.

2. A hexagonal pyramid first placed in such a way its axis is perpendicular to V.P and one edge AB parallel to H.P and then next this is turned about its axis so the base AB is now making some angle with H.P. The top view for previous and later one will be having different shapes.
a) True
b) False

View Answer

Answer: b [Reason:] For given positions of solid the solid is just rotated around itself and given the axis is perpendicular to V.P so the top view gives the true shape and size of its base but the base is just rotated to its given angle shape will not change.

3. A regular cone first placed in such a way its axis is perpendicular to H.P and next this is tilted such that its base is making some acute angle with H.P. The top view for previous and later one will be ____________
a) triangle, triangle
b) irregular shape of circle and triangle, triangle
c) circle, irregular shape of circle and triangle
d) circle, triangle

View Answer

Answer: c [Reason:] For given positions of solid the solid is just tilted to some angle with H.P and previously given the axis is perpendicular to H.P so the top view gives the triangle and next with some given angle shape will change to irregular shape of circle and triangle.

4. A regular cone first placed in such a way its axis is perpendicular to H.P and next this is tilted such that its base is making some acute angle with H.P. The front view for previous and later one will be having same shape.
a) True
b) False

View Answer

Answer: a [Reason:] For given positions of solid the solid is just tilted to some angle with H.P and previously given the axis is perpendicular to H.P so the front view gives the triangle and next with some given angle shape will not change but just rotate.

5. A regular pentagon prism first placed in such a way its axis is perpendicular to H.P and one edge is parallel to V.P and next this is tilted such that its axis is making some acute angle with H.P. The front view for previous and later one will be _____________
a) pentagon, rectangle
b) rectangle, pentagon
c) rectangle, rectangle
d) irregular hexagon, pentagon

View Answer

Answer: c [Reason:] For given positions of solid the solid is made acute angle with H.P and previously given the axis is perpendicular to H.P so the front view gives the rectangle and next with some given angle shape will rotate totally.

6. A cylinder first placed in such a way its axis is perpendicular to H.P and next this is tilted such that its axis is making some acute angle with H.P. The top view for previous and later one will be ____________
a) circle, rectangle with circular ends
b) rectangle, rectangle
c) rectangle with circular ends, rectangle
d) circle, rectangle

View Answer

Answer: a [Reason:] For given positions of solid the solid is made acute angle with H.P and previously given the axis is perpendicular to H.P so the front view gives the circle and next with some given angle shape will change to rectangle with circular ends.

7. A cylinder first placed in such a way its axis is perpendicular to H.P and next this is tilted such that its axis is making some acute angle with H.P. The front view for previous and later one will be __________
a) circle, rectangle with circular ends
b) rectangle, rectangle
c) rectangle with circular ends, rectangle
d) circle, rectangle

View Answer

Answer: b [Reason:] For given positions of solid the solid is made acute angle with V.P and previously given the axis is perpendicular to V.P so the top view gives the rectangle and next with some given angle shape will not change but just tilt to given angle.

8. A triangular pyramid is placed such that its axis is perpendicular to H.P and one of its base’s edges is parallel to H.P the front view and top view will be _________________
a) Triangle of base, triangle due to slanting side
b) Triangle due to slanting side, triangle of base
c) Triangle of base, rhombus
d) Rhombus, triangle of base

View Answer

Answer: b [Reason:] Given a triangular pyramid which means the projection to its base gives triangle of base and other orthographic views give triangle due to slanting sides. Here given is pyramid whose axis is perpendicular to H.P so its front view will be triangle due to sides and top view will be triangle of base.

9. A square pyramid is placed such that its axis is inclined to H.P and one of its base’s edges is parallel to V.P the front view and top view will be ______________
a) Square, Isosceles triangle
b) Irregular pentagon, square
c) Isosceles triangle, irregular pentagon
d) Pentagon, equilateral triangle

View Answer

Answer: c [Reason:] Given a square pyramid which means the projection to its base gives square shape and other orthographic views give triangle. Here given is pyramid whose axis is inclined to H.P so its front view will be isosceles triangle and top view will be square.

