Units And Measurements

The volume of cube of side 1 cm is equal to 10^-6 m³

Here we have to find the volume of a cube whose side is given.

We know that in a cube all sides are equal.

Let ‘a’ be the side of a cube.

Therefore ‘a’ = 1 cm

Here we have to find the volume of the cube in m³_.

Convert 1 cm into m a = 1 cm = .01 m

Volume of a cube having side ‘a’ = a³

V = (0.01 m) ³

= 10^-6 m³

Here we have to find the surface area of a solid cylinder whose radius and height is given.

Surface area of a solid cylinder having radius ‘r’ and height ‘h’

S.A = Curved surface area + Area of base and upper portion

In a solid cylinder Base & upper portions are in the form of a circle of radius ‘r’

S.A = 2πrh + 2πr² . -------------------- (1)

Where r = radius of the upper and lower portions

h = height of the cylinder.

Here values of ‘r’ and ‘h’ are given.

Note: Here we have to convert the values of ‘r & ‘h’ from cm to mm.

r = 2 cm = 20 mm

h = 10 cm = 100 mm

Substituting the values of ‘r’ and ‘h’ in equation (1).

S.A = 2πrh + 2πr²

= 2πr(r +h) π

= 2 × (22/7) × 20(20+100) mm².

= 1.5 × 10⁴ mm²

The surface area of the solid cylinder = 1.5 × 10⁴ mm²

In this question, we have to find the distance covered by a vehicle in 1 s moving with a particular speed.

We know that distance ‘x’ covered by a vehicle moving with a particular speed ‘v’

x = v × t ------------------- (1)

Where x = distance covered

v = speed of the vehicle

t = time taken to cover distance ‘x’

Here v = 18 km/h = 5 m/s

To convert km/h to m/s multiply the value of km/h with

t = 1 s

Substituting the values of ‘v’ &‘t’ in equation (1)

x = v × t

x = 5 ms^-1 × 1 s

x = 5 m

The distance covered by the vehicle in 1s = 5m

Here the relative density of lead is given. We have to find the density of lead.

Relative density of lead = -------------- (1)

Density of lead = Relative density of lead × Density of water

Relative density of lead = 11.3

Density of water = 1 g/cm³ = 1000 kg/m³

Substituting the above density values in equation (1).

Density of lead = 11.3 × 1 g/cm³

= 11.3 g/cm³

We have to find the density of lead in kg/m³ also.

Substituting the value of density of water in kg/m³ in equation (1)

Density of lead = 11.3 × 1000 kg/m³

= 1.13 × 10⁴ kg/m³

The density of lead = 11.3 g/cm³ or 1.13 × 10⁴ kg/m³

Here we have to convert kg to g & m to cm.

1 kg = 1000 g

1 m = 100 cm

Therefore 1 kg m² s^-2 = 1000g × (100 cm)² s^-2

= 10³ g× 10⁴ cm² s^-2

= 10⁷ g cm² s^-2

1 kg m² s^-2 = 10⁷ g cm² s^-2

Here we have to find 1m is how many light years.

We know that 1 ly = 9.46 × 10¹⁵ m

1 m = 1/ (9.46 × 10¹⁵⁾ m

= 1.057 × 10^-16 ly

1 m = 1.057 × 10^-16 ly

Here we have to convert m/s² into km/h²

1 km = 1000 m

1 m = 10^-3 km

1 hour = 60 min = 60 × 60 sec
or 1 sec = h

Substituting the above values

3 m s^-2 = 3 × 10^-3 km × h^-2

= 38880 km-h^-2h

= 3.88 × 10⁴ km h^-2

Here the value of a constant ‘G’ is given. We have to express it given unit.

Note: ‘G’ is known as Universal Gravitational Constant. Its value is constant all over the universe.

