NCERT Solutions Class 11 Physics Chapter 2 Units And Measurements

Question :1.
Fill in the blanks
(a) The volume of a cube of side 1 cm is equal to   …..m³
(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to …(mm)²
(c) A vehicle moving with a speed of 18 km h^–1 covers….m in 1 s
(d) The relative density of lead is 11.3. Its density is ….g cm^–3 or ….kg m^–3.

Answer :

• Volume of cube = (a)³    (equation 1)

Length of edge 1cm= (1/100) m
Volume of cube = (1/100) m³
Using equation (1); 1cm x 1cm x 1cm
= (1/100) m x (1/100) m x (1/100) m
Therefore 1cm³ = 10^-6m³
Hence, the volume of a cube of side 1cm is equal to =10^-6m³.

• The total surface area of a cylinder of radius r and height h is:

Given that:-
r=2cm = (2×1) cm = (2×10) mm r=20mm
h=10cm= (10 x 10) mm =100mm
Surface area of the cylinder S = 2 πr (h+r)
= (2×3.14x20x (20+100))
=15072
=1.5 x 10⁴(mm) ²
Therefore, the surface area of a solid cylinder of radius 2.0cm and height 10.0cm is equal to =1.5 x 10⁴mm²

• Given Time, t = 1sec.

Speed s = 18 kmh^-1
Using, 1km = 1000m
1hour = 3600sec
Speed s = 18 x (1000/3600) = 5m/sec.
Therefore speed = (distance/time)
Distance = (speed x time) = 5 x 1 =5m.
Therefore, a vehicle moving with speed of 18kmh^-1 covers 5m in 1s.

• Density of lead = (Relative density of lead) x (Density of water)

Density of water = 1g/cm³
     = (11.3 x 1) = 11.3g/cm³
             Again, 1g = (1/1000) kg = (10^-3) kg
            1cm³ = (1/100) = 10^-2m³
 Therefore, 11.3g/cm³ = (11.3 x 10^-3) kg/ (10^-2) m^-3 = (11.3 x 10^-3x10⁶) kgm^-3
= (1.13 x 10³) kgm^-3
Therefore; the relative density of lead is 11.3. Its density is 11.3 g cm^–3 or
1.13 x 10³ kg m^–3

Question : 2.
Fill in the blanks by suitable conversion of units
(a) 1 kg m² s^–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) ³ s^–2 g^–1

Answer :

• 1 kg m² s^–2 = 1 x 10³g(10² cm)²s^-2 = 10⁷ g cm² s^-2
• 1 light year = 9.46 x 10¹⁵m

Therefore 1m = (1/9.46 x 10¹⁵) = (1.053 x 10^-16) light year.

• 3ms^-2 = 3 x 10^-3 km( 1 h/60×60)^-2

= 3 x10^-3 x 3600 x 3600 kmh^-2
=3.0 m s^–2 = 3.888 x 10^-4 kmh^-2

• Given that

G= 6.67 × 10^–11 N m² (kg)^–2
We know that; 1 N = 1 kg m s^–2
1 kg = 10³ g
1 m = 100 cm = 10² cm
Therefore, we get 
=> 6.67 × 10^–11 N m² kg^–2 = 6.67 × 10^–11 × (1 kg m s^–2) (1 m²) (1kg^–2)
=> 6.67 × 10–11 × (1 kg^–1 × 1 m³ × 1 s^–2)
Converting Kg to g and m to cm;
=> 6.67 × 10^–11 × (10³ g)^-1 × (10² cm) ³ × (1 s^–2)
=>  6.67 × 10^–11 × 10^-3 g^-1 × 10⁶ cm³ × (1 s^–2)
=> 6.67 × 10^–8 cm³ s^–2 g^–1
Therefore G = 6.67 × 10^–11 N m² (kg)^–2 =6.67 × 10^–8 (cm) ³ s^–2 g^–1.

Question : 3.
A calorie is a unit of heat or energy and it equals about 4.2 J where 1J = 1 kgm²s^–2 .
Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β m, the unit of time is γ s.
Show that a calorie has a magnitude 4.2 α^–1 β^–2 γ² in terms of the new units.

Answer :
Given that,
1 calorie = 4.2 J   (1J = 1 kg m²s^–2)
Therefore, 1 calorie =4.2 kg m²s^–2 = 4.2 × (1 kg) × (1 m²) × (1 s^–2)     … (1)
Given that;
Unit of mass equals α kg => 1 kg mass = (1/α) mass
Unit of length equals β m => 1 m length = (1/ β) length
Unit of time is γ s        => 1 s time = (1/ γ) time
Putting the values in equation (1), we get
 New unit of mass = α kg
=4.2 × (1 kg) × (1 m²) × (1 s^–2)
= 4.2 × (1/ α) × (1/ β) ² × (1/ γ) ^–2
= 4.2 × (α^–1) × (β^–1)² × (γ^–1) ^–2
=4.2 α^–1 β^–2 γ² unit of energy.

