Class 11 Physics MCQ – Units and Measurements -

Practice Questions: Class 11 Physics MCQ – Units and Measurements

1. The number of base units for measuring physical quantities according to the SI system are ____
a) 1
b) 10
c) 7
d) 8
Answer: c
Explanation: There are 7 base units for measuring physical quantities according to the SI system. They are kilogram (kg) for mass, second (s) for time, kelvin (K) for temperature, ampere (A) for electric current, mole (mol) for the amount of a substance, candela (cd) for luminous intensity, and meter (m) for distance.

2. What is the unit for measuring intensity of light?
a) Candela
b) Lightyear
c) Meter
d) Mol

Answer: a
Explanation: The unit for measuring luminous intensity is Candela (cd). Lightyear and Meter are units of distance. Mol is unit of amount of a substance.

3. What is the full form of SI?
a) Standard International
b) System International
c) Scientific International
d) Science International

Answer: b
Explanation: The full for of SI is System International. It was first adopted in 1960 in the General Conference on Weights and Measures.

4. What is the standard unit of measurement of Force?
a) Newton
b) Joule
c) Meter
d) Newton-meter

Answer: a
Explanation: Newton is the SI unit for force. Joule and Newton-meter are the SI units for Energy and work. Meter is the SI unit for distance.

5. 89 Mega Joules can also be expressed as _____
a) 8900 Joules
b) 89000000 Joules
c) 89000 Joules
d) 890000 Joules

Answer: b
Explanation: 1 Mega Joule = 1000000 Joules. Therefore 89 Mega Joules = 89000000 Joules.

6. Convert 5 N to Dynes.
a) 500000 Dynes
b) 500 Dynes
c) 5 Dynes
d) 0.5 Dynes

Answer: a
Explanation: One Newton equals to 100000 Dynes, hence 5 N equal to 500000 Dynes. Dyne is unit of force in the CGS system.

7. Which of the following is not a system of units?
a) MKS
b) CGS
c) SI
d) Decibel

Answer: d
Explanation: MKS stands for meter-kilogram-second. CGS stands for centimeter-gram-second. SI stands for System international. Decibel is a unit of sound intensity.

8. What does a Voltmeter measure?
a) Voltage
b) Current
c) Length
d) Speed

Answer: a
Explanation: A voltmeter is used for measuring voltage. An ammeter is used for measuring current. A combination of both can be used to determine the resistance in a simple circuit.

9. What is the unit of measurement of solid angles?
a) Steradians
b) Degrees
c) Radians
d) Grades

Answer: a
Explanation: AS plane angles are measured in radians, in the same way solid angles are measured in steradians. A solid angle can be interpreted as the 3-dimensional form of a plane angle.

10. Which of the following system matches with the SI unit system?
a) FPS
b) MKS
c) CGS
d) American system

Answer: b
Explanation: The MKS system uses the same fundamental units as the SI system. It uses meter for distance, kilogram for mass, and second for time. These are all present as fundamental units in the SI system.

11. Which of the following is a unit of temperature?
a) Degree
b) Meter
c) Second
d) Fahrenheit

Answer: d
Explanation: Fahrenheit is the unit for temperature. Other units include Celsius and Kelvin.

12. Distance between two cities, in standard units will be measured in ______
a) Meter
b) Kilometer
c) Mile
d) Centimeter

Answer: a
Explanation: The standard unit of measurement is meter; hence the answer is meter. The distance between two cities is usually measured in kilometer but the standard unit is meter.

13. Which of the following devices cannot measure distance?
a) Vernier Calipers
b) Micrometer
c) Ruler
d) Protractor

Answer: d
Explanation: Protractor is used to measure angle. Vernier Calipers and Micrometer can measure distance very accurately. Ruler can measure accuracy up to one tenth of a millimeter.

14. How many inches are there in 1 yard?
a) 12
b) 36
c) 6
d) 18

Answer: b
Explanation: 36 inches make one yard. One Feet is made up of 12 inches. One-inch equals to 25.4 mm.

15. How many kilometers make one mile?
a) 1
b) 2.5
c) 0.5
d) 1.6

Answer: d
Explanation: One Kilometer is equal to 0.621371 miles. Inversely, one mile is equal to 1.60934 kilometers. Hence the closest option to the actual answer is 1.6.

16. How many kilometers make one nautical mile?
a) 1
b) 1.536
c) 1.852
d) 1.756

Answer: c
Explanation: One Nautical mile consists of 1.852 Kilometers. This unit is often used in air and water navigations.

17. Usually, what is the least count of a screw gauge?
a) 0.01 cm
b) 0.001 cm
c) 0.1 cm
d) 1 mm

Answer: b
Explanation: Usually the least count of a screw gauge is 0.001 cm or 0.01 mm. For more accurate results one might use a micrometer.

