Step 1: Understanding the Concept:

The Quality factor (\(Q\)-factor) of a series \(LCR\) circuit represents the sharpness of resonance.

It is defined as the ratio of the resonant frequency to the bandwidth.

A higher \(Q\)-factor indicates a lower rate of energy loss relative to the stored energy.


Step 2: Key Formula or Approach:

The mathematical expression for the Quality factor in a series \(LCR\) circuit is:
\[ Q = \frac{1}{R} \sqrt{\frac{L}{C}} \]

Where:
\(R = \) Resistance
\(L = \) Inductance
\(C = \) Capacitance


Step 3: Detailed Explanation:

Let the initial Quality factor be \( Q_1 = \frac{1}{R} \sqrt{\frac{L}{C}} \).

According to the problem, the new resistance is \( R' = \frac{R}{2} \).

The values of \(L\) and \(C\) remain unchanged.

Substituting the new resistance into the formula, we get the new Quality factor \( Q_2 \):
\[ Q_2 = \frac{1}{R'} \sqrt{\frac{L}{C}} \]
\[ Q_2 = \frac{1}{(R/2)} \sqrt{\frac{L}{C}} \]
\[ Q_2 = 2 \times \left( \frac{1}{R} \sqrt{\frac{L}{C}} \right) \]

Comparing this with the initial value:
\[ Q_2 = 2 Q_1 \]


Step 4: Final Answer:

If the resistance \(R\) is halved while keeping \(L\) and \(C\) constant, the Quality factor becomes doubled.
Quick Tip: The Quality factor is inversely proportional to the resistance (\(Q \propto 1/R\)).
Reducing resistance always increases the sharpness of resonance in a circuit.

Step 1: Understanding the Concept:

Damped oscillations occur when an oscillating system experiences resistive forces that dissipate energy over time.

The damping constant (\(b\)) determines how quickly the amplitude decays.

Resonance occurs when the frequency of an external driving force matches the natural frequency of the system.


Step 2: Detailed Explanation:

The amplitude of a driven, damped oscillator at resonance is inversely proportional to the damping constant (\(b\)).

The formula for the amplitude \(A\) at resonance is approximately:
\[ A_{max} \approx \frac{F_0}{b \omega_0} \]

Where \(F_0\) is the driving force and \(\omega_0\) is the natural frequency.

When the damping constant decreases:

1. The dissipation of energy per cycle decreases.

2. The peak amplitude at resonance increases significantly.

3. The resonance curve becomes narrower and "sharper," which is often referred to as an increase in the resonance factor or Quality factor.


Step 3: Final Answer:

If the damped oscillation constant decreases, the resonance factor increases, meaning the resonance becomes much sharper and the peak amplitude increases.
Quick Tip: Low damping \(\rightarrow\) High Quality factor \(\rightarrow\) Sharp Resonance.
High damping \(\rightarrow\) Low Quality factor \(\rightarrow\) Flat Resonance.

Step 1: Understanding the Concept:

Newton's Law of Universal Gravitation states that the force of attraction between two masses is proportional to the product of their masses and inversely proportional to the square of the distance between them.


Step 2: Key Formula or Approach:

The formula is:
\[ F = G \frac{m_1 m_2}{r^2} \]

Rearranging to find the Gravitational constant (\(G\)):
\[ G = \frac{F \cdot r^2}{m_1 \cdot m_2} \]


Step 3: Detailed Explanation:

Substitute the dimensional formulas for Force (\(F\)), Distance (\(r\)), and Mass (\(m\)):

The dimension of Force \([F] = [M L T^{-2}]\).

The dimension of Distance \([r] = [L]\).

The dimension of Mass \([m] = [M]\).

Now, substitute these into the expression for \(G\):
\[ [G] = \frac{[M L T^{-2}] \cdot [L]^2}{[M] \cdot [M]} \]
\[ [G] = \frac{[M L^3 T^{-2}]}{[M^2]} \]
\[ [G] = [M^{1-2} L^3 T^{-2}] \]
\[ [G] = [M^{-1} L^3 T^{-2}] \]


Step 4: Final Answer:

The dimensional formula for the Gravitational constant \(G\) is \([M^{-1} L^3 T^{-2}]\).
Quick Tip: Always derive dimensions from the base formula if you forget them.
Units for \(G\) are \(N \cdot m^2 / kg^2\), which helps in verifying the dimensions.

