GATE EE 2021

This sentence has a main idea ("The people... were from all sections of society") and a descriptive clause in the middle. The blank requires a pronoun to begin this clause and refer back to "the people."

Because "the people" are human, we can immediately rule out "which," which is used for things. The choice is now between "who" and "whom." The pronoun here is the subject of the verb "were" (ask yourself: who were at the demonstration?). "Who" is the pronoun used for subjects, while "whom" is used for objects. Therefore, "who" is the correct choice.

When the transparent sheet is folded along the dotted line, the right half is laid over the left half. Think of the dotted line as a mirror. The image on the right side will be reflected onto the left side. The angled part of the figure, which currently opens to the left (<), will flip horizontally to open to the right (>). This reflected shape will then appear superimposed over the left half, which already contains the vertical line. This combination creates the final figure.

First, we find the sum of all interior angles for a polygon with $n$ sides using the formula $(n-2) \times 180^\circ$. For a polygon with 10 sides, this sum is $(10-2) \times 180^\circ = 8 \times 180^\circ$, which equals $1440^\circ$.

A regular polygon has interior angles that are all equal in measure. To find the size of a single angle, we divide the total sum by the number of sides.

Therefore, the measure of one interior angle is $\frac{1440^\circ}{10} = 144^\circ$.

Let's test the options for divisibility by letting $x = 11^{13}$. The number we are dividing by is then $x+1$.

Now, let's examine the expression in option D, $11^{52}-1$. We can rewrite this in terms of $x$. Since $52 = 13 \times 4$, we have $11^{52} = (11^{13})^4 = x^4$. This makes the expression $x^{4-}1$.

We can factor $x^{4-}1$ by applying the difference of squares formula ($a^{2-}b^2 = (a-b)(a+b)$) twice:
$x^4 - 1 = (x^2 - 1)(x^2 + 1) = (x-1)(x+1)(x^{2+}1)$.

The factored form clearly contains the term $(x+1)$, which means $x^{4-}1$ is exactly divisible by $x+1$. By substituting back, we confirm that $11^{52}-1$ is exactly divisible by $11^{13}+1$.

This analogy is based on the relationship between a specific place and its surroundings. An oasis is a small area of water and fertile land located within a vast desert of sand. The sand is the defining environment that encompasses the oasis.

Following this pattern, an island is a piece of land. We need to find the vast environment that similarly encompasses an island. An island, by definition, is a landmass completely bordered on all sides by water.

The passage establishes a clear cause-and-effect relationship: lack of sleep harms your ability to recall what you've studied. The key takeaway is that studying and sleeping are not opposing activities, but partners in effective exam preparation.

Options A, B, and C all fail this logic. Option A wrongly dismisses studying, while options B and C dangerously dismiss the need for sleep, directly contradicting the passage's warning.

Therefore, the only statement that logically follows from the text is that adequate sleep is a necessary component of the entire preparation process, ensuring that the hard work of studying pays off during the exam.

Let's examine the geometric pattern. When a new square is formed by connecting the midpoints of a larger one, its side length is the original side divided by $\sqrt{2}$. We start with a side of $10 \text{ cm}$ for the largest square.

The shaded square is the fifth square inscribed inside the original, meaning this division process happens five times. So, the side length of the smallest square is $\frac{10}{(\sqrt{2})^5}$.

To calculate the area, we square this side length:
$\text{Area} = \left(\frac{10}{(\sqrt{2})^5}\right)^2 = \frac{10^2}{((\sqrt{2})^5)^2} = \frac{100}{2^5}$
This simplifies to $\frac{100}{32}$, which is $3.125 \text{ cm}^2$.

The constant probability density function, $f(x)=0.01$ over the interval $[0, 100]$, tells us this is a continuous uniform distribution. This means any temperature in the range is equally likely. The mean (or expected value) of a uniform distribution is simply the midpoint of its interval.

For an interval $[a, b]$, the mean is calculated as $\frac{a+b}{2}$.
With $a=0$ and $b=100$, the mean temperature is:
$\text{Mean} = \frac{0 + 100}{2} = 50$
Alternatively, we can compute the mean by its formal definition, integrating $x \cdot f(x)$ over the range:
$E[X] = \int_{0}^{100} x(0.01) \,dx = 0.01 \left[ \frac{x^2}{2} \right]_{0}^{100} = 0.01 \left( \frac{100^2}{2} - 0 \right) = 50$

The number of first-time test takers in a given year is the total number of students who took the exam, minus the students who are repeating it from the previous year.

