Step 1: Understanding the Concept:

The rotation of the Earth is governed by the principle of conservation of angular momentum.

Angular momentum (\( L \)) of a rotating body is the product of its moment of inertia (\( I \)) and its angular velocity (\( \omega \)), expressed as \( L = I \omega \).

If no external torque acts on the system, the total angular momentum remains constant.


Step 2: Key Formula or Approach:

The relationship between angular momentum, moment of inertia, and time period (\( T \)) is:
\[ L = I \omega = I \left( \frac{2\pi}{T} \right) = constant \]

From this, we can see that the time period \( T \) is directly proportional to the moment of inertia \( I \) (\( T \propto I \)).


Step 3: Detailed Explanation:

During a massive earthquake like the one in Japan in 2011, there is a significant shifting and redistribution of Earth's mass.

In this specific event, a large amount of Earth's crustal mass shifted closer to the Earth's rotation axis.

The moment of inertia \( I = \sum mr^2 \) depends on how mass is distributed relative to the axis of rotation.

When mass moves closer to the axis (decreasing \( r \)), the moment of inertia \( I \) decreases.

To conserve angular momentum (\( L \)), as \( I \) decreases, the angular velocity \( \omega \) must increase.

An increase in angular velocity means the Earth rotates faster, which results in a shorter time period (\( T = 2\pi / \omega \)).

Scientists calculated that this redistribution of mass shortened the day by approximately 1.8 microseconds.


Step 4: Final Answer:

The decrease in the time period was due to the redistribution of mass toward the axis of rotation, which caused the Moment of Inertia (MOI) to decrease.
Quick Tip: Think of a figure skater: when they pull their arms in (reducing their moment of inertia), they spin faster (increasing angular velocity). The Earth behaves the same way when mass moves toward its center.

Step 1: Understanding the Concept:

The total mechanical energy (\( E \)) of a conservative system is the sum of its kinetic energy (\( K \)) and its potential energy (\( U \)).


Step 2: Key Formula or Approach:

The law of conservation of mechanical energy states:
\[ E = K + U \]

where:
\( K = \frac{1}{2}mv^2 \) (Kinetic Energy)
\( U = U(x) \) (Potential Energy)


Step 3: Detailed Explanation:

The question asks for the potential energy at a point where the velocity of the body is zero.

When velocity \( v = 0 \), the kinetic energy \( K \) becomes:
\[ K = \frac{1}{2}m(0)^2 = 0 \]

Substituting this into the total energy equation:
\[ E = 0 + U \]
\[ U = E \]

This indicates that at the turning points of motion (where velocity is zero), the entire mechanical energy of the body is stored as potential energy.


Step 4: Final Answer:

When the velocity is zero, the potential energy is equal to the total mechanical energy \( E \).
Quick Tip: At maximum displacement (like a pendulum at its highest point), kinetic energy is zero and all energy is potential energy (\( U = E \)).

Step 1: Understanding the Concept:

Gibbs free energy (\( G \)) is a thermodynamic potential that can be used to calculate the maximum reversible work that may be performed by a thermodynamic system at a constant temperature and pressure.


Step 2: Key Formula or Approach:

The standard definition of Gibbs free energy is the enthalpy (\( H \)) of the system minus the product of the temperature (\( T \)) and the entropy (\( S \)).
\[ G = H - TS \]


Step 3: Detailed Explanation:

We can also express this in terms of internal energy (\( U \)). Since enthalpy is defined as \( H = U + PV \), we can substitute this into the Gibbs formula:
\[ G = (U + PV) - TS \]
\[ G = U + PV - TS \]

Where:
\( G \) = Gibbs free energy
\( H \) = Enthalpy
\( U \) = Internal energy (SI unit: joule)
\( P \) = Pressure (SI unit: pascal)
\( V \) = Volume (SI unit: \( m^3 \))
\( T \) = Temperature (SI unit: kelvin)
\( S \) = Entropy (SI unit: joule/kelvin)


Step 4: Final Answer:

The formula for Gibbs free energy is \( G = H - TS \).
Quick Tip: A process is spontaneous if the change in Gibbs free energy (\( \Delta G \)) is negative (\( \Delta G < 0 \)).

Step 1: Understanding the Concept:

In Bohr's model of the Hydrogen atom, the time period (\( T \)) of an electron in a circular orbit is the time taken to complete one revolution around the nucleus.


Step 2: Key Formula or Approach:

The time period \( T \) is given by the ratio of the circumference of the orbit to the orbital velocity:
\[ T = \frac{2\pi r}{v} \]

According to Bohr's postulates:

The radius of the \( n \)-th orbit is \( r_n \propto \frac{n^2}{Z} \).

The velocity in the \( n \)-th orbit is \( v_n \propto \frac{Z}{n} \).

Thus, the time period \( T_n \) is:
\[ T_n \propto \frac{n^2/Z}{Z/n} \implies T_n \propto \frac{n^3}{Z^2} \]


Step 3: Detailed Explanation:

For a Hydrogen atom, the atomic number \( Z = 1 \). Therefore:
\[ T \propto n^3 \]

We need the ratio of the time periods for the 1st orbit (\( n_1 = 1 \)) and the 2nd orbit (\( n_2 = 2 \)).
\[ \frac{T_1}{T_2} = \left( \frac{n_1}{n_2} \right)^3 \]

Substituting the values:
\[ \frac{T_1}{T_2} = \left( \frac{1}{2} \right)^3 = \frac{1}{8} \]

So, the ratio is \( 1 : 8 \).


Step 4: Final Answer:

The ratio of time periods of electrons in the 1st and 2nd orbits of Hydrogen is \( 1 : 8 \).
Quick Tip: Remember the dependencies in Bohr's Model: Radius \( \propto n^2 \), Velocity \( \propto 1/n \), and Time Period \( \propto n^3 \). This helps solve ratio questions quickly!

Step 1: Understanding the Concept:

The transformation of a planet into a denser star (like a neutron star) involves a change in size but generally conserves total angular momentum, assuming no external torque is applied.

The angular momentum (\( L \)) is given by \( L = I\omega \), where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.


Step 2: Key Formula or Approach:

Conservation of angular momentum: \( I_1 \omega_1 = I_2 \omega_2 \)

For a sphere, \( I = \frac{2}{5}MR^2 \). Since mass \( M \) is conserved:
\[ R_1^2 \omega_1 = R_2^2 \omega_2 \]

Substituting \( \omega = \frac{2\pi}{T} \):
\[ \frac{R_1^2}{T_1} = \frac{R_2^2}{T_2} \implies T_2 = T_1 \left( \frac{R_2}{R_1} \right)^2 \]


Step 3: Detailed Explanation:

Given:
\( R_1 = 8000 km \)
\( R_2 = 8 km \)
\( T_1 = 15 h \)

Calculate the ratio of radii: \( \frac{R_2}{R_1} = \frac{8}{8000} = 10^{-3} \).

Now, calculate the new time period in hours:
\[ T_2 = 15 \times (10^{-3})^2 = 15 \times 10^{-6} hours \]

To convert this into seconds:
\[ T_2 = 15 \times 10^{-6} \times 3600 s \]
\[ T_2 = 15 \times 3.6 \times 10^{-3} s \]
\[ T_2 = 0.054 s \]


Step 4: Final Answer:

The new time period of the star is \( 0.054 s \).
Quick Tip: When radius decreases by a factor of \( k \), the time period decreases by a factor of \( k^2 \). Here, radius decreased by 1000, so time period decreased by \( 1,000,000 \).

Step 1: Understanding the Concept:

To find the next term in a sequence, we look for a mathematical pattern involving exponents, factorials, or differences between terms.


Step 2: Detailed Explanation:

Let's analyze the terms of the sequence \( a_n \) starting from index \( n = 0 \):
\( a_0 = 1 \)
\( a_1 = 0 \)
\( a_2 = 0 \)
\( a_3 = 3 \)
\( a_4 = 20 \)

The pattern observed relates the term to its factorial minus its index: \( a_n = n! - n \).