Step 1: Understanding the Concept:

An ionic bond is formed by the complete transfer of one or more electrons from one atom (usually a metal) to another atom (usually a non-metal).

This creates oppositely charged ions that are held together by electrostatic forces of attraction.


Step 2: Detailed Explanation:

The formation of a stable ionic bond depends on three primary energetic factors:

1. Low Ionization Energy: The metal atom should have a low ionization energy so that it can easily lose electrons to form a cation.

2. High Electron Affinity: The non-metal atom should have a high electron affinity so that it can easily gain electrons to form an anion while releasing energy.

3. High Lattice Energy: When the gaseous ions combine to form a solid crystal lattice, a large amount of energy (Lattice Energy) should be released. Higher lattice energy ensures the stability of the ionic compound.

4. Electronegativity Difference: Usually, the difference in electronegativity between the two atoms should be greater than \(1.7\).


Step 3: Final Answer:

The main conditions are low ionization energy of the electropositive element, high electron affinity of the electronegative element, and high lattice energy of the resulting crystal.
Quick Tip: Ionic bonds typically form between elements from Group 1 or 2 (Metals) and Group 16 or 17 (Non-metals).

Step 1: Understanding the Concept:

The Reimer-Tiemann (spelled as "Riemen-Tiemann" in the question) reaction is used for the ortho-formylation of phenols.

When phenol is treated with chloroform (\(CHCl_3\)) in the presence of a strong base like \(KOH\) or \(NaOH\), salicylaldehyde is formed.


Step 2: Key Formula or Approach:

The reaction involves the generation of an electrophilic intermediate.

The base (\(OH^-\)) abstracts a proton from chloroform:
\[ OH^- + CHCl_3 \rightleftharpoons H_2O + :CCl_3^- \]

The trichloromethyl anion (\(:CCl_3^-\)) then loses a chloride ion to form the neutral, electron-deficient intermediate:
\[ :CCl_3^- \rightarrow :CCl_2 + Cl^- \]

The intermediate \(:CCl_2\) is called Dichlorocarbene.


Step 3: Detailed Explanation:

- Aldol Condensation: Involves an enolate ion intermediate.

- Cannizzaro Reaction: Involves a hydride transfer mechanism.

- Kolbe's Reaction: Involves \(CO_2\) as the electrophile to form salicylic acid.

- Reimer-Tiemann Reaction: Specifically uses dichlorocarbene as the reactive electrophile that attacks the phenoxide ring.


Step 4: Final Answer:

The Riemen-Tiemann reaction involves the dichlorocarbene intermediate (\(:CCl_2\)).
Quick Tip: Remember: \(CHCl_3 + Base = :CCl_2\) (Dichlorocarbene).
This intermediate is an electrophile because the carbon atom has only six valence electrons.

Step 1: Understanding the Concept:

A standard deck contains 52 cards.

The number of Aces in a standard deck is 4.

"At least one Ace" means the outcome could be exactly 1 Ace or exactly 2 Aces.

It is easier to calculate this using the complement rule:
\[ P(At least 1 Ace) = 1 - P(No Aces) \]


Step 2: Key Formula or Approach:

Probability \(P(E) = \frac{Number of favorable outcomes}{Total number of outcomes}\).

Total number of ways to draw 2 cards from 52 is given by combinations: \(\binom{52}{2}\).


Step 3: Detailed Explanation:

1. Total Outcomes:
\[ \binom{52}{2} = \frac{52 \times 51}{2 \times 1} = 26 \times 51 = 1326 \]

2. Favorable Outcomes for "No Aces":

There are \(52 - 4 = 48\) cards that are not Aces.

The number of ways to draw 2 cards from these 48 cards is:
\[ \binom{48}{2} = \frac{48 \times 47}{2 \times 1} = 24 \times 47 = 1128 \]

3. Probability of No Aces:
\[ P(No Aces) = \frac{1128}{1326} \]

4. Probability of At Least One Ace:
\[ P(At least 1 Ace) = 1 - \frac{1128}{1326} \]
\[ P(At least 1 Ace) = \frac{1326 - 1128}{1326} = \frac{198}{1326} \]

Simplify the fraction by dividing both numerator and denominator by 6:
\[ \frac{198 \div 6}{1326 \div 6} = \frac{33}{221} \]


Step 4: Final Answer:

The probability that at least one of the drawn cards is an Ace is \(\frac{33}{221}\).
Quick Tip: When a question asks for "at least one," always try the "1 - P(none)" method; it's usually much faster and less prone to errors.