Question : 4.
Explain this statement clearly:
“To call a dimensional quantity ‘large’ or ‘small’ is meaningless without specifying a standard for comparison”.
In view of this, reframe the following 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 :
The statement is true, because a dimensionless quantity can be large or small when compared with the unit of measurement.

• The size of atom is much smaller than the sharp tip of a pin.
• A jet plane moves faster than a superfast train.
• The mass of Jupiter is very large compared to the mass of earth.
• The air inside this room contains a large number of molecules than in one mole of air.
• A proton is heavier than an electron.

Mass of a proton is 1.67 × 10^-24 g while mass of electron 9.11 × 10^-28 g, therefore statement is true.

• The speed of sound is less than the speed of light.

Speed of sound 346 m/ sec while speed of light is 3 x 10⁸ m / sec in air at 25 ⁰C, therefore statement is true.

Question : 5.
A new unit of length is chosen such that the speed of light in vacuum is unity.
What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance?

Answer :
Given;
Speed of light in vacuum = unity
Distance between sun and earth = (speed of light in vacuum) x (time)
= (3 x10⁸ms^-1) x 500s
Therefore new unit of length = speed of light in vacuum
= (3 x 10⁸) ms^-1
Therefore distance between sun and earth in terms of new unit
= (3 x 10⁸ x 500)/ (3 x 10⁸)
=500 units according to new unit.

Question : 6.
Which of the following is the most precise device for measuring length?
(a) a Vernier callipers with 20 divisions on the sliding scale
(b) a screw gauge of pitch 1 mm and 100 divisions on the circular scale
(c) an optical instrument that can measure length to within a wavelength of light

Answer :

• Least count of Vernier callipers =(1/20)= 0.05cm
• Least count of screw gauge = (1/100) = 0.01cm
• Least count of an optical wave =(1/10⁵) = 0.00001cm

Optical instrument is the most precise device for measuring length.

Question : 7.
A student measures the thickness of a human hair by looking at it through a microscope of magnification 100.
He makes 20 observations and finds that the average width of the hair in the field of view of the microscope is 3.5 mm.
What is the estimate on the thickness of hair?

Answer :
Given;
Average width of the hair in the field of view of the microscope = 3.5mm
Magnification produced by microscope = 100
Using (magnification m) = (size of image) (v)/ (size of object) (O)
=> Size of object (O) = (size of image)/ (magnification m)
Therefore thickness of hair = (Thickness of image)/ (magnification)
= (3.5mm)/ (100)
=0.035mm.
                                                                                                                                                                                

Question : 8.
Answer the following:
(a)You are given a thread and a metre scale. How will you estimate the diameter of the thread?
(b)A screw gauge has a pitch of 1.0 mm and 200 divisions on the circular scale.
Do you think it is possible to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale?
(c)The mean diameter of a thin brass rod is to be measured by Vernier callipers.
Why is a set of 100 measurements of the diameter expected to yield a more reliable estimate than a set of 5 measurements only?

Answer :

• As the diameter of a thread is so small that it cannot be measured using a metre scale.
• Number of turns of the thread to be wound to get turns closely one another.

Let measure of length (l) of the windings on the scale.
Number of turns = ‘n’.
Therefore diameter of thread = (1/n)

• Least count = (pitch) /(number of divisions in circular scale)

i.e. least count decreases when number of divisions on the circular scale increases. Thereby accuracy would increase,
but it is impossible to take precise reading due to low resolution of human eye.
(c)    By taking a large number of measurements of the same quantity, it is quite possible that the majority of
these measurements will have small errors which might be positive or negative. When average values are taken positive and negative errors are likely to cancel each other

Question : 9.
The photograph of a house occupies an area of 1.75 cm² on a 35 mm slide. The slide is projected on to a screen,
and the area of the house on the screen is 1.55 m². What is the linear magnification of the projector-screen arrangement?

Answer :
Given:
Area of photograph of house =     1.75 cm²
Area of image of house projected on screen = 1.55m² = (1.55 x 10⁴) cm²
Therefore, areal of magnification of the projector screen, m_areal;
m_areal = (area of image)/(area of object)
= (1.55 x 10⁴) cm²/ (1.75cm²)
The linear magnification will be m_linear = √ ( m_areal) = √ (8857) = 94.1

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 :

• The given quantity is 0.007 m².

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

• The given quantity is (2.64 × 10²⁴) kg.

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

• 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 is also a significant figure.

• 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.

• The given quantity is 6.032 Nm^–2

The given quantity is 0.0006032 m²
All zeroes between two non-zero digits are always significant.

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

Question : 11.
The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively.
Give the area and volume of the sheet to correct significant figures.

Answer :
Given:
Length of sheet, l = = 4.234 m
Breadth of sheet, b = 1.005 m
Thickness of sheet, h = 2.01 cm = 0.0201 m
The given table lists the respective significant figures: