Linear Measurement MCQ Quiz in मल्याळम - Objective Question with Answer for Linear Measurement - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക
Last updated on Apr 8, 2025
Latest Linear Measurement MCQ Objective Questions
Top Linear Measurement MCQ Objective Questions
Linear Measurement Question 1:
The length of a line measured by a 30 m chain was found to be 450 m. If the chain was 0.2 links short, then find the true length of the line.
Answer (Detailed Solution Below)
Linear Measurement Question 1 Detailed Solution
Explanation:
As per Indian Standard, 30 m chain has 150 links, and the length of one link is 20 cm.
Let True length of the line = L
True or designated length of chain (l) = 150 links
Wrong length of the line measured (L’) = 450 m
Wrong length of the chain (l’) = 150 - 0.2 = 149.8 links
l’ × L’ = l × L
\(L = \frac{{149.80 \times 450}}{{150}} = 449.4\;m\)
Additional Information
Following are the various types of chain in common use in surveying:
| Type | Length | No. of links | Length of one link |
| Metric chain | 20 m or 30 m | 100 or 150 | 20 cm |
| Engineering chain | 100 ft | 100 | 1 ft |
| Gunter chain | 66 ft | 100 | 0.66 ft |
| Revenue chain | 33 ft | 16 | \(2\frac{1}{{16}}\) |
Linear Measurement Question 2:
What is the true area (in acres), if the area calculated by a chain which is found to be 0.8 m too long is 100 acres? (Take chain length = 100 m)
Answer (Detailed Solution Below)
Linear Measurement Question 2 Detailed Solution
Concept:
True area = measured area × (R.F.)
For area,
\({\rm{Representative\;factor\;}}\left( {{\rm{R}}.{\rm{F}}.} \right) = {\left( {\frac{{{\rm{true\;length}}}}{{{\rm{standard\;length}}}}} \right)^2}\)
Calculation:
Given:
Standard chain length = 100 m, (a 100m chain). Area measured = 100 acres.
Now,
Actual length = 100 + 0.8 = 100.8 m
\({\rm{Representative\;factor\;}}\left( {{\rm{R}}.{\rm{F}}.} \right) = {\left( {\frac{{{\rm{true\;length}}}}{{{\rm{standard\;length}}}}} \right)^2}\)
\(\left( {{\rm{R}}.{\rm{F}}.} \right) = {\left( {\frac{{100.8}}{{100}}} \right)^2}\)=1.01606
True area = 100 × 1.01606
True area = 101.606 acres.
Tips and tricks:
For finding out the true length measured by a false chain-
True length × True chain length = Measured length × Actual length,
Linear Measurement Question 3:
The distance between two points measured using 20 m chain was recorded as 327 m. It was found that the chain is 3 cm too long. The true length of the line is
Answer (Detailed Solution Below)
Linear Measurement Question 3 Detailed Solution
Concept:
\(True\;length\;line = \frac{{l'}}{l} \times measured\;length\\ \)
Where,
l' is corrected length of chain
l is actual length of chain
Calculation:
True length of line (L) = ?
Actual length of chain (l) = 20 m
Measured length of line (L') = 327 m
Corrected length of chain (l') = 20.03 m
So, we know
\(True\;length\;line = \frac{{l'}}{l} \times measured\;length\\ \)
\(\Rightarrow L = \frac{{20.03}}{{20}} \times 327\)
∴ L= 327.49 m
Linear Measurement Question 4:
Among the following corrections applied for tape measurement, identify the correction which is NOT always negative in sign.
Answer (Detailed Solution Below)
Linear Measurement Question 4 Detailed Solution
Concept:
Types of Error and Correction:
|
Error due to |
Error |
Correction |
|
Standardization |
+ve or -ve |
-ve or +ve respectively |
|
Temperature |
+ve or -ve |
-ve or +ve respectively |
|
Pull |
+ve or -ve |
-ve or +ve respectively |
|
Slope |
Always +ve |
Always -ve |
|
Sagging |
Always +ve |
Always -ve |
|
Wrong Alignment |
Always +ve |
Always -ve |
Pull correction may be Positive or negative. Hence, option 2 is incorrect.
Linear Measurement Question 5:
The shrunk scale and original of a plan are 1 cm = 20 m and 1 cm = 19 m respectively. Then the values of shrinkage factor and RF of shrunk scale is
Answer (Detailed Solution Below)
Linear Measurement Question 5 Detailed Solution
Concept:
Representative Fraction (RF): The scale expressed in the same units. For example, 1 cm = 50 m is represented in RF as 1: 5000 or 1/5000. Here 50 m is converted as 5000 cm.