10. A square prism is placed such that its axis is inclined to H.P and one of its base’s edges is parallel to V.P the front view and top view will be ____________
a) square, irregular polygon
b) irregular polygon, square
c) square, rectangle
d) rectangle, irregular polygon

View Answer

Answer: d [Reason:] Given a square prism which means the projection to its base gives square shape and other orthographic views give rectangle. Here given is prism whose axis is inclined to H.P so its front view will be rectangle and top view will be irregular polygon.

11. A regular cone having its axis parallel to V.P and perpendicular to H.P at first but then the cone’s axis keeping parallel to V.P and rotated such that its new axis is perpendicular to previous axis. The front view of the previous and later one is ______________
a) circle, triangle
b) circle, triangle with circular base
c) triangle, triangle
d) circle, circle

View Answer

Answer: c [Reason:] Given a regular cone which means the projection to its base gives circle shape and other orthographic views give triangle. But here given is inclination it may give irregular shape in its top view if the angle give is acute but given angle is 90 degrees so it gives perfect shapes.

12. A regular cone having its axis parallel to V.P and perpendicular to H.P at first but then the cone’s axis keeping parallel to V.P and rotated such that its new axis is perpendicular to previous axis. The top view of the previous and later one is ___________
a) circle, triangle
b) circle, triangle with circular base
c) triangle, triangle
d) circle, circle

View Answer

Answer: a [Reason:] Given a regular cone which means the projection to its base gives circle shape and other orthographic views give triangle. But here given is inclination of 90 degrees so previous ones will be circle and later one will be triangle.

13. A tetrahedron is made to place on H.P that is with its axis perpendicular to it and one of the edges of base parallel to V.P and then the tetrahedron is made to rotate w.r.t to H.P up to an acute angle. The top view of previous and later one is _____________
a) isosceles triangle, Isosceles triangle
b) equilateral triangle, isosceles triangle
c) equilateral triangle, square
d) square, irregular polygon of 4 sides

View Answer

Answer: b [Reason:] As normal a tetrahedron gives equilateral triangle for project to its base and isosceles triangle for other view when placed without inclination but here inclination is given but given view is top view so the shape will change to isosceles triangle.

Set 2

1. Unit for moment of inertia is _____________
a) m3
b) m4
c) m2/sec
d) m3/sec

View Answer

Answer: b [Reason:] The moment of inertia is sum of products of areas and squares of perpendicular distances from center of gravity. Since here we are considering the planar objects the area will have unit of m2 and square of distance gives another m2 product is m4.

2. Moment of inertia of circle about the axis which is passing through its center and perpendicular to that circle is _____________
a) πd4/32
b) πd3/32
c) πd3/16
d) πd4/16

View Answer

Answer: a [Reason:] The moment of inertia is sum of products of areas and squares of perpendicular distances from center of gravity. And this gives the engineering-drawing-questions-answers-moments-inertia-areas-q2 moment of inertia of any planar section.

3. Moment of inertia of a rectangle whose base is b and height is h and axis is along base is __________
a) b*h3/3
b) h*b3/3
c) b*h4/6
d) b*h3/12

View Answer

Answer: a [Reason:] The moment of inertia is sum of products of areas and squares of perpendicular distances from center of gravity. And this gives the engineering-drawing-questions-answers-moments-inertia-areas-q2 moment of inertia of any planar section.

4. Moment of inertia of a rectangle whose base is b and height is h and axis is alongside of height is ________
a) b*h3/3
b) h*b3/3
c) b*h4/6
d) b*h3/12

View Answer

Answer: b [Reason:] The moment of inertia is sum of products of areas and squares of perpendicular distances from center of gravity. And this gives the engineering-drawing-questions-answers-moments-inertia-areas-q2 moment of inertia of any planar section.

5. Moment of inertia of a triangle having base as b and height as h and axis is along the centroid and parallel the base.
a) b*h3/12
b) b*h3/24
c) b*h3/36
d) b*h3/6

View Answer

Answer: c [Reason:] The moment of inertia is sum of products of areas and squares of perpendicular distances from center of gravity. And this gives the engineering-drawing-questions-answers-moments-inertia-areas-q2 moment of inertia of any planar section.