18. How many kilometers make one light year?
a) 9.4607 × 10¹² km
b) 9.4607 × 10⁹ km
c) 9.4607 × 10¹¹ km
d) 9.4607 × 10¹⁰ km

Answer: a
Explanation: One Light year is equal to 9.4607 × 10¹² km. In miles, it can be represented as nearly 6 million million miles.

19. Interplanetary distances are measured in ____
a) Light year
b) Kilometer
c) Megameter
d) Watt

Answer: a
Explanation: Such huge distances are measured in Light years. One light year is the distance travelled by light in one year.

20. In an experiment, it is found that the experimental value is very close to actual value, hence the experimental value can be called _____
a) Accurate
b) Precise
c) Suitable
d) Mean

Answer: a
Explanation: Accuracy is the degree to which a value is near to the actual or standard value. Hence if the experimental value is very close to the actual value, the measurement can be called accurate.

21. What is the reason for the occurrence of systematic errors in an instrument ?
a) No use for a long time
b) High use
c) Manufacturing fault
d) Delivery fault

Answer: b
Explanation: Systematic errors arise due to high use of the instrument. It can also be present because of careless handling.

22. How are systematic errors removed usually for an instrument?
a) By replacing it
b) By re-calibrating it
c) By using a repairing service
d) By not using it for some time

Answer: b
Explanation: Systematic errors arise due to careless or overuse of an instrument. It can easily be removed by re-calibrating the instrument and maintaining it properly.

23. In 5 experiments with the same objective, the values obtained are very near to each other. These values can be called ____
a) Precise
b) Accurate
c) Average
d) Invalid

Answer: a
Explanation: Precision refers to the how close the data points or numbers in a set are. In this case the values obtained are very close to each other, hence they are precise.

24. Range of an instrument is _____
a) The minimum value that can be measured
b) The maximum value that can be measured
c) All values starting from the minimum to the maximum that can be measured
d) The average of all values that can be measured

Answer: c
Explanation: Range refers to all values starting from the minimum to the maximum that can be measured. Range is sometimes also referred to as span.

25. Usually how many types of errors are present in scientific measurements?
a) 2
b) 3
c) 4
d) 5

Answer: b
Explanation: There are three types of scientific errors – Random errors, Systematic errors and Blunders. Out of the three, systematic errors are the easiest to remove.

26. What is least count of an instrument?
a) The lowest value that can be measured accurately
b) The greatest value that can be measured accurately
c) Half of the lowest value that can be measured accurately
d) Mean of the lowest and highest values

Answer: a
Explanation: The least count of an instrument is the lowest value that can be measured accurately. For example, if the lowest value that can be measured by a ruler is 1 mm, then the least count is 1 mm.

27. The length and breadth of a rectangle are 4.5 mm and 5.9 mm. Keeping the number of significant figures in mind, its area in mm² is ____
a) 22.55
b) 26.55
c) 26.6
d) 27

Answer: d
Explanation: Area of rectangle is length multiplied by breadth. The area of this rectangle is 4.5×5.9 mm² = 26.55 mm². However, when multiplying/dividing, the answer should have the same number of significant figures as the limiting term. The limiting term is the number with the least number of significant figures. In our example, both the numbers have 2 significant figures. So, the result should also have 2 significant figures. Hence, 26.55 will be rounded to 2 significant figures and it will become 27.

28. The number of significant digits in 1559.00 is ____
a) 6
b) 5
c) 3
d) 4

Answer: a
Explanation: The number of significant digits is 6. The zeros after the decimal are also significant.

29. The number 0.005900, in standard scientific form can be expressed as ______
a) 5.9×10³
b) 59×10⁴
c) 5.9×10²
d) 5.9×10⁴

Answer: a
Explanation: Here the number of significant digits is 4. In standard scientific form, we need not consider the zeros after 9 and need to bring the value between 1 and 9 in the first place.

30. Number of significant digits in 0.0028900 is ______
a) 5
b) 6
c) 7
d) 8

Answer: a
Explanation: The correct answer is 5. While calculating the number of significant digits, we need not consider the zero preceding 2.

31. What is the number 75.66852 rounded off to 5 significant digits?
a) 75.67
b) 75.669
c) 75.668
d) 75.667

Answer: b
Explanation: According to the rounding off rules, if the digit to be removed is 5 and the following digit is not zero, then the last remaining digit is increased by one.
Given number = 75.66852
After applying rounding rules, 852 rounds off to 9
Hence, the final answer will be “75.669”.

32. The length of a cube is 2.3 cm. What is its volume rounded off to 4 significant digits in cm³?
a) 12.67
b) 12.167
c) 12.17
d) 13

Answer: c
Explanation: Volume of a cube is equaled to its length cubed, therefore, required volume = 12.167 cm³. The 67 in the end will be rounded off to 7, hence the final number becomes 12.17.