G = 6.67 × 10^-11 N m² kg

Where, N = Newton

m = meter

Kg = Kilogram

‘N’ can be rewritten as

N = kg × ms^-2

Substitute this value in ‘G’

G = 6.67 × 10^-11 N m² kg^-2

= 6.67 × 10^-11 kg × ms^-2 × kg^-2

= 6.67 × 10^-11 kg^-1 m³ s^-2

= 6.67 × 10^-11 × (10²)³ × (10^-3)²

= 6.67 × 10^-11 cm³ g^-1 s^-2

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

In this question, we have to apply one of the uses of Dimensional analysis

i.e. Conversion of one system of units into another

Where, n₂ = n₁ × [M₁M₂ ]^a × [L₁/L₂]^b × [T₁/T₂]^c ----------(1)

M_1, L₁, T_{1 =} Fundamental units of one system

M₂, L₂, T₂ = Fundamental units of another system

a,b,c are the dimensions of the quantity in mass, length, and time.

n₁ = numerical value of quantity in one system

n₂ = numerical value of quantity in the other system

The value of 1 calorie in S.I System of units is given. We have to express the value of calorie in another given arbitrary units.

1 calorie = 4.2 J

1 J = 1 kg m² s^–2

Substituting the above values in equation (1).

n₂ = 4.2 []¹ × []² × []^-2

= 4.2 ^-1-22

1 calorie = 4.2 α^-1 β^-2 γ² in new units.

A. We know that size of an atom is smaller than the tip of a pencil.

B. We know that a jet plane moves faster than a car.

C. The mass of Jupiter is very large when compared with the mass of Earth.

D. The air inside this room contains a large number of molecules when compared with air inside a bottle.

E. There is no need to rephrase the statement. The statement is true.

F. No need to rephrase the above statement. The statement is right.

We know that speed of light in vacuum, c = 3 × 10⁸ m/s

Relation between speed & distance travelled

Distance travelled ‘x’ = speed × time taken ------- (1)

Here the value of time taken is given

t = 8 min 20 s = 500 s

So, x = s × t

x = c × t

Where c = speed of light

t = time taken

= 3 × 10⁸ ms^-1 × 500 m

= 1.5 × 10¹¹ m

So this is the distance between the sun and the earth.

We have to express the above distance value in terms of new units. In new units, the value of the speed of light is unity.

i.e, c = 1 unit

t = 500 s

Substituting the above values in equation (1)

Distance between sun & Earth x = c × t

= 1 unit × 500 s

= 500 new units

Distance between Sun & Earth = 500 new units

In this question, three measuring/ calibrating instruments are given. Among these devices, we have to find the most precise device.

To find the most precise device we have to calculate the Least Count of these devices.

(a). Least count of Vernier calliper =

= = 0.05 mm

= 0.05 cm

(b). Least count (L.C) of screw gauge =

Pitch = 1 mm

No. of division on circular scale = 100

L.C = 1 × 10^-3 ×

= 0.01 cm

(c). Least count of optical instrument that can measure length to within a wavelength of light is about 0.00001 cm

Here optical instrument has least count among given instruments. So Optical instrument is the most precise device for measuring length.

Here given details are

Magnification of Microscope = 100

Average field of view of microscope = 3.5 mm

Thickness of Hair = ------------- (1)

Substituting the above values in equation (1)

Thickness of hair =

= 0.035 mm

Estimate thickness of hair = 0.035 mm

A. Take the thread and wrap it around the metre scale. It should be wrapped without giving any space between the coils. Assume that the diameter of the thread is d.

Then number of turns, take n (coils) multiplied with diameter d will give the length up to which you’ve used the thread (t take it l). So, d = l/n

B. It is possible theoretically to increase the accuracy of the screw gauge arbitrarily by increasing the number of divisions on the circular scale as the least count is given by pitch/no. of divisions. So, on increasing the no. of divisions, least count will decrease and thus, increasing the accuracy. But practically, it depends on the resolution of eye (to see two different things distinctly and clearly). You can increase no. of divisions upto a certain limit only.

C. The mean diameter of a thin brass rod measured by vernier callipers from a set of 100 measurements is more reliable a more reliable estimate than a set of 5 measurements because the chances of making a negative error and positive error are equally likely and thus they cancel each other giving results with less error.

Given: Area of photograph of house (A_object) - 1.75 cm²

Area of image formed on screen (A_image)- 1.55 m² = 1.55 x 10⁴ cm²

Formula: Areal magnification (m) = A_image / A_object ,

where A_imageand A_object are the areas of image and object respectively.

