Given Variables & Initial Setup

To determine the distance of the enemy submarine, we define the parameters provided by the SONAR system:

• Total time delay ($\Delta t$): $77.0\text{ s}$
• Speed of sound in water ($v$): $1450\text{ m s}^{-1}$
• Distance to be determined ($d$): $?$

[Per the principle of echo-location, the signal travels from the source to the target and then reflects back to the source, covering the distance $d$ twice.]

Step 1: Establishing the Kinematic Equation

The total distance ($D$) covered by the ultrasonic wave is the round trip to the target and back. This can be expressed as:

$D = 2d$

Given the constant velocity ($v$) of the wave in the medium, the relationship between distance, velocity, and time is defined by the basic kinematic formula:

$D = v \cdot \Delta t$

[By the definition of average velocity in uniform motion: $v = \frac{\Delta x}{\Delta t}$]

Step 2: Substitution and Algebraic Manipulation

By equating the two expressions for total distance ($D$):

$2d = v \cdot \Delta t$

To solve for the one-way distance ($d$), we rearrange the equation:

$d = \frac{v \cdot \Delta t}{2}$

Step 3: Numerical Computation

Substituting the given numerical values into the derived expression:

$d = \frac{1450\text{ m s}^{-1} \cdot 77.0\text{ s}}{2}$

$d = \frac{111650\text{ m}}{2}$

$d = 55825\text{ m}$

Step 4: Significant Figure Adjustment

Per the rules of significant figures, the calculation involves values with three significant figures ($77.0$ and $1450$). The final result should be expressed to maintain scientific precision:

$d = 5.58 \times 10^4\text{ m}$

$d = 55.8\text{ km}$

Visual Representation of the Echo Path

Final Solution: The distance of the enemy submarine is 55,825 meters or 55.8 km.

Step 1: Establishing the Total Time Interval

To determine the accuracy, we first define the total time duration in seconds over which the discrepancy is measured. Given the period of $T = 100 \text{ years}$.

Calculations for the total number of seconds ($T_{total}$):

• $1 \text{ year} \approx 365.25 \text{ days}$ (accounting for leap years)
• $1 \text{ day} = 24 \text{ hours} \times 60 \text{ minutes} \times 60 \text{ seconds} = 86,400 \text{ s}$
• $T_{total} = 100 \times 365.25 \times 86,400 \text{ s}$
• $T_{total} = 3.15576 \times 10^9 \text{ s}$

Step 2: Defining the Error and Fractional Uncertainty

The total error ($\Delta T$) accumulated over this duration is given as $0.02 \text{ s}$. The accuracy of a clock is defined by its fractional error (relative error) in measuring a time interval.

The fractional uncertainty ($\frac{\Delta T}{T}$) is defined as:

$\text{Fractional Error} = \frac{\Delta T}{T_{total}}$

Substituting the values:

$\text{Fractional Error} = \frac{0.02}{3.15576 \times 10^9}$

$\text{Fractional Error} \approx 6.337 \times 10^{-12} \text{ s}$

Step 3: Calculating Accuracy per Second

The question asks for the accuracy in measuring a time interval of $1 \text{ s}$. The uncertainty associated with $1 \text{ s}$ of measurement is represented by the fractional error derived above.

$\text{Accuracy} \approx 6 \times 10^{-12} \text{ s per second}$

Visual Representation of Relative Precision

Summary of Implications

This implies that the standard cesium clock is extremely precise. For every second measured, the uncertainty introduced by the clock mechanism is on the order of $10^{-12}$ seconds. This level of precision is fundamental for GPS synchronization, deep space navigation, and high-frequency telecommunications.

Final Solution: The accuracy of the standard cesium clock in measuring a time interval of 1 s is approximately $6 \times 10^{-12} \text{ s}$.

Step 1: Identifying the Given Parameters

To determine the accuracy of a cesium clock, we must first define the total elapsed time ($T$) and the total uncertainty (drift) accumulated over that duration ($\Delta T$).

• Total time duration ($T$): $100 \text{ years}$
• Maximum uncertainty in $100 \text{ years}$ ($\Delta T$): $0.02 \text{ s}$

Step 2: Converting Units to SI Base Units

To perform a consistent calculation, we convert the total time ($T$) into seconds ($s$).