First, let's analyze Year 2. The bar chart shows a total of $60 (\text{pass}) + 5 (\text{fail}) = 65$ students took the exam. The students repeating the exam are the ones who failed in Year 1, which is $10$. So, the number of first-time test takers in Year 2 is $65 - 10 = 55$.

Next, we apply the same logic to Year 3. The total number of students who took the exam was $50 (\text{pass}) + 3 (\text{fail}) = 53$. The number of repeaters are the $5$ students who failed in Year 2. Thus, the number of first-time students in Year 3 is $53 - 5 = 48$.

Let's focus on the most restrictive rules. We are told that Q and S must be parked together, and S and V must also be parked together. The only way for S to be next to both Q and V is for it to be in the middle, forming a three-car block. This gives us two possibilities: the sequence $QSV$ or $VSQ$.

Next, we are given that R is parked to the immediate right of V, creating the pair $VR$. If we apply this to our two possibilities, the $QSV$ block becomes $QSVR$. However, the $VSQ$ block is invalid because S is to the right of V, not R.

Therefore, the arrangement of these four cars must be the block $QSVR$. In this required sequence, there is exactly one car, S, parked between Q and V. Consequently, the statement that there are two cars between Q and V is demonstrably false.

To find the eigenvalues, we solve the characteristic equation of the given matrix. For a 2x2 matrix, the characteristic equation is $\lambda^2 - (\text{trace})\lambda + (\text{determinant}) = 0$.

First, let's compute the trace, which is the sum of the main diagonal elements:
$\text{Trace} = p + (-p) = 0$.

Next, let's compute the determinant:
$\text{Determinant} = (p)(-p) - (q)(q) = -p^2 - q^2 = -(p^{2+}q^2)$.

Substituting these into the characteristic equation, we get $\lambda^2 - (0)\lambda - (p^{2+}q^2) = 0$.
Since the problem states that $p^{2+}q^2 = 1$, the equation simplifies to $\lambda^2 - 1 = 0$.
Solving for $\lambda$ yields the eigenvalues $\lambda = 1$ and $\lambda = -1$.

For a cubic polynomial of the form $z^3 + bz^2 + cz + d = 0$, the sum of its roots is given by $-b$. In this case, the polynomial is $p(z) = z^3 + (1+j)z^2 + (2+j)z + 3$, so the sum of its roots is $-(1+j)$.

This sum is a non-real complex number. If all three roots of the polynomial were real numbers, their sum would also have to be a real number. However, since the sum of the roots is the complex value $-(1+j)$, it's logically impossible for all the roots to be real. Therefore, at least one root must be a non-real complex number.

We are given that $f'(x_0) = 0$ for some $x_0$ in the interval $(0, 1)$, which by definition makes $x_0$ a critical point.

To classify this critical point, we use the Second Derivative Test. The problem states that $f''(x) > 0$ for all $x$ in $(0, 1)$. This means the function is strictly concave up on the entire interval. Since $x_0$ is in this interval, we know specifically that $f''(x_0) > 0$. A critical point where the second derivative is positive corresponds to a local minimum.

Furthermore, because $f''(x)$ is always positive on the interval, the first derivative $f'(x)$ must be strictly increasing. A strictly increasing function can equal zero at most once. Therefore, $x_0$ is the only critical point in the interval, and the function has exactly one local minimum.

To find the Thevenin equivalent circuit, we need to calculate the Thevenin voltage ($V_{TH}$) and Thevenin resistance ($R_{TH}$).

First, let's find the Thevenin voltage, which is the open-circuit voltage across terminals 'a' and 'b'. With the terminals open, the entire 5 A from the current source flows through the 10Ω resistor. This means the controlling current is $i_1 = 5 \text{ A}$. The voltage from the dependent source is $3i_1 = 3(5) = 15 \text{ V}$. By Kirchhoff's voltage law, $V_{TH}$ is the sum of the voltage across the 10Ω resistor and the dependent source, so $V_{TH} = (5 \text{ A} \times 10 \Omega) + 15 \text{ V} = 50 \text{ V} + 15 \text{ V} = 65 \text{ V}$.

