# NCERT Solution , Units & Measurements

Question:16.The unit of length convenient on the atomic scale is known as an angstrom and is denoted by A° (1A = 10-10 m ) The size of a hydrogen atom is about 0.5 Å. What is the total atomic volume in m3 of a mole of hydrogen atom?

Solution. Given r = 0.5 Å = 0.5 × 10-10 m

Volume of each atom hydrogen $\large = \frac{4}{3}\pi r^3$

$\large = \frac{4}{3}\times 3.14 \times (0.5 \times 10^{-10})^3$

= 5.236 × 10-31 m3

No of atoms in 1 mole of H2 = 6.023 × 10-23 (Avogadro Number)

Atomic volume = 5.236 × 10-31 × 6.023 × 1023

= 3.154 × 10-7 m3.

Question:17 . One mole of an ideal gas at N.T.P occupies 22.4 litres (molar volume) . What is the ratio of molar volume to atomic volume of a mole of hydrogen . Take size of hydrogen molecule to be 1A° . Why is this ratio so large ?

Solution: $\large Atomic \; volume = \frac{4}{3}\pi R^3 \times N$

$\large = \frac{4}{3}\pi (0.5 \times 10^{-10})^3 \times 6.023 \times 10^{23}$

= 3.154 × 10-7 m3

Molar volume = 22.4 litre = 22.4 × 10-3

$\large \frac{Molar \; volume}{Atomic \; volume} = \frac{22.4 \times 10^{-3}}{3.154 \times 10^{-7}}$

= 7.1 × 103

This ratio is large due to large intermolecular separation .

Question: 18. Explain this common observation clearly . If you look out of the window of a fast moving trains , the nearby trees , house etc . seem to move rapidly in a direction opposite to the train’s motion , but the distant objects (hill tops , the moon , the stars etc. ) seem to be stationary .

Solution : The line joining the object to the eye is called the Line of sight . When a train moves rapidly , the line of sight of a nearby tree changes its direction of motion rapidly . Therefore the trees appear to run in opposite direction .

On contrary , the line of sight of far off objects (hill tops , the moon , the stars etc. ) does not change its direction so much , due to extremely large distance from the eye . Hence distant hill tops , moon , the stars etc appear stationary .

Question:19. The principle of paradox is used in the determination of distances of very distant starts. The base line AB is the line joining the Earth’s two locations six months apart on its orbit around the sun. That is, the base line is about the diameter of the Earth’s orbit = 3 × 1011 However even the nearest starts are so distant that with such a long base line of length on the astronomical scale. It is the distance of on object that will show a parallax of 1” (second) of arc . A parsec is a convenient unit of length on the astronomical scale it is the distance of an object that will show a parallax of 1” (second) of arc from opposite ends of a base line equal to the distance from the Earth to the sun. how much is a parsec in terms of metres?

Solution: Length of base line = distance from earth of sun = 1 A.V. = 1.5 × 1011 m

θ = parallax angle = 1” = 1’/60 = 1°/(60×60)

$\displaystyle = \frac{\pi}{180}\times \frac{1}{60\times 60}$ radian

Now l = r θ hence , r =l/θ

$\displaystyle = \frac{1.5 \times 10^{11}}{\frac{\pi}{180}\times \frac{1}{60\times 60}}$

= 3.1 × 1016 m

1 Parsec = 3.1 × 1016 m.

Question:20.The nearest star to our solar system is 4.29 light years away. How much is this distance in terms of par sec ? . How much parallex  would this star show when viewed from location of the earth six months apart in its orbit around the sun ?

Solution: x = 4.29 l y = 4.29 × 9.46 × 1015 m

$\large= \frac{4.29 \times 9.46 \times 10^{15}}{3.1 \times 10^{16}} parsec$

4.29 l y = 1.323 parsec

$\theta = \frac{l}{r} = \frac{2 AU}{x}$

$\large = \frac{2 \times 1.496 \times 10^{11}}{4.29 \times 9.46 \times 10^{15}}$

= 1.512 sec

Question:21 Precise measurement of a physical quantities are a need of Science . For example , to ascertain the speed of an aircraft , one must have an accurate method to find its position at closely separated instants of time . This was the actual motivation behind the discovery of radar in World War II . Think of different examples in modern science where precise measurements of length , time , mass etc. are needed . Also wherever you can , give a quantitative idea of the precision needed .

Solution: Precise measurements of Physical quantities like length , mass and time are the primary requirements for development of quantities laws of physics or any other science .
For example : In the measurement of distance of moon from earth by Laser beam , very accurate measurement of time is required .
Similarly , for measurement of distance , velocity of an aeroplane by Radar method , time measurement to be accurate . For measuring distances of nearby stars , accurate measurement of parallex angle is required .

Question:23.The sun is a hot plasma (ionised matter) with its inner core at a temperature exceeding 107 and its outer surface at a temperature of about 6000 K. At such high temps no substance remains in a solid or liquid phase. In what range do you expect the mass density of the sum to be? In the range of densities of solids, liquids or gases? Check if your guess is correct from the following data mass of sun = 2 × 1030 kg, radius of the sun = 7 × 108 m.

Solution: M = 2 × 1030 kg, r = 7 × 108 m

Density, $\large \rho = \frac{M}{V} = \frac{M}{\frac{4}{3}\pi r^3}$

$= \frac{3 M}{4 \pi r^3}$

$\large = \frac{3 \times 2 \times 10^{30}}{4 \times 3.14 \times (7 \times 10^8)^3}$

= 1.392 × 103 kg/m3.

Which is a order of density of solids and liquids , not gases.