• $1 \text{ year} = 365.25 \text{ days}$ (accounting for leap years)
• $1 \text{ day} = 24 \text{ hours} \times 60 \text{ minutes} \times 60 \text{ seconds} = 86,400 \text{ s}$
• $T = 100 \text{ years} \times 365.25 \text{ days/year} \times 86,400 \text{ s/day}$
• $T = 3.1536 \times 10^9 \text{ s}$ [Per the standard conversion constants for time]

Step 3: Calculating Relative Accuracy

The accuracy of a measurement device is defined by its relative error or fractional uncertainty. The relative accuracy for the total duration $T$ is given by:

$\text{Relative Accuracy} = \frac{\Delta T}{T}$

Substituting the values:

$\text{Relative Accuracy} = \frac{0.02 \text{ s}}{3.1536 \times 10^9 \text{ s}}$

$\text{Relative Accuracy} \approx 6.34 \times 10^{-12}$ [By applying the ratio of error to the total measured magnitude]

Step 4: Interpreting Accuracy for a 1-second Interval

The accuracy found in Step 3 represents the drift per second. If a clock has an accuracy of $\delta t$ for a time interval of $1 \text{ s}$, it implies that for any $1 \text{ s}$ measurement, the potential error is approximately $6 \times 10^{-12} \text{ s}$.

Visual Representation of Fractional Error

Summary Analysis

The calculation demonstrates that the cesium clock is exceptionally precise. An accuracy of $10^{-12}$ implies that the device is reliable to within 1 part in roughly 150 billion. This confirms that the standard cesium clock provides a highly stable timekeeping reference, as the deviation is negligible over standard experimental time scales.

Final Solution: The accuracy of the standard cesium clock in measuring a time interval of 1 s is approximately $6 \times 10^{-12} \text{ s}$.

Step 1: Dimensional Analysis of the Proposed Relation

To determine the validity of the relation $\tan \theta = v$, we must evaluate the dimensions of both sides of the equation. According to the Principle of Homogeneity of Dimensions, both sides of a physical equation must have identical dimensions.

• Left-Hand Side (LHS): The expression is $\tan \theta$. Since $\theta$ is an angle (ratio of arc length to radius), it is a dimensionless quantity $[M^0 L^0 T^0]$. Therefore, $\tan \theta$ is dimensionless.
• Right-Hand Side (RHS): The variable $v$ represents the speed of the man, which has dimensions of length per unit time $[M^0 L^1 T^{-1}]$.

[Per the Principle of Dimensional Homogeneity, a dimensionless quantity cannot be equated to a quantity with dimensions of speed.] Thus, the relation $\tan \theta = v$ is dimensionally inconsistent.

Step 2: Deriving the Correct Physical Relation

Let the velocity of the rain be $\vec{v}_r$ (acting vertically downward) and the velocity of the man be $\vec{v}_m$ (acting horizontally). The velocity of the rain relative to the man is given by:

$\vec{v}_{rm} = \vec{v}_r - \vec{v}_m$

To protect himself from the rain, the man must tilt his umbrella at an angle $\theta$ with the vertical, which is the angle of the relative velocity vector $\vec{v}_{rm}$ with the vertical axis.

Step 3: Calculating the Tangent of the Angle

From the velocity vector triangle formed by $\vec{v}_r$ and $\vec{v}_m$:

$\tan \theta = \frac{|\vec{v}_m|}{|\vec{v}_r|}$

Substituting the magnitudes, where $|\vec{v}_m| = v$ (the speed of the man) and $|\vec{v}_r| = u$ (the terminal speed of the rain):

$\tan \theta = \frac{v}{u}$

Step 4: Verification of Consistency

By including the term $u$ (speed of rain), the dimensions are now balanced:

• RHS dimensions: $\frac{[L T^{-1}]}{[L T^{-1}]} = [M^0 L^0 T^0]$ (Dimensionless).

This satisfies the condition of dimensional homogeneity. As $v \to 0$, $\tan \theta \to 0$, which implies $\theta \to 0$, consistent with the observed limit.

Final Solution: The relation $\tan \theta = v$ is incorrect because it is dimensionally inconsistent. The correct relation is $\tan \theta = \frac{v}{u}$, where $v$ is the speed of the man and $u$ is the speed of the rain.

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

Final Solution: The approximate diameter of the moon is $3.57 \times 10^6 \text{ m}$ (or $3570 \text{ km}$).

Step 1: Given Variables & Initial Setup

To determine the distance between the Earth and the Moon (radius of the lunar orbit), we utilize the principles of wave propagation and the constant speed of electromagnetic radiation in a vacuum.

