McqMate
These multiple-choice questions (MCQs) are designed to enhance your knowledge and understanding in the following areas: Civil Engineering .
| 151. |
A hexagonal nut is placed on a horizontal plane such that the axis is perpendicular to profile plane. The top view and side view will be |
| A. | rectangle, hexagon |
| B. | hexagon, rectangle |
| C. | rectangle, rectangle |
| D. | rectangle, circle |
| Answer» A. rectangle, hexagon | |
| Explanation: given a hexagonal nut is placed on horizontal plane such that the axis is perpendicular to profile plane. as per position given the front view, back view, top view and bottom view will be rectangle and side view will be hexagon. | |
| 152. |
Two points are placed in 1st quadrant of projection planes such that the line joining the points is perpendicular to profile plane the side view and top view will be |
| A. | single point, two points |
| B. | two points, single point |
| C. | single point, single point |
| D. | two points, two points |
| Answer» A. single point, two points | |
| Explanation: here given the two points such that the joining line is perpendicular to profile | |
| 153. |
A point is 5 units away from the vertical plane and 4 units away from profile plane and 3 units away from horizontal plane in 1st quadrant then the projections are drawn on paper the distance between the front view and top view of point is |
| A. | 7 units |
| B. | 8 units |
| C. | 9 units |
| D. | 5 units |
| Answer» B. 8 units | |
| Explanation: since the point is 3 units away from the horizontal plane the distance from the point to xy reference line will be 3 units. and then the point is at a distance of 5 units from the vertical plane the distance from reference line and point will be 5, sum is 8. | |
| 154. |
A point is 8 units away from the vertical plane and 2 units away from profile plane and 4 units away from horizontal plane in 1st quadrant then the projections are drawn on paper the distance between the side view and front view of point is |
| A. | 12 units |
| B. | 6 units |
| C. | 10 units |
| D. | 8 units |
| Answer» C. 10 units | |
| Explanation: since the point is 2 units away from the profile plane the distance from the point to reference line will be 2 units. and then the point is at a distance of 8 units from the vertical plane the distance from reference line and point will be 8, sum is 10. | |
| 155. |
A point is 20 units away from the vertical plane and 12 units away from profile plane and 9 units away from horizontal plane in 1st quadrant then the projections are drawn on paper the distance between the side view and front view of point is |
| A. | 29 units |
| B. | 21 units |
| C. | 32 units |
| D. | 11 units |
| Answer» C. 32 units | |
| Explanation: since the point is 12 units away from the profile plane the distance from the point to reference line will be 12 units. and then the point is at a distance of 20 units from profile plane the distance from reference line and point will be 20 units, sum is 32. | |
| 156. |
A point is 2 units away from the vertical plane and 3 units away from profile plane and 7 units away from horizontal plane in 1st quadrant then the projections are drawn on paper the shortest distance from top view and side view of point is |
| A. | 10.29 |
| B. | 5.14 |
| C. | 9 |
| D. | 7 |
| Answer» C. 9 | |
| Explanation: since here distance from side view and top view is asked for that we need the distance between the front view and side | |
| 157. |
If a point P is placed in between the projection planes. The distance from side view to reference line towards front view and the distance between top view and reference line towards top view will be same. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: the projection will be drawn by turning the other planes parallel to a vertical plane in clockwise direction along the lines of intersecting of planes. and so as we fold again the planes at respective reference lines and then drawing perpendiculars to the planes at those points the point of intersection gives the point p. | |
| 158. |
A point is 20 units away from the vertical plane and 12 units away from profile plane and 9 units away from horizontal plane in 1st quadrant then the projections are drawn on paper the distance between the side view and top view of point is |
| A. | 29 units |
| B. | 21 units |
| C. | 35.8 units |
| D. | 17.9 units |
| Answer» C. 35.8 units | |
| Explanation: since here distance from side view and top view is asked for that we need the distance between the front view and side view (12+9); front view and top view (9+20)and these lines which form perpendicular to each other gives needed distance, answer is square root of squares of both the distances √(212+292 ) = 35.80 units. | |