High density of sun is due to inward gravitational force on outer layers due to inner layers of the sun .

Question:24. When a planet Jupiter is a distance of 824.7 million km from Earth, its angular diameter is measured to be 35.72“ of arc. Calculate the diameter of Jupiter?

Solution. r = 824.7 × 106 Km

θ = 35.72”

$= \frac{35.72}{60×60} \times \frac{\pi}{180}$ radian

l  = r θ

$= 824.7 \times 10^6 \times \frac{35.72}{60×60} \times \frac{\pi}{180} km$

= 1.429 × 105

Question: 25. A man walking briskly in rain with speed v must slant his umbrella forward making an angle (with the vertical ) . A student derives the following relation between θ and v : tanθ = v and checks that the relation has a correct limit : as v → 0 , θ → 0 as expected . (We are assuming there is no strong wind and that the rain falls vertically for stationary man) . Do you think this relation can be correct ? If not , guess the correct relation .

Solution : The Relation , tanθ = v

R.H.S = tanθ = [M0L0T0]

L.H.S = v = [M0L1T-1]

L.H.S ≠ R.H.S

Hence , Relation is not correct dimensionally .

Question:26.It is claimed that two cesium clocks , if allowed to run for 100 years, free from any disturbance, may differ by only about 0.025 s. what does this imply for the accuracy of the standard cesium clock in measuring a time interval of 1 s?

Solution:  $100 years = 100 \times 365 \frac{1}{4} \times 24 \times 60 \times 60$

Error in 100 years = 0.025 s

Error in 1 sec $\large = \frac{0.025}{100 \times 365 \frac{1}{4} \times 24 \times 60 \times 60}$

= 7.9 × 10-13 s  ≈ 10-12 s

Question:27.Estimate the average atomic mass density of a sodium atom, assuming its size to be 2.5 Å. Company it with density of sodium in its crystalline phase (970 kgm-3). Are the densities of the same order of magnitude? If so why?

Solution: $\large Atomic \;volume = \frac{4}{3} \pi R^3 \times N$

$\large = \frac{4}{3} \times \frac{22}{7} (1.25 \times 10^{-10})^3 \times 6.023 \times 10^{23} m^3$

= 4.93 × 10-6 m3

Average mass density = Mass/Volume

$\large = \frac{23 \times 10^{-3}}{4.93 \times 10^{-6}}$

= 4.67 × 103 kg/m3

The two densities are of different order, which is due to interatomic spacing in the crystalline phase.

Question:28. The unit of length convention on nuclear scale is a Fermi 1 f = 10-15 Nuclear sizes obey roughly the following empirical relation, r = r0 where r is radius of the nucleus and r0 is a constant equal to 1.2 f. show that the rule implies that nuclear mass density is constant for different of sodium nucleus. Compare it with average mass density of sodium atom (4.67 × 103 kg/m3).

Solution: Let m = average mass of proton or nucleon

Nucleus mass M = mA and radius of nucleus, r = r0A1/3

Nuclear density, $\large \rho = \frac{mass}{volume}$

$\large \rho = \frac{M}{\frac{4}{3}\pi r^3}$

$\large \rho = \frac{m A}{\frac{4}{3}\pi (r_0 A^{1/3})^3}$

$\large \rho = \frac{3 m}{4 \pi r_0^3}$

As r0 and m are constant, hence nuclear density be same for all nuclei .

Now, m = 1.60 × 10-27 kg

and r0 = 1.2 f = 1.2 × 10-15 m

$\large \rho = \frac{3 \times 1.6 \times 10^{-27}}{4 \times 3.14 \times (1.2 \times 10^{-15})^3}$

= 2.29 × 1017 kg m-3

Question:29. A LASER is source of very intense, monochromatic and unidirectional beam of light. These properties of a of a laser light can be to measure long distances. The distance of the Moon from the Earth has determined very precisely using a laser as a source of light. A laser light beamed at the Moon takes 2.56 s to return after reflection at the Moon’s surface. How much is the radius of the luner orbit around the Earth?

Solution: Given, t = 2.56 s;

Velocity of laser is vacuum, c = 3 × 108 m/s

Let, x =  distance of moon from earth (i.e. radius of lunar orbit)

$\large x = \frac{c \times t}{2}$

$\large x = \frac{3 \times 10^8 \times 2.56}{2}$

= 3.84 × 108 m.

Question : 30 . A SONAR (Sound Navigation & Ranging) uses ultrasonic waves to detect and locate objects under water . In a submarine equipped with a SONAR , the time delay between generation of a probe wave and the reception of its echo after reflection from an enemy submarine is found to be 77.0 s. What is the distance of the enemy submarine ? (Speed of sound in water = 1450 m/s)

Sol: t = 77.0 sec , x = ? , v = 1450 m/s

$\large x = \frac{v \times t}{2}$

$\large x = \frac{1450 \times 77.0}{2}$

x = 55825 m

Question:31. The farthest objects in our universe discovered by modern astronomers are so distant that light emitted by them takes billions of years to reach the earth . These objects (known as quasars )  have many puzzling features , which have yet not be satisfactorily explained . What is the distance in km of a quasar from which light takes 3.0 billion years to reach us ?

Solution: Time t = 3 billion years = 3 × 109 yr = 3 × 109 × 365 × 24 × 60 × 60 sec

Speed of light in vacuum , c = 3 × 108 m/s = 3 × 105 km/s

Since , Distance = Speed × time

Distance = ( 3 × 105 ) × (3 × 109 × 365 × 24 × 60 × 60)

= 2.84 × 1022 km

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