• Total time taken for the round trip ($t_{total}$): $2.56\text{ s}$
• Speed of light ($c$): $3 \times 10^8\text{ m/s}$ [Standard physical constant]

Step 2: Calculating One-Way Transit Time

The laser pulse travels from the Earth to the Moon and then back to the Earth. Since the distance is covered twice, the time taken for a one-way trip ($t$) is exactly half of the total recorded time.

$t = \frac{t_{total}}{2}$

$t = \frac{2.56\text{ s}}{2} = 1.28\text{ s}$ [By the principle of symmetry in travel time]

Step 3: Calculating Distance (Radius of Lunar Orbit)

The distance ($d$) is defined by the product of the speed of the wave ($v$) and the time taken ($t$). In this context, $v = c$.

$d = c \times t$

$d = (3 \times 10^8\text{ m/s}) \times (1.28\text{ s})$

$d = 3.84 \times 10^8\text{ m}$

Step 4: Visualization of the Measurement Process

Step 5: Tabular Summary of Calculation

Final Solution: The radius of the lunar orbit around the Earth is $3.84 \times 10^8\text{ m}$ (or $3.84 \times 10^5\text{ km}$).

Step 1: Theoretical Proof of Constant Nuclear Density

To prove that the nuclear mass density $\rho_n$ is constant regardless of the mass number $A$, we define the mass and volume of a nucleus:

• Let $m_p$ be the average mass of a nucleon (proton or neutron), $m_p \approx 1.67 \times 10^{-27}\text{ kg}$.
• Total nuclear mass $M = A \cdot m_p$. [Assumption: Neglecting binding energy mass defect for density estimation].
• Nuclear radius $r = r_0 A^{1/3}$.
• Nuclear volume $V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (r_0 A^{1/3})^3 = \frac{4}{3}\pi r_0^3 A$.

Now, calculate density $\rho_n$:

$\rho_n = \frac{M}{V} = \frac{A \cdot m_p}{\frac{4}{3}\pi r_0^3 A}$

Since the term $A$ cancels out in the numerator and denominator, we are left with:

$\rho_n = \frac{m_p}{\frac{4}{3}\pi r_0^3}$ [Result: Density is independent of $A$].

Step 2: Calculation of Nuclear Density for Sodium ($A=23$)

Given $r_0 = 1.2 \text{ f} = 1.2 \times 10^{-15} \text{ m}$:

$\rho_n = \frac{1.67 \times 10^{-27} \text{ kg}}{\frac{4}{3} \cdot 3.14 \cdot (1.2 \times 10^{-15} \text{ m})^3}$

$\rho_n = \frac{1.67 \times 10^{-27}}{4.188 \cdot 1.728 \times 10^{-45}} \approx \frac{1.67 \times 10^{-27}}{7.239 \times 10^{-45}}$

$\rho_n \approx 2.3 \times 10^{17} \text{ kg/m}^3$

Step 3: Comparative Analysis of Nuclear vs. Atomic Density

From the previously referenced Exercise 2.27, the atomic radius of Sodium is $R_a \approx 1.9 \text{ \AA} = 1.9 \times 10^{-10} \text{ m}$. The mass of a sodium atom $M_a \approx 23 \times 1.67 \times 10^{-27} \text{ kg} = 3.84 \times 10^{-26} \text{ kg}$.

$\rho_a = \frac{M_a}{\frac{4}{3}\pi R_a^3} = \frac{3.84 \times 10^{-26}}{\frac{4}{3} \cdot 3.14 \cdot (1.9 \times 10^{-10})^3} \approx 2.24 \times 10^3 \text{ kg/m}^3$

Step 4: Summary Table of Findings

The ratio $\frac{\rho_n}{\rho_a} \approx 10^{14}$, demonstrating that nuclear matter is significantly more dense than atomic matter due to the vast empty space between the nucleus and orbital electrons.

Final Solution: The nuclear mass density is approximately $2.3 \times 10^{17} \text{ kg/m}^3$, which is roughly $10^{14}$ times greater than the average atomic mass density of $2.3 \times 10^3 \text{ kg/m}^3$.

Step 1: Identifying Given Variables and Physical Constants

To calculate the distance $d$ of the quasar, we utilize the relationship between speed, time, and distance: $d = v \times t$.

• Speed of light ($c$): $3.0 \times 10^8 \text{ m/s}$
• Time taken ($t$): $3.0 \text{ billion years} = 3.0 \times 10^9 \text{ years}$
• Conversion factor (1 year to seconds): $1 \text{ year} \approx 365.25 \text{ days} \times 24 \text{ hours/day} \times 3600 \text{ seconds/hour} \approx 3.15576 \times 10^7 \text{ seconds}$

Step 2: Performing Unit Conversions

First, convert the time $t$ from years into seconds:

Step 3: Calculating Distance in Meters

Apply the formula $d = c \times t$:

Step 4: Converting Meters to Kilometers

Since $1 \text{ km} = 10^3 \text{ m}$, we adjust the exponent:

Visual Representation of Light Propagation

Final Solution: The distance of the quasar from Earth is approximately $2.84 \times 10^{22} \text{ km}$.

Step 1: Identifying Known Parameters

To calculate the average mass density of a single sodium atom, we utilize the following standard physical constants:

• Atomic mass of sodium ($M$) = $23.0 \text{ g/mol} = 0.023 \text{ kg/mol}$
• Avogadro's number ($N_A$) = $6.022 \times 10^{23} \text{ atoms/mol}$
• Radius of sodium atom ($r$) = $2.5 \text{ \AA} = 2.5 \times 10^{-10} \text{ m}$
• Density of crystalline sodium ($\rho_{c}$) = $970 \text{ kg/m}^3$

Step 2: Calculation of Mass of a Single Sodium Atom

The mass of one sodium atom ($m$) is derived by dividing the molar mass by Avogadro's number [Per the definition of the mole]:

$m = \frac{M}{N_A} = \frac{0.023 \text{ kg/mol}}{6.022 \times 10^{23} \text{ atoms/mol}}$

$m \approx 3.82 \times 10^{-26} \text{ kg}$

Step 3: Calculation of Volume of a Sodium Atom

Assuming the sodium atom is a sphere, the volume ($V$) is calculated using the geometric formula $V = \frac{4}{3}\pi r^3$:

$V = \frac{4}{3} \times 3.14159 \times (2.5 \times 10^{-10} \text{ m})^3$

$V \approx 1.333 \times 3.14159 \times 15.625 \times 10^{-30} \text{ m}^3$

$V \approx 6.545 \times 10^{-29} \text{ m}^3$

Step 4: Calculation of Average Atomic Mass Density

The average mass density of the atom ($\rho_{a}$) is the ratio of mass to volume [By the definition of density, $\rho = \frac{m}{V}$]:

$\rho_{a} = \frac{3.82 \times 10^{-26} \text{ kg}}{6.545 \times 10^{-29} \text{ m}^3}$

$\rho_{a} \approx 583.65 \text{ kg/m}^3 \approx 5.8 \times 10^2 \text{ kg/m}^3$

Step 5: Visual Representation of Atomic Volume

Step 6: Comparative Analysis and Conclusion

Evaluation of Order of Magnitude:

Both densities are of the order of $10^2 \text{ kg/m}^3$. This indicates that in the crystalline phase, the atoms are packed relatively close to each other without significant empty space beyond the intrinsic atomic volume. The discrepancy arises because the crystalline density accounts for the inter-atomic spacing within the crystal lattice structure, whereas the atomic density calculation assumes a hard-sphere model for an isolated particle.

Final Solution: The estimated atomic density is $5.8 \times 10^2 \text{ kg/m}^3$. Yes, both values are of the same order of magnitude ($10^2 \text{ kg/m}^3$), implying that the mass is distributed with similar density in both the individual atom and the crystalline solid state.

Step 1: Identification of Physical Variables

To determine the distance of the enemy submarine, we define the following variables provided in the problem statement:

• Total time interval for the round trip of the ultrasonic wave, $\Delta t = 77.0\text{ s}$
• Speed of sound in water, $v = 1450\text{ m s}^{-1}$
• Distance to the target (enemy submarine), $d = ?$

Step 2: Theoretical Formulation

The SONAR system operates on the principle of wave reflection. The total distance $S$ covered by the ultrasonic pulse is the distance to the target and back to the source. Therefore, the total path length is $2d$.

Per the basic kinematic equation for constant velocity:

[Justification: Based on the definition of speed for non-accelerated motion in a uniform medium].

Step 3: Algebraic Derivation for Distance

To isolate $d$, we rearrange the equation:

Step 4: Numerical Computation

Substituting the given values into the derived formula:

Step 5: Significant Figures and Unit Conversion

The given data ($77.0$ s and $1450$ m s$^{-1}$) involve three significant figures. Expressing the result in scientific notation for precision:

Geometric Representation of the Wave Propagation

Final Solution: The distance of the enemy submarine is $5.58 \times 10^{4}\text{ m}$ or $55.8\text{ km}$.

Step 1: Dimensional Analysis of the Proposed Relation

The student proposes the relation: $\tan \theta = v$.

