• +91 9971497814
  • info@interviewmaterial.com

Chapter 2- Units and Measurements Interview Questions Answers

Question 1 :

Fill in the blanks

(a) Thevolume of a cube of side 1 cm is equal to.....m3

(b) Thesurface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equalto ... (mm)2

(c) Avehicle moving with a speed of 18 km h–1covers....min 1 s

(d) Therelative density of lead is 11.3. Its density is ....g cm–3or . ...kg m–3.

Answer 1 :

(a) 1 cm =

Volume of the cube = 1 cm3

But,1 cm3 = 1 cm × 1 cm × 1 cm =

1 cm3 = 10–6 m3

Hence, the volume of a cube of side 1 cm is equal to 10–6 m3.

(b) The total surface area of a cylinder of radius r andheight h is

S = 2πr (r + h).

Given that,

r = 2 cm = 2 × 1 cm = 2 × 10mm = 20 mm

h = 10 cm = 10 × 10 mm = 100mm

 = 15072 = 1.5 × 104 mm2

(c) Using the conversion,

1km/h =

Therefore, distance can be obtained using the relation:

Distance = Speed × Time = 5 × 1 = 5 m

Hence, the vehicle covers 5 m in 1 s.

(d) Relative density of a substance is given by the relation,

Relativedensity =

Density of water = 1 g/cm3

Again, 1g =

1 cm3 = 10–6 m3

1g/cm3 =

11.3 g/cm3 =11.3 × 103 kg/m3

Question 2 :

Fill in the blanks by suitable conversion of units:

(a) 1 kg m2s–2= ....g cm2 s–2

(b) 1 m =..... ly

(c) 3.0 m s–2=.... km h–2

(d) G= 6.67 × 10–11 N m2 (kg)–2=....(cm)3s–2 g–1.

Answer 2 :

(a) 1 kg = 103 g

1 m2 = 104 cm2

1 kg m2 s–2 = 1 kg × 1 m2 ×1 s–2

=103 g × 104 cm×1 s–2 = 107 g cm2 s–2

(b) Light year is the total distance travelled by light in oneyear.

1 ly = Speed of light × One year

= (3 × 108 m/s) × (365 × 24 × 60 × 60 s)

= 9.46 × 1015 m

(c) 1 m = 10–3 km

Again,1 s =

1 s–1 = 3600 h–1

1 s–2 = (3600)2 h–2

3 m s–2 = (3 × 10–3 km)× ((3600)2 h–2) = 3.88 × 104 kmh–2

(d) 1 N = 1 kg m s–2

1 kg = 10–3 g–1

1 m3 = 106 cm3

6.67 × 10–11 N m2 kg–2 =6.67 × 10–11 × (1 kg m s–2) (1 m2) (1 s–2)

= 6.67 × 10–11 × (1 kg × 1 m3 × 1s–2)

= 6.67 × 10–11 × (10–3 g–1)× (106 cm3) × (1 s–2)

= 6.67 × 10–8 cm3 s–2 g–1

Question 3 :

A calorie is a unit of heat or energy and it equals about 4.2 Jwhere 1J = 1 kg m2s–2. Suppose we employ a system ofunits in which the unit of mass equals α kg, the unit of length equals β m, theunit of time is γ s. Show that a calorie has a magnitude 4.2 α–1 β–2 γinterms of the new units.

Answer 3 :

Given that,

1 calorie = 4.2 (1 kg) (1 m2) (1 s–2)

New unit of mass = α kg

Hence,in terms of the new unit, 1 kg =
In terms of the new unit of length,
And, in terms of the new unit of time,

1 calorie = 4.2 (1 α–1)(1 β–2) (1 γ2) = 4.2 α–1 β–2 γ2

Question 4 :

Explain this statement clearly:

“To call a dimensional quantity ‘large’ or ‘small’ is meaninglesswithout specifying a standard for comparison”. In view of this, reframe thefollowing statements wherever necessary:

(a) atoms are very small objects

(b) a jet plane moves with great speed

(c) the mass of Jupiter is very large

(d) the air inside this room contains a large number of molecules

(e) a proton is much more massive than an electron

(f) the speed of sound is much smaller than the speed of light.

Answer 4 :

The given statement is true because a dimensionless quantity maybe large or small in comparision to some standard reference. For example, thecoefficient of friction is dimensionless. The coefficient of sliding frictionis greater than the coefficient of rolling friction, but less than staticfriction.

(a) An atom is a very small object in comparison to a soccer ball.

(b) A jet plane moves with a speed greater than that of a bicycle.

(c) Mass of Jupiter is very large as compared to the mass of acricket ball.

(d) The air inside this room contains a large number of moleculesas compared to that present in a geometry box.

(e) A proton is more massive than an electron.