33. How many significant digits are there in 25.33600?
a) 7
b) 8
c) 5
d) 6

Answer: a
Explanation: The correct answer is 7. The zeros after 6 in the decimal part are also significant.

34. The number of significant digits in 5.002 is ____
a) 5
b) 4
c) 3
d) 6

Answer: b
Explanation: The correct answer is 4. All the digits in the given number are significant.

35. The volume of a box, 10m wide, 4.5m long and 2.3m high up to 3 significant digits in m³ is ____
a) 104
b) 103.5
c) 103
d) 100

Answer: a
Explanation: As per the rule of multiplication and division, the result should have the same significant figure as the number with least significant figure. Among the 3 numbers: 10, 4.5 and 2.3, the number 10 has only 1 significant figure. The volume of the box is length x width x height = 103.5.

As per the actual rule, 103.5 would be rounded up to 100 and the answer would be 100 (as the number 100 has only 1 significant figure). However, we are asking for 3 significant figures in the question, 103.5 will be rounded off to 3 significant figures and the final answer will become 104.

36. How many significant digits are there in 002.5001?
a) 1
b) 5
c) 7
d) 6

Answer: b
Explanation: The correct answer is 5. The zeros preceding 2 are not significant.

37. [MLT^-2] matches with the dimensional formula of ____
a) Force
b) Modulus of elasticity
c) Displacement
d) Strain

Answer: a
Explanation: The given dimensional formula matches with that of force. Force = mass x acceleration.

38. What is the maximum number of unknowns that can be found through a simple dimensional equation?
a) 6
b) 5
c) 3
d) 2

Answer: c
Explanation: A simple dimensional equation uses three basic parameters – mass, length and time. Hence the maximum number of unknowns that can be found is 3.

39. If, v stands for velocity, L for length, T for time and M for mass, what is the value of x in the equation –
L = (vT/M)^x?
a) 3
b) 2
c) 1
d) 0

Answer: d
Explanation: The dimensions of velocity are LT^-1. The dimensions of the RHS are LM^-x and that of LHS is L. hence on equating we get x = 0.

40. Assuming standard notations, which of the following quantities is dimensionless?
a) v/a
b) P/Fv
c) FE/L
d) V²/g

Answer: b
Explanation: P/Fv is a dimensionless quantity. This is because, power(P) = Force(F) x Instantaneous Velocity(v).

41. If the units of length and time are doubled, what will be the factor by which the unit of acceleration will change?
a) 1
b) 0.25
c) 0.5
d) 2

Answer: c
Explanation: The dimensions of acceleration are LT^-2. On doubling both length and time, the new dimensions become (0.5)*LT^-2.

42. What are the dimensions of a light year?
a) L
b) T
c) LT^-1
d) M

Answer: a
Explanation: Lightyear is the unit of measurement of distance; hence it has the dimensions of distance. Lightyear is the distance travelled by light in one year. It finds its use mostly in astronomy.

43. What are the dimensions of coefficient of friction?
a) MLT^-2
b) LT^-1
c) L
d) It is dimensionless

Answer: d
Explanation: Coefficient of friction is the ratio of Frictional force to Normal Reaction force; hence it is dimensionless. The formula for calculating this is, f = µN.

44. Identify the primary quantity from the following.
a) Mass
b) Density
c) Speed
d) Volume

Answer: a
Explanation: Mass is one of the 7 primary quantities in the SI system. Density is dependent on mass and volume. Speed is dependent on length and time. And volume is dependent on length as its cube.

45. Which of the following is a use of dimensional analysis?
a) To check the dimensional correctness of an equation
b) To solve the equation dimensionally
c) To get the number of dimensional constants
d) To understand the dimensional equation

Answer: a
Explanation: Dimensional analysis is basically used for two purposes. First, to check the dimensional correctness of an equation. Second, to convert a physical quantity from one system of units to another.

46. The dimension whose unit does not depend on any other dimension’s unit is known as ________
a) Fundamental dimension
b) Dependent dimension
c) Independent dimension
d) Absolute dimension

Answer: a
Explanation: A fundamental dimension is one, whose units do not depend on the units of any other dimension. There are 7 fundamental dimensions in the SI system and rest all are derived from them.

47. Which one of the following is a dimensionless quantity?
a) Mass
b) Weight
c) Specific weight
d) Reynold’s number

Answer: d
Explanation: Reynold’s number is a dimensionless quantity. Its formula is Re = ρvD/μ. On solving the associated dimensional equation, it will turn out that, Reynold’s number is dimensionless.