To evaluate the validity of this equation, we must perform a dimensional analysis on both sides of the equation [Per the Principle of Homogeneity of Dimensions].

• Left-Hand Side (LHS): The term $\tan \theta$ represents a trigonometric function of an angle. Since an angle is the ratio of arc length to radius ($s/r$), it is a dimensionless quantity ($[M^0 L^0 T^0]$). Therefore, $\tan \theta$ is dimensionless.
• Right-Hand Side (RHS): The variable $v$ represents the speed of the man, which has dimensions of distance per unit time ($[L T^{-1}]$).

Comparing the two: $[M^0 L^0 T^0] \neq [L T^{-1}]$. Since the dimensions do not match, the relation $\tan \theta = v$ is physically incorrect, despite satisfying the boundary condition as $v \to 0$.

Step 2: Deriving the Correct Physical Relation

Let $v_r$ be the velocity of the rain falling vertically relative to the ground, and $v_m$ be the velocity of the man walking horizontally. To stay dry, the man must point his umbrella in the direction of the relative velocity of the rain with respect to himself ($\vec{v}_{rm}$).

$\vec{v}_{rm} = \vec{v}_r - \vec{v}_m$

Vectorially, if $\vec{v}_r$ is directed downward and $\vec{v}_m$ is directed forward, the resultant vector $\vec{v}_{rm}$ makes an angle $\theta$ with the vertical. From the geometry of the velocity vector triangle:

Using trigonometry in the right-angled triangle formed by these vectors [By Trigonometric Ratios in Right Triangles]:

$\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{v_m}{v_r}$

Step 3: Justification and Verification

The correct relation is $\tan \theta = \frac{v}{v_r}$, where $v_r$ is the constant speed of the falling rain.

Final Solution: The relation $\tan \theta = v$ is incorrect because it is not dimensionally consistent. The correct relation is $\tan \theta = \frac{v}{v_r}$, where $v_r$ is the constant speed at which the rain falls.

Step 1: Dimensional Analysis and Parameter Identification

To construct a quantity with the dimension of time $[T]$ from the fundamental constants $G$ (gravitational constant), $c$ (speed of light), $m_p$ (mass of proton), $m_e$ (mass of electron), and $e$ (elementary charge), we must analyze their dimensions:

• $G = [M^{-1} L^3 T^{-2}]$
• $c = [L T^{-1}]$
• $m_p, m_e = [M]$
• $e^2 / (4\pi\epsilon_0) = [M L^3 T^{-2}]$ (using the electrostatic force dimension)

Step 2: Deriving the Time Constant

Dirac observed that the ratio of the electrostatic force to the gravitational force between a proton and an electron is a dimensionless number of enormous magnitude:

$$N = \frac{e^2}{4\pi\epsilon_0 G m_p m_e} \approx \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{6.67 \times 10^{-11} \times 1.67 \times 10^{-27} \times 9.1 \times 10^{-31}} \approx 2.3 \times 10^{39}$$

To obtain a quantity with the dimension of time, we combine the gravitational constant $G$, the mass of the proton $m_p$, and the speed of light $c$. The time constant related to the scale of the universe (the light-travel time across the range defined by these constants) is given by:

$$T \approx \frac{G m_p}{c^3} \times N \approx \frac{G m_p}{c^3} \times \left( \frac{e^2}{4\pi\epsilon_0 G m_p m_e} \right)$$

Step 3: Numerical Calculation

Substituting the values ($G \approx 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2$, $m_p \approx 1.67 \times 10^{-27} \, \text{kg}$, $c \approx 3 \times 10^8 \, \text{m/s}$):

$$t \approx \frac{6.67 \times 10^{-11} \times 1.67 \times 10^{-27}}{(3 \times 10^8)^3} \times 2.3 \times 10^{39}$$

$$t \approx 1.37 \times 10^{-52} \times 2.3 \times 10^{39} \approx 3.15 \times 10^{-13} \, \text{s} \times 2.3 \times 10^{39} \approx 7.2 \times 10^{26} \, \text{s}$$

Converting to years ($1 \, \text{yr} \approx 3.15 \times 10^7 \, \text{s}$):

$$T \approx \frac{7.2 \times 10^{26}}{3.15 \times 10^7} \approx 2.2 \times 10^{19} \, \text{years}$$

Step 4: Logical Implications

The magnitude calculated depends heavily on the chosen parameters (the "large number hypothesis"). If $N$ grows with the age of the universe, and $G$ is held constant, $e$ or $m$ would need to evolve over cosmological timescales. Modern measurements have constrained such variations to be extremely small, placing doubt on the direct dynamical coupling between these microphysical constants and the expansion of the universe.

