29 Jan 2024 Shift 2

We need to find the percentage error in $Q = \frac{a^4 b^3}{c^2}$. Recall the error propagation formula. For a quantity $Q = a^m b^n c^p$, the maximum percentage error is: $\frac{\Delta Q}{Q} \times 100 = |m| \cdot \frac{\Delta a}{a} \times 100 + |n| \cdot \frac{\Delta b}{b} \times 100 + |p| \cdot \frac{\Delta c}{c} \times 100$ Apply to the given expression. $Q = a^4 b^3 c^{-2}$, so $m = 4$, $n = 3$, $p = -2$: $\% \text{error \in } Q = 4 \times 3\% + 3 \times 4\% + 2 \times 5\%$ $= 12\% + 12\% + 10\% = 34\%$ The correct answer is Option (3): 34% .

We need to find the velocity of a particle when its acceleration is zero, given $x = t^3 - 6t^2 + 20t + 15$. Find the velocity. $v = \frac{dx}{dt} = 3t^2 - 12t + 20$ Find the acceleration. $a = \frac{dv}{dt} = 6t - 12$ Find when acceleration is zero. $6t - 12 = 0 \implies t = 2$ s Find velocity at $t = 2$. $v(2) = 3(4) - 12(2) + 20 = 12 - 24 + 20 = 8$ m/s The correct answer is Option (2): 8 m/s .

The mass of the stone is $m = 900\text{ g} = 0.9\text{ kg}$ and the radius of the vertical circle is $r = 1\text{ m}$. First convert the speed of revolution from revolutions per minute (rpm) to radians per second (rad s−1). A body that makes $N$ revolutions per minute has an angular speed $\omega = 2\pi N \text{ rad per minute}$. Given $N = 10$ rpm: $\omega = 2\pi \times 10 = 20\pi \text{ rad min}^{-1}$. To express this \in seconds, divide by $60$: $\omega = \frac{20\pi}{60} = \frac{\pi}{3} \text{ rad s}^{-1}$. The linear speed $v$ of the stone is related to angular speed by $v = \omega r$. Hence $v = \frac{\pi}{3} \times 1 = \frac{\pi}{3} \text{ m s}^{-1}$. The square of the speed is therefore $v^{2} = \left(\frac{\pi}{3}\right)^{2} = \frac{\pi^{2}}{9}$. Given $\pi^{2} = 9.8$, we get $v^{2} = \frac{9.8}{9} = 1.0889 \text{ m}^{2}\text{ s}^{-2} \approx 1.09 \text{ m}^{2}\text{ s}^{-2}$. At the lowest point of the vertical circle, the tension $T$ \in the string must provide the centripetal force towards the centre \in addition to balancing the weight acting downward. Therefore $T - mg = \frac{mv^{2}}{r}$. Rearranging gives $T = mg + \frac{mv^{2}}{r}$. Substituting the values: $mg = 0.9 \times 9.8 = 8.82 \text{ N}$, $\frac{mv^{2}}{r} = 0.9 \times 1.09 = 0.98 \text{ N}$. Hence $T = 8.82 + 0.98 = 9.80 \text{ N}$. Thus, the tension \in the string at the lowest point is approximately $9.8 \text{ N}$. Option B is correct.