Checking the pattern:

For \( n = 0 \): \( 0! - 0 = 1 - 0 = 1 \)

For \( n = 1 \): \( 1! - 1 = 1 - 1 = 0 \)

For \( n = 2 \): \( 2! - 2 = 2 - 2 = 0 \)

For \( n = 3 \): \( 3! - 3 = 6 - 3 = 3 \)

For \( n = 4 \): \( 4! - 4 = 24 - 4 = 20 \)

Following this logic, the next term (\( n = 5 \)) is:
\( a_5 = 5! - 5 = 120 - 5 = 115 \)


Step 4: Final Answer:

The next number in the sequence is 115.
Quick Tip: If standard differences don't work in a sequence that grows rapidly (like 3 to 20), always check for factorial-based patterns.

Step 1: Understanding the Concept:

On a frictionless surface, only conservative forces (gravity) do work. Therefore, the total mechanical energy of the block is conserved throughout its motion.


Step 2: Detailed Explanation:

Let the starting point be at height \( h = 0 \).

Initial Kinetic Energy (\( K_i \)) = \( \frac{1}{2}mv^2 \).

Initial Potential Energy (\( U_i \)) = \( 0 \).

Total Energy (\( E \)) = \( \frac{1}{2}mv^2 \).

As the block moves up, it slows down until its velocity is zero at some height \( H \). At this point, all energy is potential: \( E = mgH \).

When the block slides back down to the starting point (\( h = 0 \)), its potential energy becomes zero again.

By the law of conservation of energy:

Final Kinetic Energy (\( K_f \)) + Final Potential Energy (\( U_f \)) = Total Energy (\( E \))
\( K_f + 0 = \frac{1}{2}mv^2 \)
\( \frac{1}{2}mv_f^2 = \frac{1}{2}mv^2 \)

This implies \( v_f = v \).


Step 4: Final Answer:

The block returns to the bottom with the same magnitude of velocity \( v \).
Quick Tip: In any frictionless system where an object returns to the same height it started from, the final speed will always equal the initial speed.

Step 1: Understanding the Concept:

To solve this differential equation, we need to integrate the function on the right-hand side with respect to \( x \).


Step 2: Key Formula or Approach:

We use the substitution method or trigonometric expansion.

Expansion of \( \sin(A - B) = \sin A \cos B - \cos A \sin B \).


Step 3: Detailed Explanation:

We have:
\[ \frac{dy}{dx} = \frac{\sin x}{\sin(x + 2)} \]

We can rewrite \( \sin x \) as \( \sin((x + 2) - 2) \):
\[ \frac{dy}{dx} = \frac{\sin(x + 2) \cos 2 - \cos(x + 2) \sin 2}{\sin(x + 2)} \]
\[ \frac{dy}{dx} = \cos 2 - \sin 2 \frac{\cos(x + 2)}{\sin(x + 2)} \]
\[ \frac{dy}{dx} = \cos 2 - \sin 2 \cot(x + 2) \]

Integrating both sides with respect to \( x \):
\[ y = \int (\cos 2 - \sin 2 \cot(x + 2)) dx \]

Since \( \cos 2 \) and \( \sin 2 \) are constants:
\[ y = x \cos 2 - \sin 2 \int \cot(x + 2) dx \]

Recall that \( \int \cot u du = \ln|\sin u| + C \):
\[ y = x \cos 2 - \sin 2 \ln|\sin(x + 2)| + C \]


Step 4: Final Answer:

The solution is \( y = x \cos 2 - \sin 2 \ln|\sin(x + 2)| + C \).
Quick Tip: Whenever you see \( \sin(x) / \sin(x+a) \), use the trick of writing \( x \) as \( (x+a) - a \) in the numerator to simplify the fraction into standard integrable forms.

Step 1: Understanding the Concept:

When forming numbers with a set of digits, each digit appears a certain number of times in each place (units, tens, hundreds, etc.). The total sum is the sum of (value of digits in each place \( \times \) its place value).


Step 2: Key Formula or Approach:

Total numbers of \( r \) digits from \( n \) digits = \( ^nP_r \).