Final Solution: The constructed time constant using fundamental constants yields a magnitude on the order of $10^{19}$ years. The existence of such large dimensionless numbers suggests that if their coincidence with the universe's age is meaningful, it would necessitate that the fundamental constants of nature are functions of cosmic time, rather than fixed values.

Step 1: Theoretical Derivation of Nuclear Density

To demonstrate that nuclear density is constant, we define the following variables:

• Nuclear radius: $r = r_0 A^{1/3}$
• Mass number: $A$
• Nuclear mass: $M_{nuc} \approx A \cdot m_p$, where $m_p \approx 1.67 \times 10^{-27} \, \text{kg}$ [Assuming mass of nucleon $\approx$ mass of proton].
• Nuclear volume: $V = \frac{4}{3} \pi r^3$

Substituting the expression for $r$ into the volume formula:

Now, define the nuclear density ($\rho_{nuc}$):

Since $m_p$, $\pi$, and $r_0$ are constants, $\rho_{nuc}$ is independent of the mass number $A$. Thus, nuclear density is constant for all nuclei.

Step 2: Estimation of Sodium Nucleus Density

Given values:

• $r_0 = 1.2 \, \text{f} = 1.2 \times 10^{-15} \, \text{m}$
• $m_p \approx 1.67 \times 10^{-27} \, \text{kg}$

Calculating the density:

Step 3: Comparison with Sodium Atom Density

From Exercise 2.27, the radius of a sodium atom is approximately $R_a \approx 1.9 \times 10^{-10} \, \text{m}$ and the mass is $M_a \approx 3.8 \times 10^{-26} \, \text{kg}$.

Density of sodium atom ($\rho_{atom}$):

Geometric Representation of Relative Scales

Final Solution: The derivation shows that nuclear density is constant because $A$ cancels out in the ratio of mass to volume. The estimated density of the sodium nucleus is $2.3 \times 10^{17} \, \text{kg/m}^3$, which is approximately $10^{14}$ times greater than the density of the sodium atom, illustrating that the mass of an atom is concentrated almost entirely in its nucleus while the volume is dominated by electron orbits.

Step 1: Identifying Given Constants and Variables

• Atomic mass of sodium ($M$): $23.0 \, \text{g/mol} = 0.023 \, \text{kg/mol}$
• Avogadro’s number ($N_A$): $6.022 \times 10^{23} \, \text{atoms/mol}$
• Atomic radius of sodium ($r$): $2.5 \, \text{Å} = 2.5 \times 10^{-10} \, \text{m}$
• Density of crystalline sodium ($\rho_{\text{crystal}}$): $970 \, \text{kg/m}^3$

Step 2: Calculating Mass of a Single Sodium Atom

The mass of a single atom ($m$) is obtained by dividing the molar mass by Avogadro's number [Per the definition of the mole]:

$m = \frac{M}{N_A} = \frac{0.023 \, \text{kg/mol}}{6.022 \times 10^{23} \, \text{atoms/mol}}$

$m \approx 3.819 \times 10^{-26} \, \text{kg}$

Step 3: Calculating the Volume of the Sodium Atom

Assuming the atom is a sphere [Geometric approximation for atomic volume]:

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

$V = \frac{4}{3} \times 3.14159 \times (2.5 \times 10^{-10} \, \text{m})^3$

$V = \frac{4}{3} \times 3.14159 \times 15.625 \times 10^{-30} \, \text{m}^3 \approx 6.545 \times 10^{-29} \, \text{m}^3$

Step 4: Calculating Average Atomic Mass Density

The density ($\rho_{\text{atom}}$) is defined as mass per unit volume [By definition of density]:

$\rho_{\text{atom}} = \frac{m}{V} = \frac{3.819 \times 10^{-26} \, \text{kg}}{6.545 \times 10^{-29} \, \text{m}^3} \approx 583.5 \, \text{kg/m}^3$

Visual Representation of Atomic Model

Step 5: Comparison and Analytical Conclusion

Comparing the values:

$\frac{\rho_{\text{crystal}}}{\rho_{\text{atom}}} = \frac{970}{583.5} \approx 1.66$

Both densities are of the same order of magnitude ($10^2 \, \text{kg/m}^3$). This occurs because in the crystalline phase, atoms are packed closely together, and the interatomic spacing is not significantly larger than the atomic dimensions, meaning the volume occupied by the atom is representative of the volume it occupies in the crystal lattice.