Next, we find the Thevenin resistance by deactivating the independent 5A source (replacing it with an open circuit) and finding the resistance looking into terminals 'ab'. Let's apply a test current $I_{test}$ into terminal 'a'. This current flows through the entire loop, so $i_1 = I_{test}$. The voltage $V_{test}$ across the terminals is the sum of voltage drops: $V_{test} = (2\Omega)I_{test} + 3i_1 + (10\Omega)i_1$. Substituting $i_1 = I_{test}$, we get $V_{test} = (2+3+10)I_{test} = 15I_{test}$. Therefore, the Thevenin resistance is $R_{TH} = V_{test}/I_{test} = 15 \Omega$.

According to a fundamental law of electromagnetism, any valid magnetic field $\vec{B}$ must have a divergence of zero. This is expressed mathematically as $\vec{\nabla} \cdot \vec{B} = 0$.

To check which vector function satisfies this condition, we calculate its divergence. For a vector field $\vec{B} = B_x\hat{X} + B_y\hat{Y} + B_z\hat{Z}$, the divergence is given by:
$\vec{\nabla} \cdot \vec{B} = \frac{\partial B_x}{\partial x} + \frac{\partial B_y}{\partial y} + \frac{\partial B_z}{\partial z}$
Let's apply this to the vector field $\vec{B} = 10x\hat{X} + 20y\hat{Y} - 30z\hat{Z}$.
$\vec{\nabla} \cdot \vec{B} = \frac{\partial}{\partial x}(10x) + \frac{\partial}{\partial y}(20y) + \frac{\partial}{\partial z}(-30z)$
$= 10 + 20 - 30 = 0$
Since the divergence is zero, this vector function can represent a magnetic field.

Let's analyze the system's properties: linearity and time-invariance.

First, we test for linearity using the superposition principle. Consider two inputs: $x_1(t) = 1$, which gives the output $y_1(t) = \text{max}(0, 1) = 1$, and $x_2(t) = -2$, which gives $y_2(t) = \text{max}(0, -2) = 0$. The output for the sum of these inputs, $x(t) = x_1(t) + x_2(t) = -1$, is $y(t) = \text{max}(0, -1) = 0$. However, the sum of the individual outputs is $y_1(t) + y_2(t) = 1 + 0 = 1$. Since these results are not equal, the system is non-linear.

Next, we test for time-invariance. Let's find the output for a delayed input, $x(t-t_0)$. The system produces $y_1(t) = \text{max}(0, x(t-t_0))$. Now, let's delay the original output $y(t)$ by $t_0$, which gives $y(t-t_0) = \text{max}(0, x(t-t_0))$. Since the output of the delayed input is identical to the delayed output, the system is time-invariant.

When two linear time-invariant (LTI) systems are connected in cascade, the overall impulse response $h[n]$ is the convolution of their individual responses, $h_1[n]$ and $h_2[n]$. We can simplify this convolution by working in the Z-domain, where convolution becomes multiplication: $H(z) = H_1(z) H_2(z)$.

First, let's find the Z-transform of each impulse response. Using the property that the Z-transform of $\delta[n-k]$ is $z^{-k}$, we get:
$H_1(z) = z + z^{-1}$
$H_2(z) = 1 + z^{-1}$

Next, we multiply these transforms to find the overall system function $H(z)$:
$H(z) = (z + z^{-1})(1 + z^{-1}) = z \cdot 1 + z \cdot z^{-1} + z^{-1} \cdot 1 + z^{-1} \cdot z^{-1}$
$H(z) = z + 1 + z^{-1} + z^{-2}$

Finally, we convert $H(z)$ back to the time domain by taking the inverse Z-transform. Each term $z^{-k}$ corresponds to an impulse $\delta[n-k]$, giving us the final impulse response:
$h[n] = \delta[n+1] + \delta[n] + \delta[n-1] + \delta[n-2]$

Here's a breakdown of the correct associations between the machine parts, their types, and the problems they solve:

• (P) Damper bars are fitted onto the rotors of (U) synchronous machines. Their purpose is to dampen rotor oscillations during transient load changes, a phenomenon known as (X) hunting. This gives the combination P-U-X.