(f) Speed of sound is less than the speed of light.

Question 5 :

A new unit of length is chosen such that the speed of light invacuum is unity. What is the distance between the Sun and the Earth in terms ofthe new unit if light takes 8 min and 20 s to cover this distance?

Answer 5 :

Distance between the Sun and the Earth:

= Speed of light × Time taken by light to cover the distance

Given that in the new unit, speed of light = 1 unit

Time taken, t = 8 min 20 s = 500 s

Distance between the Sun and theEarth = 1 × 500 = 500 units

Question 6 :

Which of the following is the most precise device for measuringlength:

(a) a vernier callipers with 20 divisions on the sliding scale

(b) a screw gauge of pitch 1 mm and 100 divisions on the circularscale

(c) an optical instrument that can measure length to within awavelength of light ?

Answer 6 :

A device with minimum count is the most suitable to measurelength.

(a) Least count of vernier callipers

= 1 standard division (SD) – 1 vernier division (VD)

(b) Least count of screw gauge = 

(c) Least count of an optical device = Wavelength of light 10–5 cm

= 0.00001 cm

Hence, it can be inferred that an optical instrument is the mostsuitable device to measure length.

Question 7 :

A student measures the thickness of a human hair by looking at itthrough a microscope of magnification 100. He makes 20 observations and findsthat the average width of the hair in the field of view of the microscope is3.5 mm. What is the estimate on the thickness of hair?

Answer 7 :

Magnification of the microscope = 100

Average width of the hair in the field of view of the microscope =3.5 mm

Actual thickness of thehair is = 0.035 mm.

Question 8 :

Answer the following:

(a) You are given a thread and a metre scale. How will youestimate the diameter of the thread?

(b) A screw gauge has a pitch of 1.0 mm and 200 divisions on the circularscale. Do you think it is possible to increase the accuracy of the screw gaugearbitrarily by increasing the number of divisions on the circular scale?

(c) The mean diameter of a thin brass rod is to be measured byvernier callipers. Why is a set of 100 measurements of the diameter expected toyield a more reliable estimate than a set of 5 measurements only?

Answer 8 :

(a) Wrap the thread on a uniform smooth rod in such a way that thecoils thus formed are very close to each other. Measure the length of thethread using a metre scale. The diameter of the thread is given by therelation,

(b) It is not possible to increase the accuracy of a screw gaugeby increasing the number of divisions of the circular scale. Increasing thenumber divisions of the circular scale will increase its accuracy to a certainextent only.

(c) A set of 100 measurements is more reliable than a set of 5measurements because random errors involved in the former are very less ascompared to the latter.

Question 9 :

The photograph of a house occupies an area of 1.75 cm2ona 35 mm slide. The slide is projected on to a screen, and the area of the houseon the screen is 1.55 m2. What is the linear magnification of theprojector-screen arrangement?

Answer 9 :

Area of the house on the slide = 1.75 cm2

Area of the image of the house formed on the screen = 1.55 m2

= 1.55 × 104 cm2

Arialmagnification, ma = 
Linearmagnifications, ml =

Question 10 :

State the number of significant figures in the following:

(a) 0.007 m2

(b) 2.64 × 1024 kg

(c) 0.2370 g cm–3

(d) 6.320 J

(e) 6.032 N m–2

(f) 0.0006032 m2

Answer 10 :

(a) Answer: 1

The given quantity is 0.007 m2.

If the number is less than one, then all zeros on the right of thedecimal point (but left to the first non-zero) are insignificant. This meansthat here, two zeros after the decimal are not significant. Hence, only 7 is asignificant figure in this quantity.

The given quantity is 2.64 × 1024 kg.

Here, the power of 10 is irrelevant for the determination ofsignificant figures. Hence, all digits i.e., 2, 6 and 4 are significantfigures.

(c) Answer: 4

The given quantity is 0.2370 g cm–3.

For a number with decimals, the trailing zeroes are significant.Hence, besides digits 2, 3 and 7, 0 that appears after the decimal point isalso a significant figure.

(d) Answer: 4

The given quantity is 6.320 J.

For a number with decimals, the trailing zeroes are significant.Hence, all four digits appearing in the given quantity are significant figures.

(e) Answer: 4

The given quantity is 6.032 Nm–2.

All zeroes between two non-zero digits are always significant.

(f) Answer: 4

The given quantity is 0.0006032 m2.

If the number is less than one, then the zeroes on the right ofthe decimal point (but left to the first non-zero) are insignificant. Hence,all three zeroes appearing before 6 are not significant figures. All zerosbetween two non-zero digits are always significant. Hence, the remaining fourdigits are significant figures.



Chapter 2- Units and Measurements Contributors


Share your email for latest updates


Our partners