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NCERT · Class 11 · Physics

Chapter 1: Units and Measurements

Practice each question, then open the answer or worked solution.

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Questions

30 questions
Q2.23Written answer

The Sun is a hot plasma (ionised matter) with its inner core at a temperature exceeding 107 K, and its outer surface at a temperature of about 6000 K. At these high temperatures, no substance remains in a solid or liquid phase. In what range do you expect the mass density of the Sun to be, in the range of densities of solids and liquids or gases? Check if your guess is correct from the following data: a mass of the Sun = 2.0 × 1030 kg, the radius of the Sun = 7.0 × 108 m.

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Worked solution

Mass = 2 x 1030 kg

Radius = 7 x 108 m

Volume V =

\(\begin{array}{l}\frac{4}{3}\pi r^{3}\end{array} \)

=

\(\begin{array}{l}\frac{4}{3}\end{array} \)
\(\begin{array}{l}\frac{22}{7}\end{array} \)
83

=

\(\begin{array}{l}\frac{88}{21}\end{array} \)
24 3243

Density =

\(\begin{array}{l}\frac{Mass}{Volume}\end{array} \)
\(\begin{array}{l}\frac{3\times 10^{30}}{2145.52\times 10^{24}}\end{array} \)
35

The density is in the range of solids and liquids. Its density is due to the high gravitational attraction on the outer layer by the inner layer of the sun.

Q2.24Written answer

When the planet Jupiter is at a distance of 824.7 million kilometres from the Earth, its angular diameter is measured to be 35.72″ of arc. Calculate the diameter of Jupiter.

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Worked solution

Distance of the planet Jupiter from Earth, D= 824.7 million kilometres  = 824.7 x 106 km

Angular diameter θ = 35.72 “= 35.72 x 4.85 x 10-6 rad
= 173.242 x 10-6 rad
Diameter of Jupiter d = θ x D= 173.241 x 10-6x 824.7 x 106 km
=142871 = 1.43 x 105 km

Q2.25Written answer

A man walking briskly in the rain with speed v must slant his umbrella forward, making an angle θ with the vertical. A student derives the following relation between θ and v: tan θ = 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 a stationary man). Do you think this relation can be correct? If not, guess the
correct relation.

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Worked solution

According to the principle of homogeneity of dimensional equations,
Dimensions of L.H.S. = Dimensions of R.H.S.

In relation v = tan θ, tan θ is a trigonometric function, and it is dimensionless. The dimension of v is  [L1 T-1]. Therefore, this relation is incorrect.
To make the relation correct, the L.H.S. must be divided by the velocity of rain, u.

Therefore, the relation becomes
v/u= tan θ

This relation is correct dimensionally

Q2.26Written answer

It is claimed that two caesium clocks, if allowed to run for 100 years, free from any disturbance, may differ by only about 0.02 s. What does this imply for the accuracy of the standard caesium clock in measuring a time interval of 1 s?

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Worked solution

Total time = 100 years = 100 x 365 x 24 x 60 x 60 s

Error in 100 years = 0.02 s
Error in 1 second=0.02/100 x 365 x 24 x 60 x 60
=6.34 x 10-12 s
The accuracy of the standard caesium clock in measuring a time-interval of 1 s is 10-12 s.

Q2.27Written answer

Estimate the average mass density of a sodium atom, assuming its size to be about 2.5 Å (Use the known values of Avogadro’s number and the atomic mass of sodium). Compare it with the mass density of sodium in its crystalline phase: 970 kg m–3. Are the two densities of the same order of magnitude? If so, why?

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Worked solution

The diameter of sodium= 2.5 A = 2.5 x 10-10 m

Therefore, the radius is 1.25 x 10-10 m

The volume of a sodium atom, V= (4/3)πr3

= (4/3) x (22/7) x (1.25 x 10-10) 3= 8.177 x 10-30 m3

Mass of one mole atom of sodium = 23 g = 23 x 10-3 kg

1 mole of sodium contains 6.023 x 1023 atoms

Therefore, the mass of one sodium atom, M= 23 x 10-3/6.023 x 1023= 3.818 x 10-26 kg

The atomic mass density of sodium, ρ= M/V
=3.818 x 10-26/8.177 x 10-30

= 0.46692 x 104= 4669.2 kg m-3
The density of sodium in its solid state is 4669.2  kg m-3, but in the crystalline phase, the density is 970 kg m-3. Hence, both are in a different order. In the solid phase, atoms are tightly packed, but in the crystalline phase, atoms arrange a sequence which contains a void. So, the density in the solid phase is greater than in the crystalline phase.