| 159. |
A point is 15 units away from the vertical plane and 12 units away from profile plane and horizontal plane in 1st quadrant then the projections are drawn on paper the distance between the front view and top view of point is |
| A. | 27 |
| B. | 15 |
| C. | 12 |
| D. | 24 |
| Answer» A. 27 | |
| Explanation: since the point is 12 units away from the horizontal plane the distance from the point to xy reference line will be 12 units. and then the point is at a distance of 15 units from the vertical plane the distance from reference line and point will be 15, sum is 27. | |
| 160. |
A point is 12 units away from the vertical plane and profile plane 15 units away from horizontal plane in 1st quadrant then the projections are drawn on paper the distance between the front view and side view of point is |
| A. | 27 |
| B. | 15 |
| C. | 12 |
| D. | 24 |
| Answer» D. 24 | |
| Explanation: since the point is 12 units away from the profile plane the distance from the point to xy reference line will be 12 units. | |
| 161. |
A point is 7 units away from the vertical plane and horizontal plane 9 units away from profile plane in 1st quadrant then the projections are drawn on paper the distance between the front view and top view of point is |
| A. | 27 |
| B. | 15 |
| C. | 16 |
| D. | 14 |
| Answer» D. 14 | |
| Explanation: since the point is 7 units away from the horizontal plane the distance from the point to xy reference line will be 7 units. and then the point is at a distance of 7 units from the vertical plane the distance from reference line and point will be 7, sum is 14 units. | |
| 162. |
A point is 16 units away from the vertical plane and horizontal plane 4 units away from profile plane in 1st quadrant then the projections are drawn on paper the distance between the side view and top view of point is |
| A. | 37.73 units |
| B. | 32.98 units |
| C. | 16 |
| D. | 8 |
| Answer» D. 8 | |
| Explanation: since here distance from side view and top view is asked for that we need the distance between the front view and side view (4+16); front view and top view (16+16)and these lines which form perpendicular to each other gives needed distance, answer is square root of squares of both the distances √202+322 ) = 37.73 units. | |
| 163. |
A point is in 2nd quadrant 20 units away from the horizontal plane and 10 units away from the vertical plane. Orthographic projection is drawn. What is the distance from point of front view to reference line, top view point to reference line? |
| A. | 20, 10 |
| B. | 10, 20 |
| C. | 0, 20 |
| D. | 10, 0 |
| Answer» A. 20, 10 | |
| Explanation: given object is point placed in 2nd quadrant the top view gives the distance from vertical plane (10) and front view gives the distance from horizontal plane (20) both are placed overlapped in orthographic projection since the object is placed in the 2nd quadrant. | |
| 164. |
A point is in 2nd quadrant 15 units away from the vertical plane and 10 units away from the horizontal plane. Orthographic projection is drawn. What is the distance from point of front view to reference line, top view point to reference line? |
| A. | 15, 10 |
| B. | 10, 15 |
| C. | 0, 15 |
| D. | 10, 0 |
| Answer» B. 10, 15 | |
| Explanation: given object is point the top view gives the distance from vertical plane | |
| 165. |
A point is in 2nd quadrant, 15 units away from the vertical plane, 10 units away from the horizontal plane and 8 units away from the profile plane. Orthographic projection is drawn. What is the distance from point of front view to point of side view? |
| A. | 25 |
| B. | 23 |
| C. | 18 |
| D. | 5 |
| Answer» B. 23 | |
| Explanation: side view is obtained by turning the profile plane along the hinge with vertical parallel to vertical plane. side view and front view have same distance from reference line. sum of distances from the point to vertical plane and profile plane gives the following that is 15+8 = 23 units. | |
| 166. |
A point in 2nd quadrant is 15 cm away from both the horizontal plane and vertical plane and orthographic projections are drawn. The distance between the points formed by front view and top view is |
| A. | 0 |
| B. | 30 |
| C. | 15 |
| D. | 15+ distance from a profile |
| Answer» A. 0 | |
| Explanation: given the point is in 2nd quadrant. while drawing orthographic projections the front view and top view overlaps and also the distance of point is same from planes of projections so the distance between them is zero. | |
| 167. |