Given, bob of pendulum is released from horizontal position, Length of Pendulum ,L= $10$ m $g = 10 \text{ m s}^{-2}$ At horizontal position, the bob is at height h=L above the lowest point. Energy of bob at Initial Position is Ei = P.E = mgh where: m = mass of the bob - $g = 10 \text{ m s}^{-2}$ (acceleration due to gravity) - h= $10$ m (length of the pendulum) So, P.E= m$\times \$g$\times \ 10$ Now Lets Calculate Energy Dissipated:- K.E=$\ \frac{\ 1}{2}\times \ m\times \ v^2$ Since 10% of the energy is dissipated due to air resistance, the energy available for conversion to kinetic energy (KE) is: K.E(final)=$90\ \%\$ of $m\times \ g\times \ h$ = $\ \frac{\ 1}{2}\times \ m\times \ v^2$ K.E(final)=$\ \frac{\ 90}{100}\times \ m\times \ 10\times \ 10$= $\ \frac{\ 1}{2}\times \ m\times \ v^2$ On cancelling m on both sides, K.E(final)=$\ \frac{\ 90}{100}\times \ 10\times \ 10$= $\ \frac{\ 1}{2}\times \ v^2$ $v^2=2\times \ \ \frac{\ 90}{100}\times \ 10\times \ 10$ $v^2=2\times \ \ \frac{\ 90}{100}\times \ 100$ $v^2=180$ $v=\sqrt{\ 180}$ $v=6\sqrt{\ 5}$ Therefore, Final velocity v=$6\sqrt{\ 5}$ Correct answer is A

Condition at the Topmost Point ($B$) For the bob to just complete the circle, the tension \in the string at point B must be zero. The centripetal force at this point is provided entirely by the gravitational force: $mg = \frac{mv_B^2}{L}$ $v_B^2 = gL$ Kinetic Energy at Point $B$ The kinetic energy at the highest point is: $(K.E.)_B=\frac{1}{2}mv_B^2=\frac{1}{2}mgL$ Conservation of Mechanical Energy $(K.E.)_A = (K.E.)_B + (P.E.)_B$ $(K.E.)_A = \frac{1}{2}mgL + mg(2L)$ $(K.E.)_A = \frac{5}{2}mgL$ The ratio is $5 : 1$ .

We need to find the new orbital period when the distance from the Sun is reduced to one-fourth. According to Kepler's Third Law, $T^2 \propto r^3$, where $T$ is the orbital period and $r$ is the orbital radius. Applying this to the initial and new conditions, we get $\frac{T_2^2}{T_1^2} = \frac{r_2^3}{r_1^3} = \left(\frac{r_2}{r_1}\right)^3 = \left(\frac{1}{4}\right)^3 = \frac{1}{64}$, which implies $\frac{T_2}{T_1} = \frac{1}{8}$. Since the original period $T_1$ is 200 days, the new period is $T_2 = \frac{T_1}{8} = \frac{200}{8} = 25$ days. The correct answer is Option (1): 25 days .

We need to find how the elongation changes when both the force and radius are halved. From Young's modulus we have $Y = \frac{F/A}{\Delta l/L} = \frac{FL}{\pi r^2 \Delta l}$ which gives $\Delta l = \frac{FL}{\pi r^2 Y}.$ When the force is halved ($F' = F/2$) and the radius is halved ($r' = r/2$), while $L$ and $Y$ remain the same, the new elongation becomes $\Delta l' = \frac{F'L}{\pi (r')^2 Y} = \frac{(F/2)L}{\pi (r/2)^2 Y} = \frac{(F/2)L}{\pi r^2 Y / 4} = \frac{4FL}{2\pi r^2 Y} = \frac{2FL}{\pi r^2 Y} = 2\Delta l.$ The new elongation is twice the original; halving the radius (which quarters the cross-sectional area) has a larger effect than halving the force. The correct answer is Option (4): 2 times .

A liquid drop of radius $R$ is divided into 27 identical smaller drops. By conserving volume, we have $\frac{4}{3}\pi R^3 = 27 \times \frac{4}{3}\pi r^3$, which simplifies to $R^3 = 27r^3$ and hence $r = \frac{R}{3}$. The initial surface area of the drop is $A_i = 4\pi R^2$. After division into 27 drops of radius $r$, the total surface area becomes $A_f = 27 \times 4\pi r^2 = 27 \times 4\pi \times \frac{R^2}{9} = 12\pi R^2$. Therefore, the change \in surface area is $\Delta A = A_f - A_i = 12\pi R^2 - 4\pi R^2 = 8\pi R^2$. The work done \in increasing the surface area against surface tension is given by $W = T \times \Delta A = T \times 8\pi R^2 = 8\pi R^2 T$. The work done is $8\pi R^2 T$, which corresponds to Option 1 .

