JEE Main 2025 Jan 24 shift 1

Transition A→C corresponds to $\lambda_1 = 2000$ Å and B\to C to $\lambda_2 = 6000$ Å. We denote the wavelength for A\to B by $\lambda_3$. Since the energy of A\to C equals the \sum of energies of A\to B and B\to C, $E_{A\to C} = E_{A\to B} + E_{B\to C}$. Expressing each energy as $hc/\lambda$ gives $\frac{hc}{\lambda_1} = \frac{hc}{\lambda_3} + \frac{hc}{\lambda_2}$. Substituting the given wavelengths yields $\frac{1}{\lambda_3} = \frac{1}{\lambda_1} - \frac{1}{\lambda_2} = \frac{1}{2000} - \frac{1}{6000} = \frac{3-1}{6000} = \frac{1}{3000}$. This gives $\lambda_3 = 3000$ Å. Therefore the correct answer is $3000$ Å .

Let us analyze each statement: A. The junction area of solar cell is made very narrow compared to a photodiode. This is false. In a solar cell, the junction area is made very large (not narrow) to capture maximum sunlight. It is the photodiode that typically has a smaller junction area. B. Solar cells are not connected with any external bias. This is true. Solar cells operate on the photovoltaic effect and generate their own EMF. They work without any external bias - they are essentially self-biased by the sunlight. C. LED is made of lightly doped p-n junction. This is false. LEDs are made of heavily doped p-n junctions to ensure high recombination rates and efficient light emission. D. Increase of forward current results in continuous increase of LED light intensity. This is false. Initially, increasing forward current increases light intensity, but beyond a certain point, the LED may reach saturation or get damaged. The increase is not continuous - there is a maximum rated current. E. LEDs have to be connected in forward bias for emission of light. This is true. LEDs emit light when electrons and holes recombine at the junction, which occurs only under forward bias. The correct statements are B and E only. The answer is Option A: B, E Only.

The given current, $I=I_A\sin\omega t + I_B\cos\omega t$, represents a sinusoidal alternating current. We can express this in the standard form $I=I_0\sin(\omega t + \phi)$, where $I_0$ is the peak amplitude of the current. The peak amplitude is found by combining the amplitudes of the sine and cosine components as vectors, resulting in $I_0 = \sqrt{I_A^{2+}I_B^2}$.

The r.m.s. value of any sinusoidal current is related to its peak value by the formula $I_{rms} = \frac{I_0}{\sqrt{2}}$.

Substituting the expression for the peak current $I_0$ into this formula gives:
$I_{rms} = \frac{\sqrt{I_A^{2+}I_B^2}}{\sqrt{2}}$

Bringing the denominator inside the square root, we get the final expression:
$I_{rms} = \sqrt{\frac{I_A^{2+}I_B^2}{2}}$

consider forces on the car on a banked road forces acting: weight = mg (vertical downward) normal reaction = N (perpendicular to surface) friction = f (along surface) for maximum speed , the car tends to slip up the incline , so friction acts down the slope resolve forces: along horizontal (towards center → provides centripetal force) N sinθ + f cosθ = m v₀² / r along vertical (no vertical acceleration) N cosθ − f sinθ = mg at limiting condition: f = μN substitute f = μN horizontal: N sinθ + μN cosθ = m v₀² / r N (sinθ + μ cosθ) = m v₀² / r ...(1) vertical: N cosθ − μN sinθ = mg N (cosθ − μ sinθ) = mg ...(2) divide (1) by (2): (sinθ + μ cosθ) / (cosθ − μ sinθ) = (v₀² / rg) cross multiply: sinθ + μ cosθ = (v₀² / rg)(cosθ − μ sinθ) expand: sinθ + μ cosθ = (v₀² / rg)cosθ − (v₀² / rg)μ sinθ collect μ terms: μ cosθ + (v₀² / rg)μ sinθ = (v₀² / rg)cosθ − sinθ factor μ: μ [cosθ + (v₀² / rg) sinθ] = (v₀² / rg)cosθ − sinθ final expression: μ = [ (v₀² / rg)cosθ − sinθ ] / [ cosθ + (v₀² / rg) sinθ ] $\mu=\frac{v_0^{2-}rg\tan\theta}{rg+v_0^2\tan\theta}$ this is the required coefficient of friction for maximum speed.