Since the dimension of areal magnification is L² and dimension of linear magnification is L,

Linear magnification (m_l) = √m

m = (1.55 × 10⁴ cm²)/(1.75 cm²)

⇒ m_l = 94.1 (report upto 3 significant figures only)

A. 1

Explanation: Significant figure- 7. If number is less than one, then 0’s before 7 are insignificant (note that there are only 0’s before 7 and numbers like 0.1007 will have 0 as significant figure)

B. 3

Explanation: Significant figure- 2, 6, 4. Powers of 10 are not taken in counting for significant figures.

C. 4

Explanation: Significant figure- 2,3, 7, 0. Trailing 0’s are significant. These 0’s increase the accuracy of the answer.

D. 4

Explanation: Significant figure- 6, 3, 2, 0. (similar to the problem C)

E. 4

Explanation: Significant figure- 6, 0, 3, 2. 0’s between 2 non-zero digits are significant.

F. 4

Explanation: Significant figure- 6, 0, 3, 2. (similar to the problem A and E)

Given:

Length (l) = 4.234 m

Breadth (b) = 1.005 m

Thickness (t) = 2.01 cm = 0.0201 m (note that significant figures remains unchanged)

Formulae:

Total surface area of rectangular sheet (S) = 2 (lb + bt + tl)

Volume of rectangular sheet (V) = l × b × h

Calculation:

S = 2×(4.234 ×1.005 + 1.005 × 0.0201 + 0.0201 × 4.234)

S = 8.72 m² (reported upto 3 significant figures as in multiplication, lowest number of significant figures in all the numbers is chosen for final answer)

V = 4.234 × 1.005× 0.0201

V = 0.0855 m³ (reported upto 3 significant figures as in multiplication, lowest number of significant figures in all the numbers is chosen for final answer)

Given:

Mass of the box (m): 2.300 Kg (2 significant figures, 1 after decimal)

Gold pieces: G₁ = 20.15 g = 0.02015 Kg (4 significant figures) and G₂ = 20.17 g = 0.02017 Kg (4 significant figures)

(a) total mass of box (M) = m + G₁ + G₂

M = 2.300 + 0.02015 + 0.02017 = 2.34032 Kg but in addition the final result should have significant digits after decimal equal to the minimum number of significant digits of all the numbers used.

Therefore, M = 2.3 Kg (upto 2 significant digits and 1 after decimal due to 1 significant digit in m)

(b) difference in the masses = G₂ – G₁ = 0.02 g (1 significant digit). In subtraction the final result should have significant digits after decimal equal to the minimum number of significant digits of all the numbers used.

Given:

The percentage error in a i.e.

The percentage error in b i.e.

The percentage error in c i.e.

The percentage error in d i.e.

(Rule: The relative error in a physical quantity raised to the power k is the k times the relative error in the individual quantity)

Multiplying both sides by 100 will make each quantity % and thus we can directly substitute the given values to find ΔP/P % or percentage error in P.

P = 3.8 (rounded off to first decimal place)

The dimensions of the quantities involved are:

y = M⁰L¹T⁰

a = M⁰L¹T⁰

T or t=T

A. correct

Explanation: t/T is unitless. Therefore, dimension of y should be the dimension of a which is true.

B. incorrect

Explanation: quantities inside the trigonometric expressions should be unitless but dimension of vt = M⁰L¹T⁰ which is having unit of length and is therefore wrong.

C. incorrect

Explanation: quantities inside the trigonometric expressions should be unitless but dimension of t/a = M⁰L^-1T¹ which is having unit of time/length and is therefore wrong.

D. correct

Explanation: t/T is unitless. Therefore, dimension of y should be the dimension of a which is true.

The dimension of the quantities involved are:

m = M¹L⁰T⁰

m₀ = M¹L⁰T⁰

v = M⁰L¹T^-1

c = M⁰L¹T^-1

For a correct dimensional equation, the dimensions of LHS and RHS should be equal. We can see that dimensions of m and m₀ are equal thus the remaining part of the equation should be constant.

should be unitless.

1 is dimensionless so should be dimensionless.

On subtraction, dimension of the 2 quantities should be same. So, dimension of 1 and v² should be same. Therefore, v² should be dimensionless and therefore v.