Step 1: Understanding the Concept:

The magnetic field at the center of a circular current-carrying loop of radius \(R\) and current \(I\) is given by \(B = \frac{\mu_0 I}{2R}\).

The direction of the magnetic field is perpendicular to the plane of the loop.

When two such loops are inclined at an angle \(\theta\), their respective magnetic field vectors \(\vec{B_1}\) and \(\vec{B_2}\) will also be inclined at an angle \(\theta\) to each other.


Step 2: Key Formula or Approach:

The magnitude of the magnetic field for each loop at the center is:
\[ B_1 = B_2 = B = \frac{\mu_0 I}{2R} \]

The resultant magnetic field \(B_{net}\) of two vectors of equal magnitude \(B\) inclined at an angle \(\theta\) is:
\[ B_{net} = \sqrt{B_1^2 + B_2^2 + 2B_1 B_2 \cos \theta} \]


Step 3: Detailed Explanation:

Substitute \(B_1 = B_2 = B\) into the resultant formula:
\[ B_{net} = \sqrt{B^2 + B^2 + 2B^2 \cos \theta} \]
\[ B_{net} = \sqrt{2B^2 (1 + \cos \theta)} \]

Using the trigonometric identity \(1 + \cos \theta = 2 \cos^2 \left( \frac{\theta}{2} \right)\):
\[ B_{net} = \sqrt{2B^2 \cdot 2 \cos^2 \left( \frac{\theta}{2} \right)} \]
\[ B_{net} = \sqrt{4B^2 \cos^2 \left( \frac{\theta}{2} \right)} \]
\[ B_{net} = 2 B \cos \left( \frac{\theta}{2} \right) \]

Substituting the value of \(B = \frac{\mu_0 I}{2R}\):
\[ B_{net} = 2 \left( \frac{\mu_0 I}{2R} \right) \cos \left( \frac{\theta}{2} \right) \]
\[ B_{net} = \frac{\mu_0 I}{R} \cos \left( \frac{\theta}{2} \right) \]


Step 4: Final Answer:

The net magnetic field at the center is \(B_{net} = \frac{\mu_0 I}{R} \cos \left( \frac{\theta}{2} \right)\).
Quick Tip: If the loops were perpendicular (\(\theta = 90^\circ\)), the resultant field would be \(\sqrt{2}B\).
Always resolve vectors using the identity \(\sqrt{2+2\cos\theta} = 2\cos(\theta/2)\) for quick calculation.

Step 1: Understanding the Concept:

In a state of free fall, all objects (regardless of their mass) accelerate downwards with the same acceleration, which is the acceleration due to gravity (\(g\)).

If two connected objects are falling with the same acceleration, there is no relative motion or tendency to change the distance between them.


Step 2: Key Formula or Approach:

Consider the system of two masses \(m\) and \(2m\). Let \(T\) be the tension in the rod and \(a\) be the downward acceleration.


Step 3: Detailed Explanation:

Let the mass \(m\) be at the top and \(2m\) be at the bottom.

For the upper mass \(m\):
\[ mg + T = ma \quad --- (i) \]

For the lower mass \(2m\):
\[ 2mg - T = 2ma \quad --- (ii) \]

Adding equations (i) and (ii):
\[ 3mg = 3ma \implies a = g \]

Now, substitute \(a = g\) into equation (i):
\[ mg + T = m(g) \]
\[ T = mg - mg = 0 \]


Step 4: Final Answer:

The tension in the rod during free fall is zero.
Quick Tip: During free fall, any internal forces like tension in a string or rod between falling masses become zero because the whole system is in a weightless state relative to itself.

Step 1: Understanding the Concept:

When an object floats, its weight is equal to the buoyant force (upthrust) acting on the submerged part.

To fully submerge a floating object, an additional downward force \(F\) must be applied to overcome the remaining upward buoyant force capacity.