48. In an experiment, it is found that the experimental value is very close to actual value, hence the experimental value can be called _____
a) Accurate
b) Precise
c) Suitable
d) Mean

49. What is the reason for the occurrence of systematic errors in an instrument ?
a) No use for a long time
b) High use
c) Manufacturing fault
d) Delivery fault

Answer: b
Explanation: Systematic errors arise due to high use of the instrument. It can also be present because of careless handling.

50. How are systematic errors removed usually for an instrument?
a) By replacing it
b) By re-calibrating it
c) By using a repairing service
d) By not using it for some time

Answer: b
Explanation: Systematic errors arise due to careless or overuse of an instrument. It can easily be removed by re-calibrating the instrument and maintaining it properly.

51. In 5 experiments with the same objective, the values obtained are very near to each other. These values can be called ____
a) Precise
b) Accurate
c) Average
d) Invalid

Answer: a
Explanation: Precision refers to the how close the data points or numbers in a set are. In this case the values obtained are very close to each other, hence they are precise.

52. Range of an instrument is _____
a) The minimum value that can be measured
b) The maximum value that can be measured
c) All values starting from the minimum to the maximum that can be measured
d) The average of all values that can be measured

Answer: c
Explanation: Range refers to all values starting from the minimum to the maximum that can be measured. Range is sometimes also referred to as span.

53. Usually how many types of errors are present in scientific measurements?
a) 2
b) 3
c) 4
d) 5

Answer: b
Explanation: There are three types of scientific errors – Random errors, Systematic errors and Blunders. Out of the three, systematic errors are the easiest to remove.

54. What is least count of an instrument?
a) The lowest value that can be measured accurately
b) The greatest value that can be measured accurately
c) Half of the lowest value that can be measured accurately
d) Mean of the lowest and highest values

55. In a screw gauge, the main scale has divisions in millimeter and circular scale
has 50 divisions. The least count of screw gauge is
(a) 2𝜇𝑚 (b) 5𝜇𝑚 (c) 20𝜇𝑚 (d) 50𝜇

Answer: c

56. In a vernier calliper, N divisions of vernier scale coincide with (N – 1) divisions
of main scale (in which 1 division represents 1mm). The least count of the
instrument in cm should be:
(a) 𝑁 (b) 𝑁 − 1 (c) 1/10𝑁 (d) 1/(𝑁−1)

Answer: c

57. A screw gauge with a pitch of 0.5 mm and a circular scale with 50 divisions is used to measure the thickness of a thin sheet of Aluminium. Before starting the measurement , it is found that when the two jaws of the screw gauge are brought in contact, the 45th division coincides with the main scale line and that the zero of the main scale is barely visible. What is the thickness of the sheet if the main scale reading is 0.5 mm and the 25th division coincides with the main scale line ?
(a) 0.50 mm
(b) 0.75 mm
(c) 0.80 mm
(d) 0.70 mm
Answer: c

Least count
= pitch / no. of division on circular scale
= 0.5 mm / 50
= 0.01 mm

Negative zero error = – 5 × LC = -5 × 0.01 = -0.05 mm

Measured value
= main scale reading + screw gauge reading – zero error
= 0.5mm + 25 × 0.01mm – (-0.05mm)
= 0.75mm + 0.05mm
= 0.80mm

58. Consider a Vernier callipers in which each 1 cm on the main scale is divided into 8 equal divisions and a screw gauge with 100 divisions on its circular scale. In the Vernier callipers, 5 divisions of the Vernier scale coincide with 4 divisions on the main scale and in the screw gauge, one complete rotation of the circular scale moves it by two divisions on the linear scale. Then:
(A) If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 mm.
(B) If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm.
(C) If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.01 mm.
(D) If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is 0.005 mm.

Answer:c

Vernier calliper:

M.S.D =1/8 cm = 0.125 cm = 1.25 mm

5 V.S.D. = 4 M.S.D

⇒ V.S.D. = 4/5 M.S.D. = 4/5 × 1.25mm = 1 mm

Least count = M.S.D. – V.S.D = 1.25 – 1 = 0.25 mm

Screw gauge:

No. of circular scale divisions = 100

1 rotation (pitch) = 2 linear scale divisions

» If the pitch of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is –

Pitch = 2 × Least Count of vernier callipers = 2 × 0.25mm = 0.5 mm

Least count of the screw gauge

= Pitch/no. of circular scale divisions = 0.5/100 =0.005mm

Option (B) is correct.

» If the least count of the linear scale of the screw gauge is twice the least count of the Vernier callipers, the least count of the screw gauge is –

Linear scale division of screw gauge

= 2 × Least count of vernier callipers

= 2 × 0.25 mm

= 0.5 mm

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Now,

Pitch = 2 linear scale divisions = 2 × 0.5 = 1mm

Least count of the screw gauge

= Pitch/ No. of circular divisions

= 1/100 mm

= 0.01 mm

Option (C) is correct.

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