Final Solution: The average mass density of a sodium atom is approximately 583.5 kg/m³. Yes, both densities are of the same order of magnitude, $10^2$ kg/m³, because the physical state of the solid is primarily determined by the tight packing of these spherical atomic volumes.

Step 1: Dimensional Analysis of Fundamental Constants

To construct a quantity with the dimension of time $[T]$ using the fundamental constants of nature: speed of light ($c$), electronic charge ($e$), mass of electron ($m_e$), mass of proton ($m_p$), and the gravitational constant ($G$), we must equate dimensions. The SI units and dimensions are as follows:

Step 2: Constructing the Dimensionless Ratio

Dirac observed the ratio of the electrostatic force to the gravitational force between an electron and a proton. [Per Coulomb's Law and Newton's Law of Gravitation]:

$F_e = \frac{e^2}{4\pi\epsilon_0 r^2}$ and $F_g = \frac{G m_e m_p}{r^2}$

The dimensionless ratio $N$ is defined as:

$N = \frac{F_e}{F_g} = \frac{e^2}{4\pi\epsilon_0 G m_e m_p}$

Using standard values: $e \approx 1.6 \times 10^{-19} C$, $G \approx 6.67 \times 10^{-11} m^3kg^{-1}s^{-2}$, $m_e \approx 9.1 \times 10^{-31} kg$, $m_p \approx 1.67 \times 10^{-27} kg$, and $\frac{1}{4\pi\epsilon_0} \approx 9 \times 10^9 Nm^2/C^2$:

$N \approx \frac{(9 \times 10^9)(1.6 \times 10^{-19})^2}{(6.67 \times 10^{-11})(9.1 \times 10^{-31})(1.67 \times 10^{-27})} \approx 2.3 \times 10^{39}$

Step 3: Deriving the Time Scale

To obtain a quantity with dimensions of time $[T]$, we multiply this dimensionless ratio by the time it takes light to traverse the classical radius of the electron ($r_e \approx \frac{e^2}{4\pi\epsilon_0 m_e c^2} \approx 2.8 \times 10^{-15} m$):

$T_{Dirac} = N \cdot \frac{r_e}{c} \approx 10^{39} \cdot \frac{2.8 \times 10^{-15}}{3 \times 10^8} \approx 10^{16} s$

Given $1 \text{ year} \approx 3.15 \times 10^7 s$, then $15 \text{ billion years} \approx 4.7 \times 10^{17} s$. The calculated $T_{Dirac}$ is of the same order of magnitude ($10^{16}$ to $10^{17} s$) as the age of the universe.

Step 4: Implications for Fundamental Constants

If the coincidence $T_{Dirac} \approx T_{Universe}$ is not accidental, it suggests that the fundamental "constants" of nature may not be strictly constant over cosmic time. [By the Large Number Hypothesis]:

• If $T_{Universe}$ increases linearly with time, then $G$ must decrease as $G \propto \frac{1}{t}$.
• Alternatively, the strength of interactions (determined by $e$ or $m_e$) might vary with the expansion of the universe.
• Modern astrophysical measurements (e.g., spectral analysis of distant quasars) generally constrain the variation of these constants to be extremely small, suggesting that Dirac's Large Number Hypothesis remains a fascinating theoretical speculation rather than a confirmed physical law.

Final Solution: The constructed time $T \approx N \frac{e^2}{4\pi\epsilon_0 m_e c^3} \approx 10^{16} - 10^{17} s$. This implies that if the coincidence is significant, the Gravitational Constant $G$ or other fundamental constants may vary inversely with the age of the universe.

Step 1: Given Variables & Physical Constants

To determine the distance $d$ of the quasar, we utilize the relationship between distance, velocity, and time in a vacuum. The following values are identified:

• Speed of light ($c$): $2.998 \times 10^8 \text{ m/s} \approx 3.0 \times 10^8 \text{ m/s}$ [Per standard physical constants]
• Time ($t$): $3.0 \times 10^9 \text{ years}$

Step 2: Conversion of Time into Seconds

We must convert the time from years into SI base units (seconds) to ensure dimensional consistency [Per the International System of Units].