• (Q) Skewed rotor slots are a design feature of (S) induction motors. By angling the rotor bars, we prevent a strong magnetic alignment with the stator teeth at startup, thus mitigating (Y) magnetic locking. This gives Q-S-Y.

• (R) Compensating windings are placed in the pole faces of large (V) DC machines. They produce a magnetic field that directly cancels out the distorting field from the armature, thereby neutralizing the effects of (Z) armature reaction. This gives R-V-Z.

The Jacobian sub-matrix $M$ represents the sensitivity of reactive power ($Q$) to changes in voltage angles ($\delta$), mathematically expressed as $\frac{\partial Q}{\partial \delta}$.

The number of rows in $M$ is determined by the number of reactive power mismatch equations ($\Delta Q$). These equations are formulated only for buses where reactive power is an unknown variable, which are the $N_L$ PQ buses.

The number of columns in $M$ corresponds to the number of unknown voltage angle variables ($\Delta \delta$). In an N-bus system, the voltage angle is known only at the single slack bus, leaving $N-1$ unknown angles to be solved.

Therefore, the dimension (rows $\times$ columns) of the sub-matrix $M$ is $N_L \times (N-1)$.

For the most economical operation, the principle of economic dispatch requires the incremental costs of both generators to be equal. This gives us our first relationship:
$40+0.2P_{1} = 32+0.4P_{2}$

The second condition is that the sum of the power generated must equal the total load demand.
$P_{1} + P_{2} = 260 \text{ MW}$

We now have a system of two linear equations with two variables, $P_1$ and $P_2$. Solving these equations simultaneously yields the required generation for each unit. From the second equation, we can write $P_{2} = 260 - P_{1}$ and substitute it into the first, which results in $P_{1}=160 \text{ MW}$ and $P_{2}=100 \text{ MW}$.

Let's derive the transfer function by analyzing the signals in the block diagram. The error signal, $E(s)$, is generated at the summing junction as the input $R(s)$ minus the feedback signal. The feedback signal is the output $C(s)$ processed by $H$. So, we can write the equation:
$E(s) = R(s) - H C(s)$

To find the ratio $\dfrac{E(s)}{R(s)}$, let's divide the entire equation by $R(s)$:
$\dfrac{E(s)}{R(s)} = \dfrac{R(s)}{R(s)} - H \dfrac{C(s)}{R(s)} = 1 - H \dfrac{C(s)}{R(s)}$

For a standard negative feedback system, the closed-loop transfer function is $\dfrac{C(s)}{R(s)} = \dfrac{G}{1+GH}$. By substituting this known relationship into our equation, we get:
$\dfrac{E(s)}{R(s)} = 1 - H \left( \dfrac{G}{1+GH} \right) = \dfrac{1+GH - GH}{1+GH}$

After simplifying the numerator, we arrive at the final expression:
$\dfrac{E(s)}{R(s)} = \dfrac{1}{1+GH}$

The Maxwell bridge is a classic AC bridge circuit specifically designed to measure an unknown inductance by balancing it against a known capacitance. Each of the other options is also a bridge circuit, but they are specialized for different tasks. The Schering bridge is used to measure capacitance and dielectric properties. The Kelvin bridge is an adaptation of the Wheatstone bridge for measuring very low resistances. Finally, the Wien bridge is most commonly used for measuring frequency.

Orthogonal intersection means the tangent lines to the circles at their intersection point are perpendicular. The product of the slopes of perpendicular lines is $-1$.

First, we find the slope of the tangent to each circle at the point $(u,v)$ using implicit differentiation.
For $x^2 + y^2 = 1$, the slope is $m_1 = \frac{dy}{dx} = -\frac{x}{y}$, so at $(u,v)$, $m_1 = -\frac{u}{v}$.
For $(x-1)^2 + (y-1)^2 = r^2$, the slope is $m_2 = \frac{dy}{dx} = -\frac{x-1}{y-1}$, so at $(u,v)$, $m_2 = -\frac{u-1}{v-1}$.