To find the temperature of a gas given its number density and pressure, we start with the ideal gas law in the form $P = nkT$, where $n$ is the number density (molecules per unit volume), $k$ is Boltzmann's constant, and $T$ is the temperature. Converting the given pressure of 1.38 atm to SI units gives $P = 1.38$ atm $= 1.38 \times 1.01325 \times 10^5$ Pa $= 1.398 \times 10^5$ Pa. Solving for temperature yields $T = \frac{P}{nk} = \frac{1.398 \times 10^5}{2.0 \times 10^{25} \times 1.38 \times 10^{-23}}$ and $= \frac{1.398 \times 10^5}{2.76 \times 10^2} = \frac{1.398 \times 10^5}{276} \approx 506.5$ K $\approx 500$ K. The correct answer is Option (1): 500 K .

N moles of a polyatomic gas (f = 6) mixed with 2 moles of a monoatomic gas (f = 3) results in a mixture with effective degrees of freedom given by the weighted average: $f_{mix} = \frac{n_1 f_1 + n_2 f_2}{n_1 + n_2}$. Since the mixture behaves as a diatomic gas (fₘᵢₓ = 5), substituting the values yields $5 = \frac{N \times 6 + 2 \times 3}{N + 2}$. Multiplying both sides by (N + 2) gives $5(N + 2) = 6N + 6$, which simplifies to $5N + 10 = 6N + 6$. Solving for N leads to $N = 4$. The value of $N$ is 4, which corresponds to Option 3 .

Electric flux is $⃗\phi=\vec{E}\cdot\vec{A}$ Given $E=\left(6i+5j^​+3k\right)$ Surface lies \in YZ-plane, so area vector is along x-axis: $A⃗=30i$ Therefore, Dot product would be $=\ 6\times \ 30$ =180

The provided circuit diagram on page 12 of the PDF can be interpreted as two parallel branches connected to a 10V source. The top branch consists of a 2Ω and a 4Ω resistor in series, giving a total resistance of $R_{top} = 2 + 4 = 6 \, \Omega$. The bottom branch consists of a 4\Omega and a 1\Omega resistor \in series, resulting \in a total resistance of $R_{bottom} = 4 + 1 = 5 \, \Omega$. The current 'i' is indicated to be flowing through the top branch. Using Ohm's law, the current \in the top branch is calculated as $i = \frac{V}{R_{top}} = \frac{10 \, \text{V}}{6 \, \Omega} = \frac{5}{3} \, \text{A} \approx 1.67 \, \text{A}$.
Note: This calculated value does not match any of the options, and contradicts the provided answer key which states 1 A. There is likely an error in the question's diagram, values, or the provided answer.

We need to find the mass ratio $\frac{m_X}{m_Y}$ of two particles with equal charges accelerated through the same potential difference, making circular paths of radii $R_1$ and $R_2$. When a particle of mass m and charge q is accelerated through a potential $V$, its kinetic energy satisfies $\frac{1}{2}mv^2 = qV \Rightarrow v = \sqrt{\frac{2qV}{m}}$. In a magnetic field, the circular radius is given by $R = \frac{mv}{qB} = \frac{m}{qB}\sqrt{\frac{2qV}{m}} = \frac{\sqrt{2mV/q}}{B} = \frac{1}{B}\sqrt{\frac{2mV}{q}}$, so that $R \propto \sqrt{m}$ when q, V, and B are constant. Hence, $\frac{R_1}{R_2} = \sqrt{\frac{m_X}{m_Y}}$ and thus $\frac{m_X}{m_Y} = \left(\frac{R_1}{R_2}\right)^2$. Therefore, the answer is Option B.