6. Moment of inertia of a triangle having base as b and height as h and axis is along the centroid and parallel the height.
a) b*h3/12
b) h*b3/36
c) b*h3/36
d) b*h3/6

View Answer

Answer: b [Reason:] The moment of inertia is sum of products of areas and squares of perpendicular distances from center of gravity. And this gives the engineering-drawing-questions-answers-moments-inertia-areas-q2 moment of inertia of any planar section.

7. A rectangle is having base b and height h. The ratio of moment of inertia when axis is passing through its base to moment of inertia of when axis passing through center of gravity and parallel to base is ___________
a) 3
b) 4
c) 12
d) 9

View Answer

Answer: b [Reason:] Moment of inertia of triangle having base b and height h when axis passing through the center of gravity is bh3/12 and moment of inertia when axis passing through base is bh3/3 and ratio is asked it gives 4.

8. A triangle is having base b and height h. The ratio of moment of inertia when axis is passing through its base to moment of inertia of when axis passing through centroid and parallel to base is ______
a) 3
b) 6
c) 12
d) 9

View Answer

Answer: a [Reason:] Moment of inertia of triangle having base b and height h when axis passing through the centroid is bh3/36 and moment of inertia when axis passing through base is bh3/12 and ratio is asked it gives 3.

9. The moment of inertia is minimum when the axis is _________
a) passing through center of object
b) passing along x-axis
c) passing along y-axis
d) passing through centroid of object

View Answer

Answer: d [Reason:] The moment of inertia is minimum when the axis is passing through the centroid because the moment of inertia is sum of products of areas and squares of perpendicular distances from center of gravity.

10. In the polar moment of inertia the axis is perpendicular to _____________
a) depends on object
b) x-axis
c) z-axis
d) y-axis

View Answer

Answer: c [Reason:] Polar moment of inertia is the sum of the moment of inertias of objects when axes are in same plane which are perpendicular to each other let us say x, y axes and polar moment of inertia is about z axis.

Set 3

1. The angle of chamfer for hexagonal and square nut is ____ degrees as per standards.
a) 30
b) 45
c) 60
d) 15

View Answer

Answer: a [Reason:] The upper corners of this nut are rounded-off or chamfered. The chamfering is generally conical. The angle of chamfer is 30-45 degrees with the base of the nut. Due to chamfering, an arc is formed on each vertical face and circle is formed on top surface.

2. The angle through which the spanner will have to be turned to get next hold is ____ degrees in case of hexagonal shape.
a) 90
b) 60
c) 30
d) 45

View Answer

Answer: b [Reason:] The angle through which the spanner will have to be turned to get next hold is 60 degrees in case of hexagonal shape as 360/6 is 60 degrees. In case of square nut the spanner can have better hold but angle is 90 degrees so it is more convenient to use hexagonal nut.

3. Width across the flats in case of hexagonal nut is___ where D is the diameter of shank.
a) 2D
b) 1.5D +3mm
c) 1.4D
d) 1.5D

View Answer

Answer: b [Reason:] Width across the flats in case of hexagonal nut is 1.5D+3mm where D is the diameter of shank, thickness or height of nut is D itself and distance across diagonally opposite corners in case of hexagonal nut is 2D.

4. Distance across diagonally opposite corners in case of hexagonal nut is ______ where D is nominal diameter.
a) 2D
b) 1.5D +3mm
c) 1.4D
d) 1.5D

View Answer

Answer: a [Reason:] Width across the flats in case of hexagonal nut is 1.5D+3mm where D is the diameter of shank, thickness or height of nut is D itself and distance across diagonally opposite corners in case of hexagonal nut is 2D.

5. What is the name of the nut which is a hexagonal nut with a washer?
a) Dome nut
b) Wing nut
c) Flanged nut
d) Cap nut

View Answer

Answer: c [Reason:] Flanged nut is hexagonal nut with a washer that is a flat circular disc attached to it. It is thus provided with a larger bearing surface. A bolt can be used in a comparatively large-size hole with help of this nut. It is widely used in automobiles.

6. What is the name of the nut which is a hexagonal nut provided with a cylindrical cap?
a) Dome nut
b) Wing nut
c) Flanged nut
d) Cap nut

View Answer

Answer: d [Reason:] Cap nut is a hexagonal nut provided with a cylindrical cap at the top to protect the end of the bolt from corrosion. It also prevents leakage through the threads. Flanged nut is hexagonal nut with a washer that is a flat circular disc attached to it.

7. What is the name of the nut which is a cap nut with spherical dome at the top?
a) Dome nut
b) Wing nut
c) Flanged nut
d) Cap nut

View Answer

Answer: a [Reason:] It is a form of cap nut with spherical dome at top. Flanged nut is hexagonal nut with a washer that is a flat circular disc attached to it. Cap nut is a hexagonal nut provided with a cylindrical cap at the top.

8. What is the name of the nut which can be easily operated by the thumb and a finger?
a) Ring nut
b) Wing nut
c) Cap nut
d) Flanged nut

View Answer

Answer: b [Reason:] Wing nut can be easily operated by thumb and a finger and is used where it is required to be adjusted frequently. It is used in a hacksaw. Cap nut is a hexagonal nut provided with a cylindrical cap at the top.

9. What is the name of the nut which is in form of a ring provided with slots in the curved surface for c-spanner?
a) Ring nut
b) Wing nut
c) Capstan nut
d) Flanged nut

View Answer

Answer: a [Reason:] Ring nut is in form of a ring provided with slots in the curved surface for a special C-spanner. These nuts are generally used in pairs, one nut acting as a lock-nut for the other. Wing nut can be easily operated by the thumb and a finger.

10. What is the outer diameter of washer when nominal diameter of bolt is 5 mm?
a) 13 mm
b) 15 mm
c) 10 mm
d) 12 mm

View Answer

Answer: a [Reason:] When nominal diameter of bolt is D the thickness of the washer is 0.12D and outer diameter is 2D+3 mm and inner diameter is D+0.5mm. Here D is 5 mm so 2D+3 mm = 2 x 5+3= 10+3 = 13 mm.

11. What is the thickness of washer when nominal diameter of bolt is D?
a) 0.12D
b) 0.1D
c) D
d) 0.09D

View Answer

Answer: a [Reason:] Washer is a cylindrical piece of metal placed below the nut to provide smooth bearing surface. The thickness of washer when nominal diameter of bolt is D is 0.12D and outer diameter is 2D+3 mm and inner diameter is D+0.5mm.

Set 4

1. Oblique projections are useful for making an assembly.
a) True
b) False

View Answer

Answer: a [Reason:] The oblique projection represents three dimensional objects on the projection plane by one view only. This type of drawing is useful for making an assembly of an object and provides directly a production drawing of the object for manufacturing purpose.

2. Lines of sights (projectors) for oblique projection will be ______________
a) Parallel to each other and perpendicular to projection plane
b) Not parallel to each other and perpendicular to projection plane
c) Parallel to each other and inclined to projection plane
d) Not parallel to each other and inclined to projection plane

View Answer

Answer: c [Reason:] When an observer looks towards an object from infinity, the lines of sight will be parallel to each other and inclined to the projection plane in oblique projection, but the lines of sight will be parallel to each other and perpendicular to projection plane in orthographic projection.

3. All the faces of the object are distorted in the shape and size.
a) True
b) False

View Answer

Answer: b [Reason:] In oblique projection the faces of object which are perpendicular to the plane of projection will be distorted and all the faces of the object are distorted in the shape and size in isometric projection.

4. Which of the following statement is wrong in case of oblique projection?
a) The object is drawn with the reduced dimensions.
b) Projectors are parallel to each other and inclined to projection plane.
c) The choice of the position of the object depends upon the shape and size.
d) The faces of object which are perpendicular to the plane of projection will be distorted.

View Answer

Answer: a [Reason:] In isometric projection the object is drawn with the reduced dimensions for about 82% but in oblique projection the object is drawn with the actual dimensions. The choice of the position of the object depends upon the shape and size.

5. When the receding lines are drawn to full size scale then the oblique projection is ___________
a) Cabinet projection
b) Isometric projection
c) Orthographic projection
d) Cavalier projection

View Answer

Answer: d [Reason:] When the receding lines are drawn to half size scale then the oblique projection is cabinet projection. When the receding lines are drawn to full size scale then the oblique projection is cavalier projection.

6. When the receding lines are drawn to half size scale then the oblique projection is __________
a) Cabinet projection
b) Isometric projection
c) Orthographic projection
d) Cavalier projection

View Answer

Answer: a [Reason:] When the receding lines are drawn to half size scale then the oblique projection is cabinet projection. When the receding lines are drawn to full size scale then the oblique projection is cavalier projection.

7. Which are not usually used as angle between the projection plane and receding lines for oblique projection?
a) 30 degrees
b) 50 degrees
c) 45 degrees
d) 60 degrees

View Answer

Answer: b [Reason:] Usually used as angles are 30, 45, 60 degrees which are between the projection plane and receding lines for oblique projection. But if needed any angle can be taken as per requirement of the view.

8. The appearance of the distortion of an object can be improved by ___________ the length of the receding lines.
a) increasing
b) doubling
c) shortening
d) dividing

View Answer

Answer: c [Reason:] The appearance of the distortion of an object can be improved by shortening the length of the receding lines. The receding lines may be inclined either upwards or downwards, or to the left or right depending upon the necessity to show the details.

9. In oblique projection, the object is assumed to be placed with one face _______________
a) parallel to plane of projection
b) parallel to adjacent edge
c) perpendicular to plane of projection
d) perpendicular to adjacent edge

View Answer

Answer: a [Reason:] In oblique projection, the object is assumed to be placed with one face parallel to plane of projection and receding lines are drawn from the faces parallel to projection plane and other parallel face.

10. The perpendicular edges of planes parallel to projection plane are drawn at an angle of 30, 45, 60 degrees with the horizontal. The inclined lines are called _____________
a) projectors
b) slanting edges
c) contour lines
d) receding lines

View Answer

Answer: d [Reason:] Projectors are the imaginary lines drawn from object to projection planes. Slanting edges can be used for edges of pyramid etc. The perpendicular edges of planes parallel to projection plane are drawn at an angle of 30, 45, 60 degrees with the horizontal. The inclined lines are called receding lines.

11. The faces parallel to projection plane are having ________ size and shape in oblique projection.
a) actual
b) double
c) half
d) increased

View Answer

Answer: a [Reason:] Oblique projection is a specified projection which is used for making an assembly of an object and provides directly a production drawing of the object for manufacturing purpose.

12. In cavalier projection the receding lines are drawn _________
a) half of its actual size
b) double of its actual size
c) full size
d) increased or decreased to particular ratio

View Answer

Answer: c [Reason:] Cavalier projection is one of the type of oblique projection in which the receding lines are drawn to full size scale and projectors are inclined at 30 degrees, 45 degrees, 60 degrees to the plane of projection.

Set 5

1. What are the two parts of Vernier scale?
a) Primary scale and secondary scale
b) Plain scale and comparative scale
c) Vernier scale and secondary scale
d) Primary scale and Vernier scale

View Answer

Answer: d [Reason:] The Vernier scale consists of two parts. They are the primary scale and the Vernier scale. Vernier scale is used to read very small dimensions with great accuracy. The Vernier part of the scale is derived from the primary scale.

2. Which of the following scale is a plain scale with fully divided minor divisions?
a) Diagonal scale
b) Vernier scale
c) Primary scale
d) Comparative scale

View Answer

Answer: c [Reason:] The primary scale is a plain scale with fully divided minor divisions. For reading very small dimensions, minor divisions cannot be further divided. Hence we use Vernier scale to read those dimensions.

