Page No 38: - Chapter 2 Units & Measurements class 11 ncert solutions Physics - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Question 2.30:

A SONAR (sound navigation and 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^–1).

Answer:

Let the distance between the ship and the enemy submarine be ‘S’.

Speed of sound in water = 1450 m/s

Time lag between transmission and reception of Sonar waves = 77 s

In this time lag, sound waves travel a distance which is twice the distance between the ship and the submarine (2S).

Time taken for the sound to reach the submarine

∴ Distance between the ship and the submarine (S) = 1450 × 38.5 = 55825 m = 55.8 km

Question 2.31:

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 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?

Answer:

Time taken by quasar light to reach Earth = 3 billion years

= 3 × 10⁹ years

= 3 × 10⁹ × 365 × 24 × 60 × 60 s

Speed of light = 3 × 10⁸ m/s

Distance between the Earth and quasar

= (3 × 10⁸) × (3 × 10⁹ × 365 × 24 × 60 × 60)

= 283824 × 10²⁰ m

= 2.8 × 10²² km

Question 2.32:

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 from the information you can gather from examples 2.3 and 2.4, determine the approximate diameter of the moon.

Answer:

The position of the Sun, Moon, and Earth during a lunar eclipse is shown in the given figure.

Distance of the Moon from the Earth = 3.84 × 10⁸ m

Distance of the Sun from the Earth = 1.496 × 10¹¹ m

Diameter of the Sun = 1.39 × 10⁹ m

It can be observed that ΔTRS and ΔTPQ are similar. Hence, it can be written as:

Hence, the diameter of the Moon is 3.57× 10⁶ m.

Question 2.33:

A great physicist of this century (P.A.M. Dirac) loved playing with numerical values of Fundamental constants of nature. This led him to an interesting observation. Dirac found that from the basic constants of atomic physics (c, e, mass of electron, mass of proton) and the gravitational constant G, he could arrive at a number with the dimension of time. Further, it was a very large number, its magnitude being close to the present estimate on the age of the universe (~15 billion years). From the table of fundamental constants in this book, try to see if you too can construct this number (or any other interesting number you can think of). If its coincidence with the age of the universe were significant, what would this imply for the constancy of fundamental constants?

Answer:

One relation consists of some fundamental constants that give the age of the Universe by:

Where,

t = Age of Universe

e = Charge of electrons = 1.6 ×10^–19 C

 = Absolute permittivity

= Mass of protons = 1.67 × 10^–27 kg

= Mass of electrons = 9.1 × 10^–31 kg

c = Speed of light = 3 × 10⁸ m/s

G = Universal gravitational constant = 6.67 × 10¹¹ Nm² kg^–2

Also, Nm²/C²

Substituting these values in the equation, we get