A point in 2nd quadrant is 10 units away from the horizontal plane and 13 units away from both the vertical plane and profile plane. Orthographic projections are drawn find the distance from side view and front view. |
| A. | 10 |
| B. | 13 |
| C. | 20 |
| D. | 26 |
| Answer» D. 26 | |
| Explanation: given the point is in 2nd quadrant. the front view and side view lie parallel to the horizontal plane when orthographic projections are drawn. the distance from side view to vertical reference is 13 and distance from front view to profile plane is 13. sum is 13+13= 26. | |
| 168. |
A point in 2nd quadrant is 25 units away from both the horizontal plane and profile plane and 15 units away from the vertical plane. Orthographic projections are drawn find the distance from side view and front view. |
| A. | 25 |
| B. | 15 |
| C. | 30 |
| D. | 40 |
| Answer» D. 40 | |
| Explanation: given the point is in 2nd quadrant. the front view and side view lie parallel to the horizontal plane when orthographic projections are drawn. the distance from side view to vertical reference is 15 and distance from front view to profile plane is 25. sum is 15+25 =40. | |
| 169. |
A point in 2nd quadrant is 12 units away from the horizontal plane and vertical plane and 13 units away from both the profile plane. Orthographic projections are drawn find the distance from side view and front view. |
| A. | 13 |
| B. | 26 |
| C. | 25 |
| D. | 24 |
| Answer» C. 25 | |
| Explanation: given the point is in 2nd quadrant. the front view and side view lie parallel to the horizontal plane when orthographic projections are drawn. the distance from side view to vertical reference is 12 and distance from front view to profile plane is 13. sum 12 + 13 =25. | |
| 170. |
A point in 2nd quadrant is 25 units away from both the horizontal plane and profile plane 15 units away from the vertical plane. Orthographic projections are drawn find the distance from the side view and top view. |
| A. | 40 |
| B. | 50.99 |
| C. | 33.54 |
| D. | 41.23 |
| Answer» D. 41.23 | |
| Explanation: given the point is in 2nd quadrant. since here distance from side view and top view is asked for that we need the distance between the front view and side view (25+15); front view and top view (25-15) and these lines which form perpendicular to each other gives needed distance, answer is | |
| 171. |
23units. |
| A. | 25.6 |
| B. | 25 |
| C. | 17.69 |
| D. | 13 |
| Answer» B. 25 | |
| Explanation: given the point is in 2nd quadrant. since here distance from side view and top view is asked for that we need the distance between the front view and side view (12+13); front view and top view (12-12) and these lines which form perpendicular to each other gives needed distance, answer is | |
| 172. |
A point in 2nd quadrant is 10 cm away from the vertical plane and 15 cm away from the horizontal plane, orthographic projections are drawn. What is the distance from a side view of point to line of vertical reference? |
| A. | 10 |
| B. | 15 |
| C. | 25 |
| D. | can’t found |
| Answer» A. 10 | |
| Explanation: given the point is in 2nd quadrant. the distance from the side view of point to line of vertical reference will be the | |
| 173. |
A point is in 2nd quadrant which is 5 meters away from horizontal and 3 meters away from profile plane. Orthographic projections are drawn. What is the distance from the top view to xy reference line? |
| A. | 5 |
| B. | 3 |
| C. | 8 |
| D. | can’t found |
| Answer» D. can’t found | |
| Explanation: given the point is in 2nd quadrant. the xy reference line is between the vertical plane and horizontal plane but distance from a vertical point is not given in question so we can’t found some given information. | |
| 174. |
A point is in 2nd quadrant which is 7 meters away from horizontal and 2 meters away from profile plane. Orthographic projections are drawn. What is the distance from the front view to xy reference line? |
| A. | 7 |
| B. | 2 |
| C. | 5 |
| D. | 9 |
| Answer» A. 7 | |
| Explanation: given the point is in 2nd quadrant. the distance from front view is given by distance between point and horizontal plane here it is given 7 meters. and distance from vertical reference will be 2 meters. | |
| 175. |
A point is in 2nd quadrant which is 8 meters away from vertical and 6 meters away from profile plane. Orthographic projections are drawn. What is the distance from the side view to vertical reference line? |
| A. | 8 |
| B. | 6 |
| C. | 2 |
| D. | can’t found |
| Answer» A. 8 | |
| Explanation: given the point is in 2nd quadrant. the distance from the side view is given by distance between point and vertical plane here it is given 8 meters. and the distance from front view will be 6 meters. | |
| 176. |
