# Surveying Questions and Answers – Chain Surveying – Obstacles in Chaining

This set of Surveying Questions and Answers for Aptitude test focuses on “Chain Surveying – Obstacles in Chaining”.

1. How many kinds of obstacles of chaining are there?
a) 2
b) 3
c) 4
d) 5

Explanation: Obstacles of chaining are of three kinds. They are obstacles to ranging, obstacles to chaining, obstacles to both chaining and ranging.

2. Which of the following is not one among the three major kinds of obstacles of chaining?
a) obstacles to ranging
b) obstacles to chaining
c) obstacles to levelling
d) obstacles to ranging and chaining

Explanation: Obstacles to levelling is not a kind of obstacles to chaining. Obstacles of chaining are of three kinds. They are obstacles to ranging, obstacles to chaining, obstacles to both chaining and ranging.

3. Both ends of the lines may be visible from intermediate points on the line. This case comes under which among the three kinds of obstacles to chaining?
a) obstacles to ranging but not chaining
b) obstacles to chaining but not ranging
c) obstacles to levelling
d) obstacles to ranging and chaining

Explanation: Obstacles to ranging but not ranging is a type of obstacle, in which the ends are not Intervisible, is quite expected in a flat country. There may be two cases of this obstacle, both ends of the lines may be visible from intermediate points on the line and both ends of the line may not be visible from intermediate points on the line.

4. When it is possible to chain round the obstacle, i.e a pond, hedge etc. This case comes under which among the three kinds of obstacles to chaining?
a) obstacles to ranging but not chaining
b) obstacles to chaining but not ranging
c) obstacles to levelling
d) obstacles to ranging and chaining

Explanation: There may be two cases of this obstacle i.e obstacle to chaining but not ranging, when it is possible to chain round the obstacle, i.e a pond, hedge etc and when it is not possible to chain round the obstacle e.g. a river.

5. Both ends of the line may not be visible from intermediate points on the line. This case comes under which among the three kinds of obstacles to chaining?
a) obstacles to ranging but not chaining
b) obstacles to chaining but not ranging
c) obstacles to levelling
d) obstacles to ranging and chaining

Explanation: There may be two cases of this obstacle, both ends of the lines may be visible from intermediate points on the line and both ends of the line may not be visible from intermediate points on the line.
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6. When it is not possible to chain round the obstacle e.g. a river. This case comes under which among the three kinds of obstacles to chaining?
a) obstacles to ranging but not chaining
b) obstacles to chaining but not ranging
c) obstacles to levelling
d) obstacles to ranging and chaining

Explanation: There may be two cases of this obstacle i.e obstacle to chaining but not ranging, when it is possible to chain round the obstacle, i.e a pond, hedge etc and when it is not possible to chain round the obstacle e.g. a river.

7. To continue a survey line AB past an obstacle, a line BC 100 m long was set out perpendicular to AB and from C angles BCD and BCE were set out at 60° and 45° respectively. Determine the lengths which must be chained off along CD in order that ED may be in AB produced?
a) 100 m
b) 200 m
c) 300 m
d) 400 m

Explanation: Here angle ABC is 90°. From, ∆ BCD, CD = BC sec 60° = 100 × 2 = 200m.

8. To continue a survey line AB past an obstacle, a line BC 100 m long was set out perpendicular to AB and from C angles BCD and BCE were set out at 60° and 45° respectively. Determine the lengths which must be chained off along CE in order that ED may be in AB produced?
a) 141.42 m
b) 282.84 m
c) 140.14 m
d) 267.7 m

Explanation: Angle ABC is 90°. From ∆BCE, and CE = BC sec 45° = 100 × 1.4142 = 141.42 m.

9. To continue a survey line AB past an obstacle, a line BC 300 m long was set out perpendicular to AB and from C angles BCD and BCE were set out at 60° and 45° respectively. Determine the obstructed length BE?
a) 250 m
b) 600 m
c) 452.28 m
d) 300 m

Explanation: Here angle ABC is 90°. BE = BC tan 45° = 300 × 1 = 300 m.

10. If we select two points A and B on either side of the obstacle and equal perpendiculars AC and BD are set out. Then AB is equal to?
a) AC
b) CD
c) DA
d) BD

Explanation: Since AC and BD are two perpendiculars set either side of the obstacle of equal length. Therefore CD is parallel and equal to AB. Therefore, AB = CD.

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