We can see that dimension of v and c are same. So, replacing v by v/c will make it dimensionless and the given equation dimensionally correct.

Hence, correct relation will be,

Given

Radius of hydrogen atom, r = 0.5 Å = 0.5 × 10^-10 m

Volume of hydrogen atom, V = (4/3)πr³

⇒

⇒ V = 5.24 × 10^-31 m³

One mole of hydrogen atom contains N_A = 6.023 × 10²³ atoms.

Where, N_A is Avogadro’s number.

So, Volume of one mole of hydrogen atom,

V_t = 6.023 × 10²³ × 5.24 × 10^-31 m³

⇒ V_t = 3.16 × 10^-7 m³

Given,

Volume of one mole of ideal gas at STP, V_m=22.4L=22.4 × 10^-3 m³

Radius of hydrogen atom, r = 1 Å / 2 = 0.5 Å = 0.5 × 10^-10 m

Volume of the hydrogen atom, V = (4/3)πr³

⇒ V = (4/3)×3.14×(0.5×10^-10 m)³

⇒ V = = 5.24 × 10^-31 m³

One mole of hydrogen atom contains N_A = 6.023 × 10²³ atoms.

Where, N_A is Avogadro’s number.

So, Volume of one mole of hydrogen atom,

V_t = 6.023 × 10²³ × 5.24 × 10^-31 m³

⇒ V_t = 3.16 × 10^-7 m³

The ratio of molar volume to atomic volume is

⇒ V_{m �/}V_t = 7.089 × 10⁴

This ratio is very high due to the fact that the inter-atomic distance is very high as compared to the size of atoms in hydrogen gas.

Due to the fast movement of the train, the line of sight of the passenger for nearby objects changes rapidly. So, the nearby trees, houses, etc. appear to move fast in the opposite direction. But, in case of distant objects, the line of sight changes very slowly. So, the distant objects like hilltops, moon, stars, etc. seem to be stationary.

NOTE: Line of sight is an imaginary line joining an object to the viewer’s eyes.

Given,

Diameter of Earth’s orbit, d = 3 × 10¹¹ m

Radius of earth’s orbit, r = d/2

⇒ r = (3×10¹¹m)/2 = 1.5×10¹¹ m

Parallax angle, θ = 1”

⇒ θ = 1’/60

⇒ θ = 1°/(60×60)

We know,

θ = Arc/Radius

⇒ θ = r/x

⇒ x = r/θ

⇒ x = 3.09 × 10¹⁶ m

Hence, 1 parsec = 3.09 × 10¹⁶ m

NOTE: The parsec is a unit of length used to measure large distances to astronomical objects outside the Solar System. Parallax is a difference in the apparent position of an object viewed along two different lines of sight, and is measured by the angle or semi-angle of inclination between those two lines of sight.

Given,

Distance of the nearest star, d = 4.29 ly

1 lightyear is the distance travelled by light in one year.

∴ 1 ly = Speed of light × 1 yr

⇒ 1 ly = 3 × 10⁸ m s^-1 × (365 × 24 × 60 × 60 s)

⇒ 1 ly = 9.46 × 10¹⁵ m

So, d = 4.29 × 9.46 × 10¹⁵ m

⇒ d = 4.058 × 10¹⁶ m

We know, 1 parsec = 3.09 × 10¹⁶ m

So,

⇒ d = 1.313 parsec

We know,

where, θ is the parallax angle

l is the diameter of the Earth’s orbit = 3 × 10¹¹ m

d is the distance of the star from Earth

∴

⇒ θ = 7.39 × 10^-6 rad

⇒

∴ θ = (7.39 × 10^-6 rad)/(π/648000 rad)

⇒ θ = 1.52”

NOTE: It is to be noted that a lightyear is a unit of distance and not time.

Parallax is a difference in the apparent position of an object viewed along two different lines of sight, and is measured by the angle or semi-angle of inclination between those two lines of sight.

It is true that the precise measurements of physical quantities is very essential for science. For instance, laser length measurements require a precision in the order of Angstrom (1 Å = 10^–10 m). In satellite launching through space rockets, time is an important factor and is measured up to a precision of the order of 1 micro second (1μs = 10^-6 s). Mass of sub-atomic particles are measured in the order of 10^-30 kg. Inter-atomic spacing is measured in the order of a few Angstroms which needs to be precise to avoid errors in spectroscopy.