Two points are placed in 3rd quadrant of projection planes such that the line joining the points is perpendicular to vertical plane the side view and top view will be |
| A. | single point, two points |
| B. | two points, single point |
| C. | single point, single point |
| D. | two points, two points |
| Answer» D. two points, two points | |
| Explanation: here given the two points such that the joining line is perpendicular to vertical plane in 3rd quadrant asked side view and top view. the views in any quadrant will remain same but the relative positions in projection will change accordingly the quadrant. | |
| 177. |
A point is 7 units away from the vertical plane and 3 units away from profile plane and 3 units away from horizontal plane in 3rd quadrant then the projections are drawn on paper the distance between the front view and top view of point is |
| A. | 10 units |
| B. | 8 units |
| C. | 9 units |
| D. | 5 units |
| Answer» A. 10 units | |
| Explanation: since the point is 3 units away from the horizontal plane the distance from the point to xy reference line will be 3 units. | |
| 178. |
A point is 9 units away from the vertical plane and 5 units away from profile plane and 4 units away from horizontal plane in 3rd quadrant then the projections are drawn on paper the distance between the side view and front view of point is |
| A. | 12 units |
| B. | 14 units |
| C. | 10 units |
| D. | 8 units |
| Answer» B. 14 units | |
| Explanation: since the point is 5 units away from the profile plane the distance from the point to a reference line will be 5 units. and then the point is at distance of 9 units from the vertical plane the distance from reference line and point will be 9, sum is 14. | |
| 179. |
A point is 7 units away from the vertical plane and 5 units away from profile plane and 7 units away from horizontal plane in 3rd quadrant then the projections are drawn on paper the distance between the front view and side view of point is |
| A. | 10 |
| B. | 5 |
| C. | 9 |
| D. | 12 |
| Answer» D. 12 | |
| Explanation: since the point is 5 units away from the profile plane the distance from the point to a reference line will be 5 units. and then the point is at distance of 7 units from the profile plane the distance from reference line and point will be 7 units, sum is 12. | |
| 180. |
8 PROJECTION OF POINTS IN THIRD QUADRANT |
| A. | 29 units |
| B. | 20 units |
| C. | 21 units |
| D. | 17 units |
| Answer» C. 21 units | |
| Explanation: since the point is 12 units away from the profile plane the distance from the point to a reference line will be 12 units. and then the point is at distance of 8 units from profile plane the distance from reference line and point will be 8 units, sum is 20. | |
| 181. |
A point is 20 cm away from the vertical plane and 8 units away from profile plane and 17 cm away from horizontal plane in 3rd quadrant then the projections are drawn on paper the shortest distance from top view and side view of point is |
| A. | 37 |
| B. | 44.65 |
| C. | 46.40 |
| D. | 37.53 |
| Answer» C. 46.40 | |
| Explanation: since here distance from side view and top view is asked for that we need the distance between the front view and side view (8+20); front view and top view (17+20)and these lines which form perpendicular to each other gives needed distance, answer is square root of squares of both the distances √(282+372 ) =46.40 units. | |
| 182. |
If a point P is placed in between the projection planes in third quadrant. The distance from side view to reference line towards front view and the distance between top view and reference line towards top view will be same. |
| A. | true |
| B. | false |
| Answer» A. true | |
| Explanation: the projection will be drawn by turning the other planes parallel to vertical plane in clockwise direction along the lines of intersecting of planes. and so as we fold | |
| 183. |
A point is 2 m away from the vertical plane and 1 m away from profile plane and 9 m away from horizontal plane in 3rd quadrant then the projections are drawn on paper the distance between the side view and top view of point is |
| A. | 21 |
| B. | 14.86 |
| C. | 11.4 |
| D. | 10.4 |
| Answer» B. 14.86 | |
| Explanation: since here distance from side view and top view is asked for that we need the distance between the front view and side view (1+9); front view and top view (9+2)and these lines which form perpendicular to each other gives needed distance, answer is square root of squares of both the distances | |
| 184. |
A point is 6 units away from the vertical plane and profile plane and 10 units away from horizontal plane in 3rd quadrant then the projections are drawn on paper the distance between the side view and top view of point is |
