Linear Measurement MCQ Quiz in मल्याळम - Objective Question with Answer for Linear Measurement - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Last updated on Apr 8, 2025

നേടുക Linear Measurement ഉത്തരങ്ങളും വിശദമായ പരിഹാരങ്ങളുമുള്ള മൾട്ടിപ്പിൾ ചോയ്സ് ചോദ്യങ്ങൾ (MCQ ക്വിസ്). ഇവ സൗജന്യമായി ഡൗൺലോഡ് ചെയ്യുക Linear Measurement MCQ ക്വിസ് പിഡിഎഫ്, ബാങ്കിംഗ്, എസ്എസ്‌സി, റെയിൽവേ, യുപിഎസ്‌സി, സ്റ്റേറ്റ് പിഎസ്‌സി തുടങ്ങിയ നിങ്ങളുടെ വരാനിരിക്കുന്ന പരീക്ഷകൾക്കായി തയ്യാറെടുക്കുക

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.

  1. 448.0 m
  2. 449.4 m
  3. 449.8 m
  4. 449.6 m

Answer (Detailed Solution Below)

Option 2 : 449.4 m

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)

  1. 100.8
  2. 99.2
  3. 98.4
  4. 101.6

Answer (Detailed Solution Below)

Option 4 : 101.6

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 

  1. 326.55 m
  2. 327.49 m
  3. 327.55 m
  4. 326.49 m

Answer (Detailed Solution Below)

Option 2 : 327.49 m

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.

  1. Slope of tape
  2. Pull
  3. Sag
  4. Wrong alignment

Answer (Detailed Solution Below)

Option 2 : Pull

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

  1. 19/20 and 1/2000
  2. 19/20 and 1/1900
  3. 20/19 and 1/2000
  4. 20/19 and 1/1900

Answer (Detailed Solution Below)

Option 1 : 19/20 and 1/2000

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.

  1. 0.97
  2. 9.7
  3. 97
  4. 1.03

Answer (Detailed Solution Below)

Option 1 : 0.97

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, _______.

  1. Horizontal and vertical
  2. Straight and oblique
  3.  Horizontal and curved oblique
  4. Perpendicular and oblique

Answer (Detailed Solution Below)

Option 4 : Perpendicular and oblique

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 line

Linear Measurement Question 8:

Identify the surveying instrument shown in the image:

F1 Shubham Ravi 21.12.21 D11

  1. Pegs
  2. Offset rods
  3. Ranging rod
  4. Arrows

Answer (Detailed Solution Below)

Option 4 : Arrows

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?

  1. 1 in 140
  2. 1 in 100
  3. 1 in 120
  4. 1 in 160

Answer (Detailed Solution Below)

Option 2 : 1 in 100

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:

  1. \(\frac{{{l^2}{w^2}}}{{24{p^2}}}\)
  2. \(\frac{{{l^2}{w^3}}}{{24{p^2}}}\)
  3. \(\frac{{{l^3}{w^2}}}{{24{p^2}}}\)
  4. \(\frac{{{l^2}{w^2}}}{{24{p^3}}}\)

Answer (Detailed Solution Below)

Option 3 : \(\frac{{{l^3}{w^2}}}{{24{p^2}}}\)

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)

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