3. The graduations of which scale is derived from the primary scale?
a) Comparative scale
b) Vernier scale
c) Plain scale
d) Diagonal scale

View Answer

Answer: b [Reason:] The graduations on the primary scale helps in making the graduations on the Vernier scale. For reading very small dimensions we need to further divide the minor divisions on the primary scale, which will to problems in reading dimensions properly. Hence Vernier scale is used.

4. Which of the following is used in checking the instruments to measure angles with great accuracy?
a) Circular Vernier scale
b) Plain scale
c) Diagonal scale
d) Comparative scale

View Answer

Answer: a [Reason:] The circular Vernier scale is used to check the instruments to measure angles with high accuracy. They are used in measuring instruments like micrometer in mechanical engineering. They work on the same principle as the linear Vernier scale.

5. Which of the following scales represent two different units having same representative fration?
a) Plain scale
b) Diagonal scale
c) Comparative scale
d) Vernier scale

View Answer

Answer: c [Reason:] Comparative scales are those in which two different units are measured by using different graduations but have same representative fraction. A drawing can be read in two units with the help of this scale.

6. What is the length of the scale, the representative fraction is 1:50000 and the scale must measure up to 25 km?
a) 5 x 10-4 cm
b) 50 cm
c) 5 cm
d) 0.5 cm

View Answer

Answer: b [Reason:] The formula for calculating the length of the scale is given by, Length of the scale = R.F. x maximum length to be measured; Hence length of the scale = 1 ÷ 50000 x 25 x 1000 x 100 = 50 cm. It is very important to note the units while calculating.

7. In a particular drawing, 40 inches is represented by a 10 cm long line. A comparative scale is constructed that measures up to 180 inches and 2 m respectively. What are the lengths of scale for both the inches as well as the meter scales? (1 inch = 0.0254 meter)
a) 7 cm and 8.3 cm
b) 35 cm and 18.35
c) 2.2 cm and 22.9 cm
d) 45 cm and 19.7 cm

View Answer

Answer: d [Reason:] The length of the scale = R.F. x maximum length to be measured. Hence for the inches scale, R.F. = 10 ÷ 40 and the maximum length to be measured is 180 inches. Hence length of inch scale = 10 ÷ 40 x 180 = 45 cm. Similarly length of meter scale = 10 ÷ 40 ÷ 0.0254 ÷ 100 x 2 x 100 = 19.7 cm.

8. Which of the following is used to set or measure angles when a protractor is not available?
a) Plain scale
b) Diagonal scale
c) Scale of chords
d) Comparative scale

View Answer

Answer: c [Reason:] The scale of chords is used to measure or set angles when there is no protractor. In this method, a line PQ is drawn and then from Q a perpendicular QR is drawn. An arc is drawn with centre Q and is made to intersect perpendicular QR. Then the arc is divided into 9 equal parts with each part denoting 10˚ from Q.

9. In a map, 30 miles is represented by 20 cm. What is the length of the kilometer scale if the maximum length to be measured is 10 km? (1 mile = 1.609 km)
a) 4.14 cm
b) 10.73 cm
c) 9.32 cm
d) 24.14 cm

View Answer

Answer: a [Reason:] The formula for calculating the length of the scale is given by; Length of the scale = R.F. x maximum length to be measured. Hence from the formula, length of the scale = 20 ÷ (30 x 1.609 x 1000 x 100) x 10 x 1000 x 100 = 4.14 cm.

10. What is the representative fraction of the kilometer scale if 20 miles is represented as 5 cm and maximum length to be measured is 5 km? (1 mile = 1.609 km)
a) 1:4
b) 1:643600
c) 1:1.287
d) 0.7768:1

View Answer

Answer: b [Reason:] The representative fraction is calculated by the formula, R.F. = length of the drawing ÷ actual length of the object. Here the length of the drawing is 5 cm and the actual length is 20 miles. In R.F. both quantities should be in same unit. R.F. = 5 ÷ (20 x 1.609 x1000 x 100) = 1:643600.