1. Horizontal angle with the Arbitrary Meridian through one of the extremities of the line is called _______
a) True bearing
b) Magnetic Bearing
c) Arbitrary bearing
d) Magnetic Declination
Explanation: Arbitrary bearing of a line is the horizontal angle which it makes with the arbitrary meridian through one of the extremities of the line. Angle measured always keeping magnetic north as reference is called magnetic bearing.
2. The Magnetic Bearing of a line is 48°24′. Calculate the true bearing if the magnetic declination is 5°38′ East.
a) 54°02′
b) 44°02′
c) 54°22′
d) 45°02′
Explanation: Magnetic Declination is the horizontal angle between true meridian and magnetic meridian. Declination = +5°38’, magnetic bearing = 48°24’, then here, true bearing is sum of both i.e 48″24′ + 5°38′ = 54°02′.
3. Horizontal angle between true meridian and magnetic meridian is __________
a) True bearing
b) Magnetic Bearing
c) Arbitrary bearing
d) Magnetic Declination
Explanation: Magnetic Declination is a horizontal angle between true meridian and magnetic meridian. Arbitrary bearing of a line is the horizontal angle which it makes with the Arbitrary Meridian through one of the extremities of the line.
4. When magnetic meridian is right side to true meridian, then Magnetic Declination is said to be _________
a) Eastern
b) Western
c) Southern
d) Northern
Explanation: When magnetic meridian is right side to true meridian, then Magnetic Declination is said to be positive or eastern. Magnetic Declination is horizontal angle between true meridian and magnetic meridian.
5. When magnetic meridian is left side to true meridian, then Magnetic Declination is said to be ________
a) Eastern
b) Western
c) Southern
d) Northern
Explanation: When magnetic meridian is left side to true meridian, then Magnetic Declination is said to be negative or western. Magnetic Declination is horizontal angle between true meridian and magnetic meridian.
6. The length of a line measured with a 20 m chain was found to be 250 m. Calculate the true length of the line if the chain was 10 cm too long.
a) 252.25 m
b) 251.25 m
c) 225.25 m
d) 221.25 m
Explanation: Incorrect length of the chain is 20 + 10/100, ie 20.1 m. Measured length is 250, hence true length of the line is 250 × (20.1/20)=251.25 m.
7. The length of a survey line was measured with a 20 m chain and was found to be equal to 1200 m. If the length again measured with 25 m chain it is 1212 m. On comparing the 20 m chain with the test gauge, it was found to be 1 decimeter too long. Find the actual length of 25 m chain used.
a) 22.25 m
b) 21.64 m
c) 24.25 m
d) 24. 88 m
Explanation: Incorrect length of 20 m line is 20+0.10 = 20.10 m. True length of line = 1200×(20.10/20) = 1206 m. Actual or True length of 25 m chain = (1206×25)/1212 = 24.88 m.
8. A surveyor measured the distance between two points on the plan drawn to a scale of 1 cm is equal 40 m and the result was 468 m. But, actual scale is 1 cm = 20 m. Find the true distance between the two points.
a) 992 m
b) 936 m
c) 987 m
d) 967 m
Explanation: Distance between two points measured with a scale of 1 cm to 20 m is 468/20 = 23.4 cm. Actual scale of a plan is 1 cm = 40 m. True distance between the points is 23.4 × 40 = 936 m.
9. If L is true length of chain and L’ is incorrect length of the chain the correction to area A is _________
(Where ∆L/L = e, e is small and A’ is measured area)
a) 1+2e A’
b) (1+2e)/A’
c) (1+2e) x A’
d) (1+ e)xA’
Explanation: By using A=A'(L’/L)2 and L’/L=(L+∆L)/L=1+e where e = ∆L/L.
10. If L is true length of chain and L’ is incorrect length of the chain the correction to Volume V is _______
(Where ∆L/L = e, e is small and V’ is measured area)
a) 1+3e)+ V’
b) (1+3e)/V’
c) (1+3e)xV’
d) (1+ e) ×V’
Explanation: By using V = V'(L’/L)3, e = ∆L/L and L’/L = (L+∆L)/L = 1+e. Then V = V’ (1+e)3 here e is small so V = (1+3e)xV’.