Applying the orthogonality condition $m_1 \cdot m_2 = -1$:
$\left(-\frac{u}{v}\right) \left(-\frac{u-1}{v-1}\right) = -1$
This simplifies to $u(u-1) = -v(v-1)$, which becomes $u^{2-}u = -v^{2+}v$.
Rearranging the terms, we get $u^{2+}v^2 = u+v$.
Since the point $(u,v)$ is on the first circle, we know that $u^{2+}v^2=1$.
Substituting this into our equation gives $1=u+v$.

For the current $I$ from the voltage source to be zero, the total impedance of the circuit must be infinite. This will occur if the parallel portion of the circuit acts as an open circuit, which is the condition for parallel resonance.

The impedance of a parallel combination becomes infinite when the denominator of the product-over-sum formula goes to zero. The two parallel branches have impedances of $Z_1 = j5 \Omega$ and $Z_2 = j5 - jX_C$. Setting their sum to zero yields:
$j5 + (j5 - jX_C) = 0$

This simplifies to $j10 = jX_C$, so the required capacitive reactance is $X_C = 10 \Omega$.

Finally, we find the capacitance from its reactance formula, $C = \frac{1}{\omega X_C}$. Given $\omega = 5 \times 10^3$ rad/s:
$C = \frac{1}{(5 \times 10^3 \text{ rad/s})(10 \Omega)} = \frac{1}{5 \times 10^4} \text{ F} = 20 \times 10^{-6} \text{ F} = 20 \mu\text{F}$

Let's determine the total conductor and insulation resistances for the series combination.

When the two cables are joined in series, their conductor resistances add directly.
$R_{\text{conductor, total}} = 0.7\;\Omega + 0.5\;\Omega = 1.2\;\Omega$

The insulation provides a leakage path from the core to the sheath for each cable. When the cables are joined, these two leakage paths are in parallel. Therefore, the equivalent insulation resistance is found using the formula for parallel resistors.
$\frac{1}{R_{\text{insulation, total}}} = \frac{1}{600\;M\Omega} + \frac{1}{900\;M\Omega} \implies R_{\text{insulation, total}} = 360\;M\Omega$

Finally, we calculate the required ratio of total insulation resistance to total conductor resistance.
$\frac{R_{\text{insulation, total}}}{R_{\text{conductor, total}}} = \frac{360 \times 10^6\;\Omega}{1.2\;\Omega} = 300 \times 10^6$

To find the value of $\beta$ that makes $V_y=0$, we will use nodal analysis.

First, let's write the Kirchhoff's Current Law (KCL) equation at node $V_x$. Summing the currents leaving the node yields $\frac{V_x-6}{1} + \frac{V_x}{4} + \frac{V_x-V_y}{2} = 0$, which simplifies to $7V_x - 2V_y = 24$.
Next, applying KCL at node $V_y$ gives us $\frac{V_y-V_x}{2} + \frac{V_y-\beta V_x}{3} = 2$, which simplifies to $5V_y - (3+2\beta)V_x = 12$.

We are given the condition $V_y = 0$. Substituting this into our first simplified equation gives $7V_x = 24$, so $V_x = \frac{24}{7}$ V.
Now, we substitute both $V_y=0$ and $V_x=\frac{24}{7}$ into our second simplified equation: $-(3+2\beta)\left(\frac{24}{7}\right) = 12$.
Solving for $\beta$, we find $3+2\beta = -12 \cdot \frac{7}{24} = -3.5$. This leads to $2\beta = -6.5$, and thus, $\beta = -3.25$.

The work done moving a charge in an electrostatic field depends only on the start and end points, not the path taken, because the electric force is conservative. The charge $q_2 = 10 \;\mu C$ starts at a distance $r_i = 10 \;\text{m}$ from the origin and ends at a distance $r_f = 5 \;\text{m}$. The intermediate point is irrelevant for the total work.