Step 2: Key Formula or Approach:

Initial floating condition: \(Weight (W) = Buoyant\ Force_{initial}\)
\(W = V_{submerged} \cdot \rho_{water} \cdot g = 0.25 V \rho_w g\)

Required condition (Fully submerged): \(W + F = Buoyant\ Force_{total}\)
\(F = V \rho_w g - W\)


Step 3: Detailed Explanation:

Given:

Initial submerged volume = \(25%\) of \(V = 0.25V\).

Density of water (\(\rho_w\)) = \(10^3\ kg/m^3\).

From the floating condition, the weight of the block is:
\[ W = 0.25 \cdot V \cdot 10^3 \cdot g \]

When the body is fully submerged, the total upthrust (\(B_{total}\)) is:
\[ B_{total} = V \cdot 10^3 \cdot g \]

The force \(F\) required to maintain this state is the difference between the total upthrust and the weight:
\[ F = B_{total} - W \]
\[ F = (V \cdot 10^3 \cdot g) - (0.25 \cdot V \cdot 10^3 \cdot g) \]
\[ F = (1 - 0.25) \cdot V \cdot 10^3 \cdot g \]
\[ F = 0.75 \cdot V \cdot 10^3 \cdot g \]


Step 4: Final Answer:

The force required is \(0.75 V \rho_w g\).
Quick Tip: The force required to submerge a floating object is simply the buoyant force acting on the portion of the volume that is currently above the water level.

Step 1: Understanding the Concept:

The capacitance \(C\) of a parallel plate capacitor is determined by the area of the plates (\(A\)), the distance between them (\(d\)), and the permittivity of free space (\(\epsilon_0\)).

The charge \(Q\) stored is given by the product of capacitance and the potential difference \(V\) applied.


Step 2: Key Formula or Approach:

1. Capacitance: \( C = \frac{\epsilon_0 A}{d} \)

2. Charge: \( Q = CV = \frac{\epsilon_0 A V}{d} \)


Step 3: Detailed Explanation:

Convert all units to SI (meters):

Area \(A = 4\ cm \times 4\ cm = 16\ cm^2 = 16 \times 10^{-4}\ m^2\).

Distance \(d = 0.1\ mm = 0.1 \times 10^{-3}\ m = 10^{-4}\ m\).

Voltage \(V = 100\ V\).

Permittivity = \(\epsilon_0\) (as required by the question).

Substitute values into the charge formula:
\[ Q = \frac{\epsilon_0 \times (16 \times 10^{-4}) \times 100}{10^{-4}} \]

The \(10^{-4}\) in the numerator and denominator cancel out:
\[ Q = \epsilon_0 \times 16 \times 100 \]
\[ Q = 1600 \epsilon_0 \]


Step 4: Final Answer:

The charge is \(1600 \epsilon_0\) in SI units.
Quick Tip: Always double-check unit conversions: \(1\ cm^2 = 10^{-4}\ m^2\) and \(1\ mm = 10^{-3}\ m\).
Keeping terms in scientific notation helps cancel powers of 10 easily.

Step 1: Understanding the Concept:

According to Raoult's law and Dalton's law of partial pressures, the mole fraction of a component in the vapour phase (\(y\)) is related to its mole fraction in the liquid phase (\(x\)) and its pure vapour pressure (\(P^\circ\)).


Step 2: Key Formula or Approach:

For component A: \( y_A \cdot P_{total} = x_A \cdot P_A^\circ \)

For component B: \( y_B \cdot P_{total} = x_B \cdot P_B^\circ \)

Taking the ratio: \( \frac{y_A}{y_B} = \frac{x_A P_A^\circ}{x_B P_B^\circ} \)


Step 3: Detailed Explanation:

Given:

1. \(P_A^\circ = 2 P_B^\circ\) (Vapour pressure of A is twice that of B).

2. \(y_A = y_B\) (Equimolar in vapour phase, so \(y_A/y_B = 1\)).

Substitute these values into the ratio equation:
\[ 1 = \frac{x_A \cdot (2 P_B^\circ)}{x_B \cdot P_B^\circ} \]

Cancel \(P_B^\circ\) from both sides:
\[ 1 = 2 \cdot \left( \frac{x_A}{x_B} \right) \]

Solve for the ratio of mole fractions in liquid phase (\(x_A/x_B\)):
\[ \frac{x_A}{x_B} = \frac{1}{2} \]


Step 4: Final Answer:

The ratio of mole fractions of A to B in the liquid phase is \(1:2\).
Quick Tip: The more volatile component (higher \(P^\circ\)) will always be more enriched in the vapour phase than in the liquid phase.