Shrinkage factor or Shrinkage ratio = Shrunk scale / Original scale or Shrunk length / Actual length
Note: The Shrinkage ratio is always less than one.
Calculation:
Given,
Original Scale, 1 cm = 19 m
Shrunk Scale, 1 cm = 20 m
Therefore,
RF of shrunk scale = 1: 2000 or 1 / 2000
RF of original scale = 1: 1900 or 1 / 1900
Shrinkage Factor = Shrunk scale / Original scale = (1/2000) / (1/1900) = 19/20
Shrinkage Factor = 19/20 and RF of shrunk scale = 1/2000
Linear Measurement Question 6:
The area of a plan of an old survey plotted on a sheet is found to have shrunk so that a line originally 10 cm long now measures 9.7 cm only. Calculate the shrinkage factor.
Answer (Detailed Solution Below)
Linear Measurement Question 6 Detailed Solution
Explanation:
Original length of line = 10 cm
Length after shrinkage of line = 9.7 cm
\({\bf{Shrinkage}}{\rm{ }}{\bf\ {Factor}} = \frac{{{\bf{Shrunk}}\ {\rm{ }}{\bf{length}}}}{{{\bf{Original}}{\rm{ }}{\bf\ {length}}}}\)
\({\bf{Shrinkage}}{\rm{ }}{\bf{Factor}} = \frac{{9.7}}{{10}}\)
Shrinkage Factor = 0.97
Linear Measurement Question 7:
Offsets are of two types, _______.
Answer (Detailed Solution Below)
Linear Measurement Question 7 Detailed Solution
Explanation:
Offsets:
- The distance measured right or left of the chain line to locate details like boundaries, culverts, etc. are called offsets.
- Offset which can be judged by the naked eye or offset less than 15 m is called short offset and offset greater than 15 m is called long offset.
- Most commonly short offsets are preferred.
These are of two types:
(i) Perpendicular offset at right angles to the chain line
(ii) oblique offsets not at right angles to the chain lineLinear Measurement Question 8:
Identify the surveying instrument shown in the image:
Answer (Detailed Solution Below)
Linear Measurement Question 8 Detailed Solution
Explanation:
The various instruments used in surveying and their purpose is given below in the tabulated form:
|
Survey Instrument |
Purpose |
|
Pegs |
To mark survey stations and endpoints of survey lines on the ground. |
|
Arrows |
To mark the position of the end of the chain or tape on the ground. |
| Telescopic Alidade | A telescopic alidade is used when it is required to take inlined sights |
|
Ranging Rods |
For locating a number of points on a long survey line. |
|
Offset Rods |
To set out offset lines at right angles. |
|
Prism Square |
Setting out right angles |
|
Plumb Bob |
To indicate whether the line is vertical or not. |
|
Clinometers |
To measure the slope of the ground. |
|
Optical Square |
To set out right angles |
|
Prismatic Square |
Advanced version of Optical square and used to set out right angles |
|
French Cross Staff |
Used to set out either 45° or 90°. |
|
Open Cross staff |
Type of cross-staff which is also used to set out 90°. |
|
Theodolite |
For measurement of all horizontal and vertical angles |
|
Dumpy Level |
Measurement of Angles and Elevation |
|
Sextant |
Used to measure the angle between an astronomical object and the horizon for the purposes of celestial navigation. |
Linear Measurement Question 9:
What is the degree of accuracy in terms of pacing type of measurement?
Answer (Detailed Solution Below)
Linear Measurement Question 9 Detailed Solution
Concept:
Measurement by Pacing:
- Pacing is a rapid means of approximately checking more precise measurements of distance.
- Pacing over rough country may be done with a precision of one in one hundred(1 in 100).
- In average conditions, a person with some experience should have little difficulty in pacing with a precision of one in two hundred.
- Obviously, there is not much precision in this method and the procedure provides only an approximation of distance.
- The natural pace of each individual normally varies from 2 ½ to 3 ft.
- A convenient relation between the pace and the foot is 40 paces approximately equal 100 ft.
- Technicians involved in surveying standardize their pace by walking over known distances on level, sloping, and uneven ground.
Linear Measurement Question 10:
A tape of length l and weight w kg/m is suspended at its ends with a pull of p kg. the sag correction is:
Answer (Detailed Solution Below)
Linear Measurement Question 10 Detailed Solution
Correction for sag is given by:
\({C_s} = \; - \frac{{{w^2}{l^3}}}{{24\;{P^2}}} = \; - \frac{{{W^2}{l}}}{{24\;{P^2}}}\)
W is the total weight of tape = w × L
Where,
W = weight of tape per unit length (N/m)
l = length of tape suspended between supports (m)
p = pull applied (N)