The work done by an external force is equal to the change in the system's electric potential energy:
$W = \Delta U = U_{\text{final}} - U_{\text{initial}}$

Using the formula for the potential energy between two point charges, $U = k \dfrac{q_1 q_2}{r}$, we get:
$W = k q_1 q_2 \left(\dfrac{1}{r_f} - \dfrac{1}{r_i}\right)$

Plugging in the given values, with $q_1 = 1 \;\mu C$ and $k = \frac{1}{4\pi \varepsilon_{0}} = 9 \times 10^9$:
$W = (9 \times 10^9) (1 \times 10^{-6}) (10 \times 10^{-6}) \left(\dfrac{1}{5} - \dfrac{1}{10}\right)$

$W = (9 \times 10^{-2}) \left(\dfrac{1}{10}\right) = 9 \times 10^{-3} \;\text{J} = 9 \;\text{mJ}$.

To find the motor's efficiency, we must trace the flow of power and subtract the losses at each stage. The initial power input is $40 \text{ kW}$.

First, the power transferred from the stator to the rotor is the input power minus the stator losses: $P_{rotor\_in} = 40 \text{ kW} - 1 \text{ kW} = 39 \text{ kW}$.

Next, we calculate the slip ($s$). The synchronous speed is $N_s = \frac{120f}{P} = \frac{120 \times 50}{6} = 1000 \text{ RPM}$. Thus, the slip is $s = \frac{N_s - N_r}{N_s} = \frac{1000 - 975}{1000} = 0.025$.

The gross mechanical power developed by the rotor is $P_{mech} = P_{rotor\_in} \times (1-s) = 39 \times (1 - 0.025) = 38.025 \text{ kW}$.

To get the final useful power at the shaft, we subtract the friction and windage losses: $P_{out} = 38.025 \text{ kW} - 2.025 \text{ kW} = 36 \text{ kW}$.

The efficiency ($\eta$) is the ratio of the final output power to the total input power:
$\eta = \frac{P_{out}}{P_{\in}} = \frac{36 \text{ kW}}{40 \text{ kW}} \times 100\% = 90\%$.

The real power transferred from the alternator to the motor is governed by the power-angle relationship. The key is to first find the total reactance connecting the two internal voltage sources. We sum the synchronous reactance of the alternator, the transmission line reactance, and the synchronous reactance of the motor.

Total Reactance, $X_{total} = 0.4 + 0.1 + 0.35 = 0.85 \text{ p.u.}$

Next, we use the formula for real power transfer: $P = \frac{|E_1||E_2|}{X_{total}}\sin(\delta_1 - \delta_2)$. Plugging in the given values:
$0.866 = \frac{(1.0)(0.85)}{0.85} \sin(\delta_1 - \delta_2)$

This simplifies the expression to $\sin(\delta_1 - \delta_2) = 0.866$.
Therefore, the angle difference is $(\delta_1 - \delta_2) = \arcsin(0.866)$, which is $60^\circ$.

First, we use the Bode plot to determine the system's damping ratio, $\xi$. The plot shows a resonant peak magnitude $M_r$ of 26 dB. In linear terms, this magnitude is $M_r = 10^{26/20} \approx 20$. For a second-order system with a large resonant peak, the magnitude is approximated by $M_r \approx \frac{1}{2\xi}$. From this, we can calculate the damping ratio: $\xi \approx \frac{1}{2 \times 20} = 0.025$.

Next, we relate this damping ratio to the physical components of the series RLC circuit. The formula for the damping ratio in this circuit configuration is $\xi = \frac{R}{2}\sqrt{\frac{C}{L}}$. We can now solve for the resistance R.

Rearranging the formula and substituting the given values:
$R = 2\xi\sqrt{\frac{L}{C}} = 2(0.025)\sqrt{\frac{1 \times 10^{-3}}{250 \times 10^{-6}}} = 0.10 \, \Omega$.

The signal generator, with its source resistance ($R = 50\:\Omega$), and the external capacitor ($C$) form a simple RC voltage divider. The open-circuit terminal voltage is the input to this divider, $V_{\in} = 10\text{ V}$. When the capacitor is connected, the voltage across it is the output, $V_{out} = 8\text{ V}$.

The magnitude ratio for this voltage divider is given by:
$\frac{|V_{out}|}{|V_{\in}|} = \frac{1}{\sqrt{1 + (\omega R C)^2}}$
Substituting the known values, we get $\frac{8}{10} = \frac{1}{\sqrt{1 + (\omega R C)^2}}$.