| A. | 15 |
| B. | 16 |
| C. | 12 |
| D. | 20 |
| Answer» D. 20 | |
| Explanation: since here distance from side view and top view is asked for that we need the distance between the front view and side view (6+6); front view and top view (10+6)and these lines which form perpendicular to each other gives needed distance, answer is square root of squares of both the distances √(122+162 ) = 20 units. | |
| 185. |
A point is 15 cm away from the vertical plane and 10 cm away from profile plane and horizontal plane in 3rd quadrant then the projections are drawn on paper the distance between the front view and top view of point is |
| A. | 27 cm |
| B. | 15 cm |
| C. | 12 cm |
| D. | 25 cm |
| Answer» D. 25 cm | |
| Explanation: since the point is 10 cm away from the horizontal plane the distance from the point to xy reference line will be 10 cm. and then the point is at distance of 15 cm from the vertical plane the distance from reference line and point will be 15, sum is 25 cm. | |
| 186. |
A point is 6 m away from the vertical plane and profile plane 5 m away from horizontal plane in 3rd quadrant then the projections is drawn on paper the distance between the front view and side view of point is |
| A. | 27 |
| B. | 15 |
| C. | 12 |
| D. | 24 |
| Answer» C. 12 | |
| Explanation: since the point is 6 m away from the profile plane the distance from the point to xy reference line will be 6 m. and then the point is a distance of 6 from the profile plane the distance from reference line and point will be 6, sum is 12. | |
| 187. |
A point is 50 cm away from the vertical plane and horizontal plane 80 cm away from profile plane in 3rd quadrant then the projections is drawn on paper the distance between the front view and top view of point is |
| A. | 130 |
| B. | 100 |
| C. | 160 |
| D. | 0 |
| Answer» B. 100 | |
| Explanation: since the point is 50 cm away from the horizontal plane the distance from the point to xy reference line will be 50 cm. and then the point is at distance of 50 cm from the vertical plane the distance from reference line and point will be 50 cm, sum is 100 cm. | |
| 188. |
A point is 5 units away from the vertical plane and horizontal plane 4 units away from profile plane in 3rd quadrant then the projections are drawn on paper the distance between the side view and top view of point is |
| A. | 13.45 |
| B. | 12.72 |
| C. | 19 |
| D. | 12.04 |
| Answer» A. 13.45 | |
| Explanation: since here distance from side view and top view is asked for that we need the distance between the front view and side view (4+5); front view and top view (5+5)and these lines which form perpendicular to each other gives needed distance, answer is square root of squares of both the distances √(102+92 | |
| 189. |
A point is 3 m away from the vertical plane and horizontal planes in 3rd quadrant then the projections are drawn on paper the distance between the side view and vertical reference line? |
| A. | 3 |
| B. | 0 |
| C. | can’t found |
| D. | 6 |
| Answer» A. 3 | |
| Explanation: the side view’s distance from reference line will be the perpendicular distance from the vertical plane and front view’s distance from reference line will be | |
| 190. |
A point is 3 m away from the vertical plane and 7 m away from profile plane in 3rd quadrant then the projections are drawn on paper the distance between the side view and vertical reference line? |
| A. | 6 |
| B. | 3 |
| C. | 14 |
| D. | 7 |
| Answer» B. 3 | |
| Explanation: the side view’s distance from vertical reference line will be the perpendicular distance from vertical plane and top view’s distance from a vertical reference line will be the perpendicular distance from profile plane. | |
| 191. |
A point is in 4th quadrant 10 units away from the horizontal plane and 20 units away from the vertical plane. Orthographic projection is drawn. What is the distance from point of front view to reference line, top view point to reference line? |
| A. | 20, 10 |
| B. | 10, 20 |
| C. | 0, 20 |
| D. | 10, 0 |
| Answer» B. 10, 20 | |
| Explanation: given object is point placed in 4th quadrant the top view gives the distance from the vertical plane (20) and front view gives the distance from horizontal plane (10) both are placed overlapped in orthographic projection since the object is placed in 4th quadrant. | |
| 192. |
A point is in 4th quadrant, 5 m away from the vertical plane, 1 m away from the horizontal plane and 8 units away from the profile plane. Orthographic projection is drawn. What is the distance from point of front view to point of top view? |
| A. | 6 |
| B. | 4 |
| C. | 10 |
| D. | 2 |
| Answer» B. 4 | |