The average power dissipated in an AC circuit is given by: $P_{avg} = \frac{V_0 I_0}{2} \cos\phi$, where $V_0$ is the peak voltage, $I_0$ is the peak current, and $\phi$ is the phase difference between voltage and current. From the given equations: $V = 100\sin(100t)$ V, so $V_0 = 100$ V $I = 100\sin\left(100t + \frac{\pi}{3}\right)$ mA $= 0.1\sin\left(100t + \frac{\pi}{3}\right)$ A, so $I_0 = 0.1$ A The phase difference is $\phi = \frac{\pi}{3} = 60°$, so $\cos\phi = \cos 60° = \frac{1}{2}$. Substituting into the formula: $P_{avg} = \frac{100 \times 0.1}{2} \times \frac{1}{2} = \frac{10}{2} \times \frac{1}{2} = \frac{10}{4} = 2.5$ W Therefore, the average power dissipated is $2.5$ W, which is Option (3).

For a plane electromagnetic wave travelling in free space, the electric field and magnetic field are related by: $B = \frac{E}{c}$, where $c = 3 \times 10^8$ m/s is the speed of light. Given: $\vec{E} = 9.6\hat{j}$ V/m. So $E = 9.6$ V/m. Substituting into the formula: $B = \frac{9.6}{3 \times 10^8} = 3.2 \times 10^{-8}$ T Now we need to find the direction of $\vec{B}$. For an EM wave, the direction of propagation is along $\vec{E} \times \vec{B}$. The wave travels along the $x$-direction, so $\vec{E} \times \vec{B}$ must be along $\hat{i}$. Since $\vec{E}$ is along $\hat{j}$, we need $\vec{B}$ along $\hat{k}$, because $\hat{j} \times \hat{k} = \hat{i}$. Therefore, $\vec{B} = 3.2 \times 10^{-8}\hat{k}$ T, which is Option (1).

For a virtual image formed by a curved mirror, the magnification $m = -v/u = 2$, which implies $v = -2u$. The object is at distance $u$ (in front, $u < 0$) and the image is at distance $v$ (behind, $v > 0$). The distance between them is $|v - u| = 15$ cm. Substituting $v = -2u$ into the distance equation:
$v - u = 15 \implies -2u - u = 15 \implies -3u = 15 \implies u = -5 \text{ cm}$
With $u = -5$ cm, we find $v = -2(-5) = 10$ cm. Using the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$:
$\frac{1}{f} = \frac{1}{10} + \frac{1}{-5} = \frac{1 - 2}{10} = -\frac{1}{10}$
Thus, $f = -10$ cm.

Path diff=7λ/4. Phase diff=2π×7/4=7π/2. cos(7π/2)=cos(3π+π/2)=0. I=I_max×cos²(δ/2)=I_max×cos²(7π/4)=I_max×(1/√2)²=I_max/2. Ratio=1/2. Option (1).

We need to find the ratio of the number of photons emitted by two light sources with the same power but different wavelengths. The energy of a single photon is given by $E = \frac{hc}{\lambda}$. If the power of each source is $P$, the number of photons emitted per second is $n = \frac{P}{E} = \frac{P\lambda}{hc}$, and since both sources have the same power $P$, it follows that $n \propto \lambda$. Therefore, the ratio of the number of photons emitted is $\frac{n_1}{n_2} = \frac{\lambda_1}{\lambda_2} = \frac{300}{500} = \frac{3}{5}$. The correct answer is Option (4): 3:5 .

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The circuit consists of two AND gates followed by an OR gate.

1. The top AND gate receives inputs $A$ and $B$, outputting $A \cdot B$.
2. The bottom AND gate receives the inverted input $\bar{A}$ and input $B$, outputting $\bar{A} \cdot B$.
3. The OR gate combines these signals to produce the final output $Y$:
   $Y = (A \cdot B) + (\bar{A} \cdot B)$
4. Factoring out $B$, we get $Y = B(A + \bar{A})$.
5. Since $A + \bar{A} = 1$, the expression simplifies to $Y = B$.
6. This means $Y$ follows $B$ regardless of $A$, resulting in the truth table where $Y$ is $0$ when $B=0$ and $1$ when $B=1$.