A. Height of water column during monsoon is recorded as 215 cm.

H = 200 cm = 2.0 m

Area of the country, A = 3.3 × 10¹² m²

Volume of water column, V = AH

⇒ V = 3.3 × 10¹² m² × 2.0 m

⇒ V = 6.6 × 10¹² m³

Density of water, ρ = 10³ kg m^-3

Mass of the rain-water, m = Vρ

⇒ m = 6.6 × 10¹² m³ × 10³ kg m^-3

⇒ m = 6.6 × 10¹⁵ kg

The mass of the rain-bearing clouds should be approximately equal to this mass.

B. Consider a boat of known base-area A. Measure the depth upto which it sinks in water when it is empty. Let this height be h.

Volume of water displaced by boat, v = Ah

Measure the depth again when elephant is kept on the boat. Let this depth be h’.

Volume of water displaced, V = Ah’

Volume of water displaced by elephant, V’ = V – v = A(h’-h)

The mass of this volume of water is equal to the mass of the elephant.

Density of water, ρ = 10³ kg m^-3

Mass of water displaced by elephant, m = V’ρ

⇒ m = A(h’-h)×10³ kg

This is approximately equal to the mass of the elephant.

C. Consider a balloon filled with air. When there is no wind, note the height of the balloon from a fixed point. When the wind blows, measure the distance of the balloon from the fixed point after a certain time. The displacement of balloon can be calculated from the angular displacement. Now, the ratio of the displacement to the time of flight is the speed of wind.

Also, an anemometer can be used to measure the speed of the wind. The number of rotations in one second gives the speed of wind.

D. Let A be the area of the head covered with hair.

If r is the radius of hair strand, the base area of hair strand,

a = πr²

So, number of hairs, n = A/a = A/πr²

This gives an approximation of the number of hair strands on a head.

E. If l, b and h are the length, breadth and height of the classroom, then its volume is v = lbh.

If r is the radius of an air molecule, then volume of the air molecule, v’ = (4/3)πr³

So, number of air molecules in the classroom, n = v/v’

⇒

⇒ n = 3lbh/4πr³

This gives an approximation of the number of air molecules in the classroom.

Given,

Mass of the sun, M = 2.0 × 10³⁰ kg

Radius of the sun, R = 7.0 × 10⁸ m

Volume of the sun, V = (4/3)πR³

⇒ V = (4/3) × 3.14 × (7.0 × 10⁸ m)³

⇒ V = 1.436 × 10²⁷ m³

Density of the sun,

⇒

⇒ ρ = 1.39 × 10³ kg m^-3

This value corresponds to the range of solids and liquids.

Given,

Distance of Jupiter from Earth, D = 824.7 million kilometres

⇒ D = 824.7 × 10⁶ × 10³ m

⇒ D = 8.247 × 10¹¹ m

Angular diameter, θ = 35.72”

⇒ θ = 35.72 × 1’/60

Let d be the diameter of Jupiter.

We know,

θ = d/D

⇒ d = θD

⇒ d = 1.43 × 10⁸ m

Dimensions of tan θ = M⁰L⁰T⁰

(∵ All trigonometric functions have no units)

Dimensions for v (velocity) = M⁰L¹T^-1

∴ The equation, tan θ = v is Dimensionally incorrect.

To balance the equation,

Let the speed of the rainfall be V, then we can make R.H.S dimensionless by dividing with V.

tan θ =

(∵ velocity dimension = M⁰L¹T^-1 always)

Now, the equation is dimensionally balanced and correct.

Time allowed to run (T) = 100 years

= 100×365×24×60×60 sec

∴ T = 3.15×10⁹ s

Difference after 100 years (d) = 0.02 s

In 1 s the time difference shown by clock,

∴ The accuracy of the standard caesium clock in measuring time interval of 1s is given by,

A = s ≈ 1.5×10¹¹ s

Given,

Diameter of the sodium atom, (d)= 2.5 Å

∴ Radius, (r) = 1.25 Å

Density of sodium in crystalline phase, (ρ_c)= 970 kg m^-3

Consider every atom of sodium to be sphere,

Volume (V) = r³ m³

∴ V = ×3.14× (1.25×10^-10)³ m³

≈ 8.18×10^-30 m³

We know that,

According to Avogadro law, a mole contains 6.023×10²³ atoms.