Number of times each digit appears in a specific position = \( ^{n-1}P_{r-1} \).

Sum of digits in one place = (Number of appearances) \( \times \) (Sum of available digits).


Step 3: Detailed Explanation:

Here, \( n = 6 \) digits (1, 2, 3, 4, 5, 6) and \( r = 4 \).

Sum of digits = \( 1 + 2 + 3 + 4 + 5 + 6 = 21 \).

Number of times each digit appears in any one position (e.g., units place):
\[ ^{6-1}P_{4-1} = ^5P_3 = 5 \times 4 \times 3 = 60 \]

Sum of values in each position:

Sum in units place = \( 60 \times 21 = 1260 \).

Sum in tens place = \( 1260 \times 10 = 12600 \).

Sum in hundreds place = \( 1260 \times 100 = 126000 \).

Sum in thousands place = \( 1260 \times 1000 = 1260000 \).

Total Sum = \( 1260 + 12600 + 126000 + 1260000 \)

Total Sum = \( 1260 \times (1111) = 1,399,860 \).


Step 4: Final Answer:

The sum of all such 4-digit numbers is 1,399,860.
Quick Tip: Formula for sum of all \( r \)-digit numbers using \( n \) distinct non-zero digits:
\( Sum = (^{n-1}P_{r-1}) \times (sum of digits) \times (11\dots1)_{r times} \).

Step 1: Understanding the Concept:

This integral involves a function and its derivative. Substituting the inverse trigonometric function simplifies the expression significantly.


Step 2: Key Formula or Approach:

Use substitution \( u = \tan^{-1} x \).

The derivative is \( du = \frac{1}{1 + x^2} dx \).


Step 3: Detailed Explanation:

Let \( u = \tan^{-1} x \). Then \( du = \frac{1}{1+x^2} dx \).

The integral becomes:
\[ I = \int u e^u du \]

Applying Integration by Parts (\( \int f g' = fg - \int f' g \)):

Let \( f = u \) (algebraic) and \( g' = e^u \) (exponential).

Then \( f' = 1 \) and \( g = e^u \).
\[ I = u e^u - \int 1 \cdot e^u du \]
\[ I = u e^u - e^u + C = e^u(u - 1) + C \]

Substituting back \( u = \tan^{-1} x \):
\[ I = e^{\tan^{-1} x}(\tan^{-1} x - 1) + C \]


Step 4: Final Answer:

The value of the integral is \( e^{\tan^{-1} x}(\tan^{-1} x - 1) + C \).
Quick Tip: Always look for the pattern \( \int e^x (f(x) + f'(x)) dx = e^x f(x) + C \).
Here, \( \int e^u (u - 1 + 1) du \) isn't exactly that, but standard integration by parts for \( u e^u \) always results in \( e^u(u-1) \).

Step 1: Understanding the Concept:

The given equation is of a shifted ellipse. We first identify the center, semi-axes lengths, and eccentricity to find the foci.


Step 2: Key Formula or Approach:

Standard form: \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \).

Foci for a horizontal ellipse (\( a > b \)): \( (h \pm ae, k) \).

Eccentricity \( e = \sqrt{1 - \frac{b^2}{a^2}} \).


Step 3: Detailed Explanation:

From the equation:

Center \( (h, k) = (3, 7) \).
\( a^2 = 25 \implies a = 5 \).
\( b^2 = 16 \implies b = 4 \).

Calculate eccentricity \( e \):
\[ e = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5} \]

The distance from the center to the foci is \( ae \):
\[ ae = 5 \times \frac{3}{5} = 3 \]

The coordinates of the foci are:
\[ (3 \pm 3, 7) \]

Focus 1: \( (3 - 3, 7) = (0, 7) \)

Focus 2: \( (3 + 3, 7) = (6, 7) \)


Step 4: Final Answer:

The foci of the ellipse are \( (0, 7) \) and \( (6, 7) \).
Quick Tip: Always check if \( a > b \) or \( b > a \) to decide whether the foci lie on a line parallel to the x-axis or y-axis. Since \( 25 > 16 \), the foci are horizontal.