Solving for the denominator gives $\sqrt{1 + (\omega R C)^2} = \frac{10}{8} = 1.25$.
Squaring both sides and isolating the $(\omega R C)$ term yields $\omega R C = \sqrt{1.25^2 - 1^2} = 0.75$.

With an angular frequency $\omega = 2\pi f = 2\pi(1000)\text{ rad/s}$, we can solve for the capacitance:
$C = \frac{0.75}{\omega R} = \frac{0.75}{2\pi(1000)(50)} \approx 2.39 \:\mu\text{F}$

The output $Y$ is formed by OR-ing the two most significant bits, $Q_{15}$ and $Q_{14}$. This means $Y$ will be high ($Y=1$) as long as at least one of these bits is a '1'. For a 16-bit up-counter, the states where both $Q_{15}$ and $Q_{14}$ are '0' range from $0$ to $2^{14}-1$.

At state $2^{14}$ (binary $0100...00$), $Q_{14}$ becomes '1' for the first time, making $Y$ high. $Y$ then remains high for all subsequent states up to the counter's maximum value of $2^{16}-1$. The total number of states where $Y=1$ is $(2^{16}-1) - 2^{14} + 1 = 2^{16} - 2^{14} = 49152$ states.

Each state corresponds to one clock cycle. Therefore, the high duration is the number of high states times the clock period, $T_{\text{CLK}}$.
$49152 \times T_{\text{CLK}} = 24 \text{ ms}$

Solving for the clock frequency, $f_{\text{CLK}} = 1/T_{\text{CLK}}$:
$f_{\text{CLK}} = \frac{49152}{24 \times 10^{-3} \text{ s}} = 2,048,000 \text{ Hz} = 2.048 \text{ MHz}$
Rounding to two decimal places, the frequency is $2.05 \text{ MHz}$.

Our starting point is the specified voltage across the collector resistor, $R_C$. We can use Ohm's law to find the collector current: $I_C = 5\,\text{V} / 3.3\,\text{k}\Omega = 1.515\,\text{mA}$.

Given the current gain $\beta=100$, we can determine the other two transistor currents. The base current is $I_B = I_C/\beta = 1.515\,\text{mA} / 100 = 0.01515\,\text{mA}$, and the emitter current is $I_E = I_C + I_B = 1.53\,\text{mA}$.

Next, we find the voltage at the base node, $V_B$, by applying KVL around the emitter-base loop. First, the emitter voltage is $V_E = 12\,\text{V} - I_E R_E = 12 - (1.53\,\text{mA})(1.2\,\text{k}\Omega) = 10.164\,\text{V}$. Since $V_{BE} = -0.7\,\text{V}$ for this PNP transistor, the base voltage is $V_B = V_E + V_{BE} = 10.164\,\text{V} - 0.7\,\text{V} = 9.464\,\text{V}$.

Now we analyze the currents in the base biasing circuit. The current through $R_1$ is $I_{R1} = (12\,\text{V} - V_B)/4.7\,\text{k}\Omega = 0.5396\,\text{mA}$. Applying KCL at the base node, the current flowing down through $R_2$ is found by summing the currents from $R_1$ and the base: $I_{R2} = I_{R1} + I_B = 0.5396\,\text{mA} + 0.01515\,\text{mA} = 0.5547\,\text{mA}$.

Finally, we can find the resistance of $R_2$ using Ohm's law: $R_2 = V_B / I_{R2} = 9.464\,\text{V} / 0.5547\,\text{mA} \approx 17.06\,\text{k}\Omega$.

The Peak-Inverse Voltage (PIV) for any diode in this bridge rectifier configuration is equal to the peak value of the AC input voltage, $V_{peak}$.

To determine the worst-case PIV, we need to use the highest possible input voltage, which occurs at the upper limit of the given tolerance.

First, let's find the maximum RMS voltage:
$V_{RMS, max} = 230 \, V + (0.20 \times 230 \, V) = 230 \times 1.2 = 276 \, V$.

Now, we convert this maximum RMS voltage to the corresponding peak voltage, which represents the worst-case PIV.
$PIV_{worst} = V_{peak, max} = V_{RMS, max} \times \sqrt{2} = 276 \times \sqrt{2} \approx 390.32 \, V$.