| Explanation: as the point is in 4th quadrant while drawing the projections the planes should rotate along the hinges such that the plane with top view overlaps the front view. so the distance between them is difference of distances from respective planes that is 5 (5- | |
| 193. |
A point in 4th quadrant is 30 mm away from both the horizontal plane and vertical plane and orthographic projections are drawn. The distance between the points formed by front view and top view is |
| A. | 0 |
| B. | 30 |
| C. | 15 |
| D. | 15+ distance from profile |
| Answer» A. 0 | |
| Explanation: given the point is in the 4th quadrant. while drawing orthographic projections the front view and top view overlaps and also the distance of point is same from planes of projections so the distance between them is zero. | |
| 194. |
A point in 4th quadrant is 13 inches away from the horizontal plane and 10 inches away from both the vertical plane and profile plane. Orthographic projections are drawn find the distance from side view and front view. |
| A. | 10 |
| B. | 13 |
| C. | 20 |
| D. | 26 |
| Answer» C. 20 | |
| Explanation: given the point is in 4th quadrant. the front view and side view lie parallel to the horizontal plane when orthographic projections are drawn. the distance from side view to vertical reference is 10 and distance from front view to profile plane is 10. sum is 10+10= 20 inches. | |
| 195. |
A point in 4th quadrant is 18 units away from the horizontal plane and vertical plane and 17 units away from both the profile plane. Orthographic projections are drawn find the distance from side view and front view. |
| A. | 1 |
| B. | 24 |
| C. | 35 |
| D. | 36 |
| Answer» C. 35 | |
| Explanation: given the point is in 4th quadrant. the front view and side view lie parallel to the horizontal plane when orthographic projections are drawn. the distance from side view to vertical reference is 12 and distance from front view to profile plane is 13. sum is 18 + 17 =35 units. | |
| 196. |
A point in 4th quadrant is 8 inches away from the horizontal plane and 20 inches away from both the vertical plane and profile plane. Orthographic projections are drawn find the distance from side view and top view. |
| A. | 41.76 |
| B. | 20 |
| C. | 43.08 |
| D. | 16 |
| Answer» A. 41.76 | |
| Explanation: given the point is in the 4th quadrant. since here distance from side view and top view is asked for that we need the distance between the front view and side view (20+20); front view and top view (20-8) and these lines which form perpendicular to each other gives needed distance, answer is | |
| 197. |
A point in 4th quadrant is 15 cm away from the vertical plane and 10 cm away from the horizontal plane, orthographic projections are drawn. What is the distance from side view of point to line of vertical reference? |
| A. | 10 |
| B. | 15 |
| C. | 25 |
| D. | can’t found |
| Answer» B. 15 | |
| Explanation: given the point is in 4th quadrant. the distance from the side view of point to line of vertical reference will be the distance from the point to the vertical plane in plane of projection that is as given 15 cm. | |
| 198. |
A point is in 4th quadrant which is 15 inches away from horizontal and 30 inches away from profile plane. Orthographic projections are drawn. What is the distance from the top view to xy reference line? |
| A. | 5 |
| B. | 3 |
| C. | 8 |
| D. | can’t found |
| Answer» D. can’t found | |
| Explanation: given the point is in 4th quadrant. the xy reference line is between the vertical plane and horizontal plane but distance from vertical point is not given in question so we can’t found some given information. | |
| 199. |
A point is in 4th quadrant which is 17 dm away from horizontal and 12 dm away from profile plane. Orthographic projections are drawn. What is the distance from the front view to xy reference line? |
| A. | 17 |
| B. | 12 |
| C. | 5 |
| D. | 29 |
| Answer» A. 17 | |
| Explanation: given the point is in 4th quadrant. the distance from front view is given by distance between point and horizontal plane here it is given 17 dm. and distance from vertical reference will be 12 dm. | |
| 200. |
A point is in 4th quadrant which is 18 mm away from vertical and 20 mm away from profile plane. Orthographic projections are drawn. What is the distance from the side view to vertical reference line? |
| A. | 18 |
| B. | 2 |
| C. | 20 |
| D. | can’t found |
| Answer» A. 18 | |
| Explanation: given the point is in 4th quadrant. the distance from side view is given by distance between point and vertical plane here it is given 18 mm. and distance from front view will be 20 mm. | |
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