The molecular weight of sodium atom, (M) = 23 g = 23×10^-3 kg

Density, (ρ) = kg/m³

∴ ρ = ≈ 4.67×10^-7 kg/m³

Thus,

The density of sodium in crystalline phase (ρ_c) is much higher than the density of general sodium in general configuration (ρ). Because, sodium atoms in crystalline phase have very less inter-atomic separation.

The equation for the radius of the nucleus is given by,

r = r₀A^1/3

∴ Volume on the nucleus, V =

=

=

Mass number is given in the problem as, A.

Thus,

M = Mass number× Mass of single Nucleus

= A×1.67×10^-27 kg

(∵ Mass of single Nucleus = 1.67×10^-27 kg)

∴ Density, ρ =

= = kg/m³

This equation doesn’t have Mass number A and it has only one variable, Radius of the atom (r). Since, every atom has same radius then the density of sodium atom nuclei remains constant.

∴ Density of sodium atom nucleus, ρ _nucleus = kg/m³

2.29×10¹⁷ kg/m³

Given,

Time taken by laser to reflect back from moon to earth = 2.56 s

We know that,

Speed of the light in vacuum = 3×10⁸ m/s

Radius of the lunar orbit (r) = distance between Earth and Moon (d)

Times taken for laser beam to reach moon = 0.5×2.56 = 1.28 s

∴ r = d = speed of light in vacuum × Time taken

= 3×10⁸ × 1.28 m

= 3.84×10⁵ km

Hence, the radius of the lunar orbit is, 3.84×10⁵ km.

Given,

Speed of sound in water = 1450 m/s

Total time taken for transmission = 77 s

Let distance between our submarine and enemy submarine be, D.

∴ Time taken to travel the distance D be, t = 0.5×77 = 38.5 s

Thus,

D = Speed of sound in water × Time taken to travel the distance D

= 1450×38.5 m

= 55.8 km

Quasar: The active galactic centre which emit huge amount of energy in the form of light.

Given,

Time taken by Quasar light to reach earth, T = 3 billion years

∴ T = 3×10⁹×365×24×60×60 s

(∵ 1 billion = 10⁹ years)

We know that,

Speed of light in vacuum, V = 3×10⁸ m/s

∴ Distance between the Earth and Quasar is given by,

D= V×T

= 3×10⁸×(3×10⁹×365×24×60×60) m

= 2.8×10²² km

Fig. The Total Solar Eclipse.

Given, from examples 2.3 and 2.4 we have,

Distance between the Sun and the Earth = 1.496×10¹¹ m

Distance between the Moon and the Earth = 3.84×10⁸ m

Diameter of the Sun = 1.39×10⁹ m

Fig. Schematic diagram of positions of Sun, Moon and Earth during Total Solar Eclipse.

From the above figure,

Δ PQT and Δ RST are similar. Thus,

=

Where,

PQ = Diameter of the Sun

RS = Diameter of the Moon

VT = Distance between the Sun and the Earth

UT = Distance between the Moon and the Earth

∴ =

⇒ RS = ×10⁶ m = 3.57×10⁶ m

Hence, The diameter of the Moon = RS = 3.57×10³ km.

One relation consists of some fundamental units that give the age of the universe is,

T = × s

Where,

T = Age of the Universe

e = charge of Electron = -1.6×10^-19 C

ϵ₀ = absolute permittivity

m_p = Mass of the Proton = 1.67×10^-27 kg

m_e = Mass of the Electron = 9.1×10^-31 kg

c = Speed of light in vacuum = 3×10⁸ m/s

G = Universal Gravitational constant = 6.67×10¹¹ Nm²kg^-2

And, = 9×10⁹ Nm² /C²

∴ T =

≈ 6×10⁹ years

≈ 6 billion years

(∵ 1 billion year = 10⁹ years)

Hence, The age of the universe, T ≈ 6 billion years.