Step 1: Understanding the Concept:

A line parallel to \( Ax + By + C = 0 \) has the form \( Ax + By + k = 0 \). For it to be a tangent to a circle, its distance from the center must equal the radius.


Step 2: Key Formula or Approach:

Circle: \( x^2 + y^2 + 2gx + 2fy + c = 0 \).

Center \( (-g, -f) \), Radius \( r = \sqrt{g^2 + f^2 - c} \).

Perpendicular distance from \( (x_1, y_1) \) to \( ax + by + c = 0 \) is \( \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \).


Step 3: Detailed Explanation:

Circle: \( x^2 + y^2 - 12x + 16y + 19 = 0 \).
\( g = -6, f = 8, c = 19 \).

Center \( C = (6, -8) \).

Radius \( r = \sqrt{(-6)^2 + 8^2 - 19} = \sqrt{36 + 64 - 19} = \sqrt{81} = 9 \).

Line parallel to \( 4x + 3y = 5 \) is \( 4x + 3y + k = 0 \).

Setting distance from \( (6, -8) \) equal to 9:
\[ \frac{|4(6) + 3(-8) + k|}{\sqrt{4^2 + 3^2}} = 9 \]
\[ \frac{|24 - 24 + k|}{5} = 9 \]
\[ |k| = 45 \implies k = \pm 45 \]

The equations are \( 4x + 3y + 45 = 0 \) and \( 4x + 3y - 45 = 0 \).


Step 4: Final Answer:

The equations of the tangents are \( 4x + 3y \pm 45 = 0 \).
Quick Tip: Tangents parallel to a given line always come in pairs. If you find one \( k \), the other is usually its negative counterpart if the center is at the origin or if terms cancel out as they did here.

Step 1: Understanding the Concept:

For a line \( y = mx + c \) to be a tangent to a standard parabola \( y^2 = 4ax \), there is a specific condition relating \( a, m, \) and \( c \).


Step 2: Key Formula or Approach:

Condition of tangency for \( y^2 = 4ax \): \( c = \frac{a}{m} \).


Step 3: Detailed Explanation:

Comparing \( y^2 = 4x \) with \( y^2 = 4ax \), we get \( 4a = 4 \implies a = 1 \).

The given line is \( y = mx + 1 \). Comparing with \( y = mx + c \), we get \( c = 1 \).

Using the condition \( c = \frac{a}{m} \):
\[ 1 = \frac{1}{m} \]
\[ m = 1 \]


Step 4: Final Answer:

The value of \( m \) is 1.
Quick Tip: For the parabola \( y^2 = 4ax \), the equation of any tangent is always \( y = mx + \frac{a}{m} \). Matching this to the given line helps find the slope instantly.

Step 1: Understanding the Concept:

Combustion analysis allows us to determine the mass of Carbon and Hydrogen in a compound from the mass of \( CO_2 \) and \( H_2O \) produced.


Step 2: Key Formula or Approach:

Moles of \( C \) = Moles of \( CO_2 \).

Moles of \( H \) = \( 2 \times \) Moles of \( H_2O \).


Step 3: Detailed Explanation:

Mass of \( CO_2 = 6.6 g \).

Moles of \( CO_2 = \frac{6.6}{44} = 0.15 mol \).

Therefore, Moles of Carbon (\( n_C \)) = 0.15 mol.

Mass of \( H_2O = 1.35 g \).

Moles of \( H_2O = \frac{1.35}{18} = 0.075 mol \).

Since each water molecule has 2 Hydrogen atoms, Moles of Hydrogen (\( n_H \)) = \( 0.075 \times 2 = 0.15 mol \).

Molar ratio of \( C : H \):
\[ 0.15 : 0.15 = 1 : 1 \]

The empirical formula is \( CH \).


Step 4: Final Answer:

The empirical formula of compound X is \( CH \). (Possible molecular formulas include \( C_2H_2 \) or \( C_6H_6 \)).
Quick Tip: In combustion, remember that for every 44g of \( CO_2 \), there are 12g of \( C \), and for every 18g of \( H_2O \), there are 2g of \( H \). Use these ratios for quick calculations.