We need to find $\alpha$ where the time for the first revolution is $\frac{1}{\alpha}(1 - e^{-2\pi})$ s. Considering radius $R = 0.5$ m and initial speed $v_0 = 4$ m/s, and using $a_t = a_n$, we have $\frac{dv}{dt} = \frac{v^2}{R}\,.$ Separating variables gives $\frac{dv}{v^2} = \frac{dt}{R}\,,$ and integrating yields $-\frac{1}{v} = \frac{t}{R} + C\,.$ At $t = 0$, $C = -\frac{1}{v_0} = -\frac{1}{4}$, so $v = \frac{4}{1 - 8t}\,.$ The distance for one revolution is $s = 2\pi R = \pi$ and also $s = \int_0^T v\,dt = \int_0^T \frac{4}{1-8t}\,dt = -\frac{1}{2}\ln(1-8T)\,.$ Setting this equal to $\pi$ gives $-\frac{1}{2}\ln(1-8T) = \pi\,,$ so $1-8T = e^{-2\pi}$ and $T = \frac{1 - e^{-2\pi}}{8} = \frac{1}{8}(1 - e^{-2\pi})\,.$ Comparing this result with the given expression $\frac{1}{\alpha}(1-e^{-2\pi})$ shows that $\alpha = 8\,.$ Option: 8

We need to find the angular momentum of a 5 kg body moving with speed $3\sqrt{2}$ m/s along the line $y = x + 4$ about the origin. The formula for angular momentum is $L = mvd$, where $d$ is the perpendicular distance from the point (origin) to the line of motion. Rewriting the line $y = x + 4$ \in standard form gives $x - y + 4 = 0$. Hence the perpendicular distance from $(0, 0)$ to this line is $d = \frac{|0 - 0 + 4|}{\sqrt{1^2 + (-1)^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$ m. Substituting into the angular momentum formula yields $L = mvd = 5 \times 3\sqrt{2} \times 2\sqrt{2} = 5 \times 3 \times 2 \times (\sqrt{2} \times \sqrt{2}) = 5 \times 6 \times 2 = 60$ kg m$^2$/s. The answer is 60 kg m$^2$ s$^{-1}$.

Two metallic wires $P$ and $Q$ have the same volume and are made of the same material (so they have the same Young's modulus $Y$). Their cross-sectional areas are \in the ratio $A_P : A_Q = 4 : 1$. Since both wires have the same volume: $V = A_P \times L_P = A_Q \times L_Q$ $\Rightarrow \frac{L_P}{L_Q} = \frac{A_Q}{A_P} = \frac{1}{4}$ Using Young's modulus formula: $Y = \frac{F \cdot L}{A \cdot \Delta l}$ Rearranging for force to produce extension $\Delta l$: $F = \frac{Y \cdot A \cdot \Delta l}{L}$ For wire P: $F_1 = \frac{Y \cdot A_P \cdot \Delta l}{L_P}$ For wire Q (same extension $\Delta l$): $F_2 = \frac{Y \cdot A_Q \cdot \Delta l}{L_Q}$ Taking the ratio: $\frac{F_1}{F_2} = \frac{A_P}{A_Q} \times \frac{L_Q}{L_P}$ Since $L = V/A$, we have $L_Q/L_P = A_P/A_Q$: $\frac{F_1}{F_2} = \frac{A_P}{A_Q} \times \frac{A_P}{A_Q} = \left(\frac{A_P}{A_Q}\right)^2 = \left(\frac{4}{1}\right)^2 = 16$ The answer is $\frac{F_1}{F_2} = 16$.