Step 1: Understanding the Concept:

The roots of a quadratic equation \(ax^2 + bx + c = 0\) are given by the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

The nature of the roots depends on the discriminant \(D = b^2 - 4ac\).


Step 2: Detailed Explanation:

We are given that \(a, b, c > 0\).

There are two possibilities for the discriminant \(D\):



Case 1: \(D \ge 0\) (Real Roots):

The roots are \( x_1 = \frac{-b + \sqrt{D}}{2a} \) and \( x_2 = \frac{-b - \sqrt{D}}{2a} \).

Since \(D = b^2 - 4ac\), it follows that \(D < b^2\), and thus \(\sqrt{D} < b\).

Therefore, \(-b + \sqrt{D} < 0\), making both roots real and negative.



Case 2: \(D < 0\) (Complex Roots):

The roots are \( x = \frac{-b \pm i\sqrt{|D|}}{2a} \).

The real part of these complex roots is \( Re(x) = -\frac{b}{2a} \).

Since \(b > 0\) and \(a > 0\), the real part is always negative.


Step 3: Final Answer:

In both cases (real or complex), the real part of the roots is always negative.
Quick Tip: Using Descartes' Rule of Signs: Since all coefficients (\(a, b, c\)) are positive, there are 0 sign changes. This implies there are no positive real roots.

Step 1: Understanding the Concept:

If two quadratic equations have the same roots, their corresponding coefficients must be proportional.

Given \(f_1(x) = ax^2 + bx + 5\) and \(f_2(x) = px^2 + qx + 10\).


Step 2: Key Formula or Approach:

For same roots: \( \frac{p}{a} = \frac{q}{b} = \frac{10}{5} = k \)

This implies that \(f_2(x) = k \cdot f_1(x)\) for all values of \(x\).


Step 3: Detailed Explanation:

Comparing the constant terms:
\[ k = \frac{10}{5} = 2 \]

Therefore, the relationship between the two functions is:
\[ f_2(x) = 2 \cdot f_1(x) \]

We need to find the ratio \( \frac{f_2(10)}{f_1(5)} \):

Substitute \(x = 10\) into the expression for \(f_2\):
\[ f_2(10) = 2 \cdot f_1(10) \]

So, the required ratio is:
\[ \frac{f_2(10)}{f_1(5)} = \frac{2 \cdot f_1(10)}{f_1(5)} \]
\[ = \frac{2(100a + 10b + 5)}{25a + 5b + 5} = \frac{10(20a + 2b + 1)}{5(5a + b + 1)} = \frac{2(20a + 2b + 1)}{5a + b + 1} \]


Step 4: Final Answer:

The value is given by the functional relationship \(\frac{f_2(10)}{f_1(5)} = \frac{2 f_1(10)}{f_1(5)}\).
Quick Tip: When roots are identical, the ratio of the functions is equal to the ratio of any pair of corresponding coefficients (like the constants).

Step 1: Understanding the Concept:

For a polynomial of the form \(P(x) = (x - a_1)(x - a_2)...(x - a_n)\), the coefficient of \(x^{n-1}\) is equal to the negative of the sum of the roots (\(a_1 + a_2 + ... + a_n\)).

This is a direct application of Vieta's formulas.


Step 2: Key Formula or Approach:

Coefficient of \(x^{n-1} = - \sum_{i=1}^{n} a_i\)

Sum of first \(n\) natural numbers \(= \frac{n(n+1)}{2}\)


Step 3: Detailed Explanation:

In the given expression \((x - 1)(x - 2)...(x - 16)\), the degree of the polynomial is \(n = 16\).

The roots are \(1, 2, 3, ..., 16\).

The term \(x^{15}\) is the \(x^{n-1}\) term.

Coefficient of \(x^{15} = -(1 + 2 + 3 + ... + 16)\)

Using the sum formula for \(n = 16\):
\[ Sum = \frac{16(16 + 1)}{2} \]
\[ Sum = 8 \times 17 = 136 \]

Applying the negative sign:

Coefficient = \(-136\)


Step 4: Final Answer:

The coefficient of \(x^{15}\) is \(-136\).
Quick Tip: In the expansion \((x-a)(x-b)(x-c)...\), the coefficient of the second-highest power is always the negative sum of the constants.