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Q2.32:
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.
Solution :
Step 1: Establishing the Geometric Model
To determine the diameter of the moon, we model the solar eclipse as a problem of similar triangles or angular diameters. Since the moon covers the Sun, their angular diameters $\alpha$ as seen from Earth must be approximately equal.
The relationship between the linear diameter ($D$), the distance from the observer ($d$), and the angular diameter ($\alpha$) in radians is given by:
$\alpha = \frac{D}{d}$ [Per the definition of radian measure for small angles]
Step 2: Identifying Given Constants and Parameters
From standard astronomical data referenced in introductory physics curricula (specifically Examples 2.3 and 2.4 of the NCERT Physics textbook):
- Distance of the Sun from Earth ($d_s$): $\approx 1.496 \times 10^{11} \text{ m}$
- Diameter of the Sun ($D_s$): $\approx 1.39 \times 10^9 \text{ m}$
- Distance of the Moon from Earth ($d_m$): $\approx 3.84 \times 10^8 \text{ m}$
Step 3: Calculating the Angular Diameter
Using the dimensions of the Sun:
$\alpha = \frac{D_s}{d_s} = \frac{1.39 \times 10^9 \text{ m}}{1.496 \times 10^{11} \text{ m}} \approx 9.29 \times 10^{-3} \text{ radians}$
Step 4: Deriving the Diameter of the Moon
Since the moon subtends the same angle $\alpha$ at Earth during the eclipse, we equate the angular diameter of the moon ($\alpha_m$) to the angular diameter of the Sun ($\alpha_s$):
$\frac{D_m}{d_m} = \frac{D_s}{d_s}$ [By the condition of total solar eclipse]
$D_m = D_s \times \left( \frac{d_m}{d_s} \right)$
$D_m = (1.39 \times 10^9 \text{ m}) \times \left( \frac{3.84 \times 10^8 \text{ m}}{1.496 \times 10^{11} \text{ m}} \right)$
$D_m = 1.39 \times 10^9 \times 0.0025668$
$D_m \approx 3.568 \times 10^6 \text{ m}$
Step 5: Geometric Visualization
Step 6: Final Tabulation of Data
| Parameter | Value |
|---|---|
| Distance to Sun ($d_s$) | $1.496 \times 10^{11} \text{ m}$ |
| Distance to Moon ($d_m$) | $3.84 \times 10^8 \text{ m}$ |
| Diameter of Sun ($D_s$) | $1.39 \times 10^9 \text{ m}$ |
| Diameter of Moon ($D_m$) | $3.57 \times 10^6 \text{ m}$ |
Final Solution: The approximate diameter of the moon is $3.57 \times 10^6 \text{ m}$ (or $3570 \text{ km}$).
More Questions from Class 11 Physics Units and Measurements Additional Exercises
- Q2.25: A man walking briskly in rain with speed $v$ must slant his umbrella forward making an angle $\theta$ with the vertical. A student derives the following relation between $\theta$ and $v$ : $\tan \theta = v$ and checks that the relation has a correct limit: as $v \to 0$, $\theta \to 0$, 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.
- Q2.25: A man walking briskly in rain with speed (v) must slant his umbrella forward making an angle (\theta) with the vertical. A student derives the following relation between (\theta) and (v) : (\tan \theta = v) and checks that the relation has a correct limit: as (v \rightarrow 0), (\theta \rightarrow 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.
- Q2.26: It is claimed that two cesium clocks, if allowed to run for 100 years, free from any disturbance, may differ by only about 0.02 s. What does this imply for the accuracy of the standard cesium clock in measuring a time-interval of 1 s ?
- Q2.26: It is claimed that two cesium clocks, if allowed to run for 100 years, free from any disturbance, may differ by only about 0.02 s. What does this imply for the accuracy of the standard cesium clock in measuring a time-interval of 1 s ?
- Q2.27: 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 ?
- Q2.27: 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 ?
- Q2.28: 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 : $r = r_0 A^{1/3}$ where $r$ is the radius of the nucleus, $A$ its mass number, and $r_o$ 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 sodium nucleus. Compare it with the average mass density of a sodium atom obtained in Exercise. 2.27.
- Q2.28: 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 = r_{0} A^{1/3}) where (r) is the radius of the nucleus, (A) its mass number, and (r_{o}) 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 sodium nucleus. Compare it with the average mass density of a sodium atom obtained in Exercise. 2.27.
- Q2.29: A LASER is a source of very intense, monochromatic, and unidirectional beam of light. These properties of a laser light can be exploited to measure long distances. The distance of the Moon from the Earth has been already 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 ?
- Q2.29: A LASER is a source of very intense, monochromatic, and unidirectional beam of light. These properties of a laser light can be exploited to measure long distances. The distance of the Moon from the Earth has been already 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 ?
- Q2.30: 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$^{–1}$).
- Q2.30: 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})).
- Q2.31: 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 ?
- Q2.31: 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 ?
- Q2.32: 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.
- Q2.33: 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 ($\sim$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?
- Q2.33: 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?
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