Given time period T=6π So angular frequency $\omega=\frac{2\pi}{T}$ $=\frac{2\pi}{6\pi}$ $=\frac{1}{3}$ Since oscillation starts from mean position, $x=A\sin\omega t$ We need time taken from $x=A$ to $x=\frac{\sqrt{3}}{2}A$ At x=A $\sin\omega t_1=1$ $t_1=\frac{\pi}{2}\omega$ At $x=\frac{\sqrt{3}}{2}A$ while returning from extreme point, $\sin\omega t_2=\frac{\sqrt{3}}{2}$ after $\pi/2$, corresponding angle is $\omega t_2=\frac{2\pi}{3}$ So required time is $\Delta t=\frac{\frac{2\pi}{3}-\frac{\pi}{2}}{\omega}$ $=\frac{\pi/6}{1/3}$ $=\frac{\pi}{2}$ Given $\Delta t=\frac{\pi}{x}$ so x=2

all three resistors (R₁, 20 Ω, 15 Ω) are in parallel between the same two nodes so total current = sum of branch currents given: current through 20 Ω = 0.3 A total current = 0.9 A so current through the other two branches: I₁ + I₃ = 0.9 − 0.3 = 0.6 A step 1: find voltage across network V = I × R = 0.3 × 20 = 6 V step 2: current through 15 Ω I₃ = 6 / 15 = 0.4 A step 3: current through R₁ I₁ = 0.6 − 0.4 = 0.2 A step 4: find R₁ R₁ = V / I₁ = 6 / 0.2 = 30 Ω

A charge $q = 4.0\;\mu\text{C} = 4.0 \times 10^{-6}$ C moves with velocity $\vec{v} = 4.0 \times 10^6\;\hat{j}$ m/s \in a magnetic field $\vec{B} = 2\hat{k}$ T. The magnetic force on a moving charge is given by $\vec{F} = q(\vec{v} \times \vec{B}).$ To compute the cross product $\vec{v} \times \vec{B},$ observe that $\vec{v} \times \vec{B} = (4.0 \times 10^6\;\hat{j}) \times (2\hat{k}) = 4.0 \times 10^6 \times 2\;(\hat{j} \times \hat{k}).$ Using the rule $\hat{j} \times \hat{k} = \hat{i},$ it follows that $\vec{v} \times \vec{B} = 8.0 \times 10^6\;\hat{i}.$ Multiplying by the charge yields $\vec{F} = q(\vec{v} \times \vec{B}) = 4.0 \times 10^{-6} \times 8.0 \times 10^6\;\hat{i},$ so $\vec{F} = 32\;\hat{i}\text{ N}.$ The answer is $x = 32$.

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When A and B are shorted, they become the same node. step 1: identify final steady state In DC steady state, capacitors act as open circuits. So no current flows through capacitors, but they can hold charge. step 2: reduce the circuit Now A and B are same node. So: • top node = 9 V • bottom node = 0 V (ground) • A = B is some intermediate node Between top and A: 6 Ω Between A and bottom: 3 Ω So it becomes a simple voltage divider. step 3: find voltage at node A (= B) $V_A=\frac{3}{6+3}\times9=\frac{3}{9}\times9=3\text{ V}$ step 4: voltage across 6 \mu F capacitor Top plate is at 9 V, bottom plate at node B = 3 V $V_C=9-3=6\text{ V}$ step 5: charge $Q=CV=6\mu F\times6=36\mu C$

In a single slit diffraction pattern, the position of the $n$th minimum is given by $y_n = \frac{n\lambda D}{a}$, where the wavelength $\lambda = 6000$ Angstrom $= 6000 \times 10^{-10}$ m $= 6 \times 10^{-7}$ m, the distance from slit to screen $D = 50$ cm $= 0.50$ m, and $a$ is the slit width. The distance between the first and third minima is $y_3 - y_1 = 3$ mm $= 3 \times 10^{-3}$ m; substituting into the expression gives $y_3 - y_1 = \frac{3\lambda D}{a} - \frac{1 \cdot \lambda D}{a} = \frac{2\lambda D}{a}.$ Rearranging yields $a = \frac{2\lambda D}{y_3 - y_1},$ and thus $a = \frac{2 \times 6 \times 10^{-7} \times 0.50}{3 \times 10^{-3}} = \frac{6 \times 10^{-7}}{3 \times 10^{-3}} = 2 \times 10^{-4} \text{ m}.$ Expressed as $\_\_\_ \times 10^{-4}$ m, the numerical result is 2.