Given: $1 \text{ year} \approx 365.25 \text{ days} \times 24 \text{ hours/day} \times 3600 \text{ seconds/hour}$

$1 \text{ year} \approx 3.156 \times 10^7 \text{ s}$

Total time in seconds ($t$):

$t = (3.0 \times 10^9 \text{ years}) \times (3.156 \times 10^7 \text{ s/year})$

$t = 9.468 \times 10^{16} \text{ s}$

Step 3: Calculating Distance in Meters

Using the kinematic formula for constant velocity: $d = c \times t$

$d = (3.0 \times 10^8 \text{ m/s}) \times (9.468 \times 10^{16} \text{ s})$

$d = 28.404 \times 10^{24} \text{ m}$

$d \approx 2.84 \times 10^{25} \text{ m}$

Step 4: Conversion to Kilometers

To convert from meters ($m$) to kilometers ($km$), we divide by $10^3$ [Unit Conversion Factor]:

$d_{\text{km}} = \frac{2.84 \times 10^{25} \text{ m}}{10^3 \text{ m/km}}$

$d_{\text{km}} = 2.84 \times 10^{22} \text{ km}$

Visual Representation of Light Propagation

Final Solution: The distance of the quasar from Earth is approximately $2.84 \times 10^{22} \text{ km}$.

Step 1: Establishing the Geometric Foundation

The phenomenon of a total solar eclipse occurring such that the Moon perfectly covers the Sun implies that the angular diameter of the Moon ($\alpha_m$) is approximately equal to the angular diameter of the Sun ($\alpha_s$) as viewed from the Earth. Mathematically, this is expressed as:

$$\alpha_m \approx \alpha_s \implies \frac{d_m}{D_m} = \frac{d_s}{D_s}$$

Where:

• $d_m$: Diameter of the Moon
• $D_m$: Distance of the Moon from Earth $\approx 3.84 \times 10^8 \text{ m}$ [Per standard astronomical data referenced in NCERT Example 2.3]
• $d_s$: Diameter of the Sun $\approx 1.39 \times 10^9 \text{ m}$ [Per standard astronomical data referenced in NCERT Example 2.4]
• $D_s$: Distance of the Sun from Earth $\approx 1.50 \times 10^{11} \text{ m}$ [Per standard astronomical data referenced in NCERT Example 2.4]

Step 2: Geometrical Visualization

Step 3: Algebraic Manipulation

We rearrange the proportionality equation to solve for the Moon's diameter ($d_m$):

$$d_m = \frac{d_s \times D_m}{D_s}$$

Substitute the provided astronomical constants into the equation:

$$d_m = \frac{(1.39 \times 10^9 \text{ m}) \times (3.84 \times 10^8 \text{ m})}{1.50 \times 10^{11} \text{ m}}$$

Step 4: Numerical Computation

Perform the multiplication in the numerator:

$$1.39 \times 3.84 = 5.3376$$

$$10^9 \times 10^8 = 10^{17}$$

Divide by the denominator:

$$d_m = \frac{5.3376 \times 10^{17}}{1.50 \times 10^{11}} = 3.5584 \times 10^6 \text{ m}$$

Rounding to significant figures consistent with the provided data (three significant figures):

$$d_m \approx 3.56 \times 10^6 \text{ m} = 3560 \text{ km}$$

Final Solution: The approximate diameter of the Moon is $3.56 \times 10^6 \text{ m}$ or $3560 \text{ km}$.

Step 1: Given Variables & Initial Setup

To determine the radius of the lunar orbit (the distance between the Earth and the Moon), we identify the following physical parameters:

• Total time taken for the laser signal to travel to the Moon and back ($t$): $2.56\ \text{s}$
• Speed of light in vacuum ($c$): $3.00 \times 10^8\ \text{m/s}$

Step 2: Theoretical Formulation

The laser beam travels from the Earth to the Moon and reflects back to the Earth. [Per the principle of rectilinear propagation of light in vacuum]. The total distance traveled ($D_{total}$) is twice the distance ($d$) between the Earth and the Moon.

The kinematic equation for constant speed is:

$D_{total} = c \times t$

Since the light covers the distance $d$ twice (once to the Moon, once back):

$2d = c \times t$

Step 3: Calculating the Distance

Rearranging the equation to solve for $d$:

$d = \frac{c \times t}{2}$

Substituting the given values:

$d = \frac{(3.00 \times 10^8\ \text{m/s}) \times (2.56\ \text{s})}{2}$

$d = (3.00 \times 10^8) \times (1.28\ \text{s})$

$d = 3.84 \times 10^8\ \text{m}$

Step 4: Visualizing the Laser Ranging

Summary of Calculation

Final Solution: The radius of the lunar orbit around the Earth is $3.84 \times 10^8\ \text{m}$ (or $3.84 \times 10^5\ \text{km}$).