Q2.28Written answer

The unit of length convenient on the nuclear scale is a fermi: 1 f = 10–15 m. Nuclear sizes obey roughly the following empirical relation :
r = r0 A1/3
where r is the radius of the nucleus, A its mass number, and r0 is a constant equal to
about 1.2 f. Show that the rule implies that nuclear mass density is nearly constant
for different nuclei. Estimate the mass density of the sodium nucleus. Compare it with the average mass density of a sodium atom obtained in Exercise. 2.27.

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Worked solution

The radius of the nucleus

r = r0 A1/3

ro = 1.2 f = 1.2 x 10-15 m

Considering the nucleus is spherical. Volume of nucleus
= 4/3 πr3 = 4/3 π [r0 A1/3]3 = 4/3 πr03A
Mass of nucleus = mA
m is the average mass of the nucleon.
A is the number of nucleons.
Nuclear mass density = Mass of nucleus/Volume of the nucleus
= mA/(4/3πr3) = 3mA/4πr3 = 3mA/4πr03A
= 3m/4πr03
Using m = 1.66 x 10-27  kg  and ro = 1.2 f = 1.2 x 10-15 m in the above equation
= 3 x 1.66 x 10-27 /4 x 3.14 x ( 1.2 x 10-15)3= 4.98 x 10-27/21. 703 x 10-45= 2.29 x 1017 kg/m3

So, the nuclear mass density is much larger than the atomic mass density for a sodium atom we got in 2.27.

Q2.29Written answer

A LASER is a source of very intense, monochromatic, and unidirectional beam of light. These properties of laser light can be exploited to measure long distances. The distance of the Moon from the Earth has already been 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 lunar orbit around the Earth?

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Worked solution

Time taken for the laser beam to return to Earth after reflection by the Moon’s surface = 2.56 s

The speed of laser light is c = 3 x 108 m/s.

Let d be the distance of the Moon from the Earth,

The time taken by laser signal to reach the Moon, t = 2d/c

Therefore, d = tc/2 = (2.56 x 3 x 108)/2 = 3.84 x 108 m

Q2.30Written answer

2. 30. A SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate objects underwater. In a submarine equipped with a SONAR, the time delay between the 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–1).

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Worked solution

Speed of sound in water,v = 1450 m s–1

The time between generation and the reception of the echo after reflection, 2t= 77.0 s

Time taken for the sound waves to reach the submarine, t = 77.0/2 = 38. 5 s

Then v = d/t

Distance of enemy submarine, d  = tv

Therefore, d=vt=(1450 x 38. 5) =55825 m=55.8 x 103 m or 55.8 km.

Q2.31Written answer

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 not yet been satisfactorily explained. What is the distance in km of a quasar from which light takes 3.0 billion years to reach us?

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Worked solution

Time taken by light from the quasar to reach the observer, t = 3.0 billion years = 3.0 x 109 years = 3.0 x 109 x 365 x 24 x 60 x 60 s

= 94608000 x 109  s

=  9.46 x 1016 m

Speed of light = 3 x 108 m/s
Distance of quasar from Earth  = 3.0 x 108 x 9.46 x 1016 m
= 28.38 x 1024 m

Q2.32Written answer

It is a well-known fact that during a total solar eclipse-the disk of the moon almost completely covers the disk of the Sun. From this fact and the information you can gather from examples 2.3 and 2.4, determine the approximate diameter of the moon.

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Worked solution

From examples 2.3 and 2.4, we get the following data

Distance of the Moon from Earth = 3.84 x 108 m

Distance of the Sun from Earth = 1.496 x 1011 m

Sun’s diameter = 1.39 x 109 m

Sun’s angular diameter,θ = 1920″ = 1920 x 4.85 x 10-6 rad = 9.31 x 10-3 rad [1″ = 4.85 x 10-6 rad]

During a total solar eclipse, the disc of the moon completely covers the disc of the sun, so the angular diameter of both the sun and the moon must be equal.

Therefore, the Angular diameter of the moon, θ = 9.31 x 10-3 rad
The earth-moon distance, S = 3.8452 x 108 m

Therefore, the diameter of the moon, D = θ x S
= 9.31 x 10-3 x 3.8452 x 108 m = 35.796 x 105 m