To observe Balmer series lines in the emission spectrum, the hydrogen atom must be excited to at least the $n = 3$ level, since Balmer series corresponds to transitions ending at $n = 2$ and the minimum transition is $n = 3 \rightarrow n = 2$. The energy of the $n$th level of hydrogen is given by $E_n = -\frac{13.6}{n^2} \text{ eV}$. Therefore, the energy required to excite the hydrogen atom from $n = 1$ to $n = 3$ is calculated as $\Delta E = E_3 - E_1 = -\frac{13.6}{9} - \left(-\frac{13.6}{1}\right) = 13.6\left(1 - \frac{1}{9}\right) = 13.6 \times \frac{8}{9} = \frac{108.8}{9} = 12.09 \text{ eV}.$ At $T = 0\text{ K}$, all hydrogen atoms are \in the ground state ($n = 1$). The electrons must be accelerated through a potential difference $V$ so that they gain at least $12.09\text{ eV}$ of kinetic energy. Since $eV = 12.09\text{ eV}$, it follows that $V = 12.09\text{ V} \approx 12.1\text{ V}.$ This potential difference is expressed as $\frac{\alpha}{10}\text{ V}$, so $\frac{\alpha}{10} = 12.1 \quad\Rightarrow\quad \alpha = 121.$ Hence, $\alpha = 121$.

The hydrogen spectral series are determined by the Rydberg formula transitions to a lower energy level $n_1$.

1. Lyman series ($n_1=1$) corresponds to the UV region (A-II).
2. Balmer series ($n_1=2$) corresponds to the visible region (B-IV).
3. Paschen series ($n_1=3$) corresponds to the infrared region (C-III).
4. Pfund series ($n_1=5$) corresponds to the infrared region (D-I).

Aligning these gives: A-II, B-IV, C-III, D-I.

We need to determine which element among Si, Al, N, and C has the highest first ionization enthalpy. The first ionization enthalpies (approximate values) are: Al (Z = 13): 577 kJ/mol Si (Z = 14): 786 kJ/mol C (Z = 6): 1086 kJ/mol N (Z = 7): 1402 kJ/mol General trend: Ionization enthalpy increases across a period from left to right. However, there are exceptions due to electronic configuration stability. Why Nitrogen has the highest IE among these elements: Nitrogen has the electronic configuration $1s^2\, 2s^2\, 2p^3$. The $2p$ subshell is exactly half-filled, with one electron \in each of the three $2p$ orbitals. This half-filled configuration provides extra stability due to: 1. Maximum exchange energy 2. Symmetrical distribution of electrons This extra stability makes it significantly harder to remove an electron from nitrogen compared to the other elements listed. The correct answer is Option 3: N (Nitrogen).

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1. Identify the principal functional group as the hydroxyl (-OH) group, which receives the lowest possible locant, assigned as C-1.
2. Number the ring to give the double bond the lowest possible locant. Moving clockwise, the double bond begins at position 2.
3. The parent ring is cyclohexane with a double bond, denoted as cyclohexene. With the -OH at C-1 and the double bond starting at C-2, the structure is named cyclohex-2-en-1-ol.
4. Counter-clockwise numbering would assign the double bond to C-5, which violates the lowest locant rule for the alkene.
5. Thus, the IUPAC name is Cyclohex-2-en-1-ol.

The acidity of a hydrogen atom is determined by the stability of the conjugate base formed after its removal as a proton.

• Removal of H_A gives an alkoxide ion, stabilized by resonance with the adjacent carbonyl group. The negative charge is on the highly electronegative oxygen.
• Removal of H_B gives a carbanion (enolate), also stabilized by resonance.
• Removal of H_D gives a carbanion on an sp² hybridized carbon.
• Removal of H_C gives a carbanion on an sp³ hybridized carbon, which is the least stable.
  The order of stability of conjugate bases is A > B > D > C. Therefore, the ascending order of acidity is C < D < B < A.

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