$T^2 \propto R^3 \implies T \propto R^{3/2}$ (Kepler's Third Law) Since the increase \in radius is small ($R$ to $1.03R$, a $3\%$ change), we can use the approximation method for small percentage changes. The percentage change \in the radius is: $\frac{\Delta R}{R} \times 100 = \frac{1.03R - R}{R} \times 100 = 3\%$ $\text{Percentage change \in } T \approx \frac{3}{2} \times (\text{Percentage change \in } R)$ $\text{Percentage change \in } T \approx \frac{3}{2} \times 3\% = 4.5\%$

Use ideal gas law: $PV=nRT$ Given pressure increases linearly with temperature, $P\propto T$ or $\frac{P}{T}=\text{constant}$ From ideal gas law, $\frac{P}{T}=\frac{nR}{V}$ Since P/T is constant, $V=\text{constant}$ So process is isochoric. Hence statement E is true. For isochoric process, $W=\int PdV=0$ So A is true. First law: $Q=\Delta U+W$ Since W=0 $Q=\Delta U$ So B ("heat added is different from change in internal energy") is false. Volume does not increase (constant), so C is false. Pressure increases linearly with temperature, so temperature increases. For ideal gas internal energy depends only on temperature, so internal energy increases. D is true. Correct statements: A, D, E

Electron with $\vec{v}=v_0\hat{i}$ enters field $\vec{E}=-E_0\hat{k}$. Find de Broglie wavelength after time $t$. The force on the electron is $\vec{F}=-e\vec{E}=eE_0\hat{k}$. Integrating acceleration gives $v_z=\frac{eE_0}{m}t$ while $v_x=v_0$ remains constant. Hence the speed is $v=\sqrt{v_0^{2+}\frac{e^2E_0^2t^2}{m^2}}$. The de Broglie wavelength is $\lambda=\frac{h}{mv}=\frac{h}{m\sqrt{v_0^{2+}e^2E_0^2t^2/m^2}}$. This can be written as $\lambda=\frac{h}{mv_0\sqrt{1+e^2E_0^2t^2/(m^2v_0^2)}}$. Therefore $\lambda=\frac{\lambda_0}{\sqrt{1+\frac{e^2E_0^2t^2}{m^2v_0^2}}}$.

Initial power is $P_1 = 2.5$ D, which gives $f_1 = 1/2.5 = 0.4$ m. When the power increases to $P_2 = 2.6$ D, the focal length becomes $f_2 = 1/2.6$ m. The relative decrease is $\frac{f_1-f_2}{f_1} = 1 - \frac{f_2}{f_1} = 1 - \frac{P_1}{P_2} = 1 - \frac{2.5}{2.6} = \frac{0.1}{2.6} \approx 0.0385 \approx 0.04$. Therefore the correct answer is Option 2: 0.04.

Given the force $F = \alpha + \beta x^2$, the work done is 5 J over a displacement of 1 m and \alpha = 1 N, so we need to find \beta . Since the work is given by the integral of the force from 0 to 1, we have $W = \int_0^1(\alpha+\beta x^2)dx = \$[\alpha x + \frac{\beta x^3}{3}]_0^1 = \alpha + \frac{\beta}{3}$. Substituting the values gives $5 = 1 + \frac{\beta}{3}$, and therefore solving for \beta yields $\beta = 12$ N/m². Thus the correct answer is Option 2: 12 N/m².

The combination of a silvered lens acts as an equivalent mirror. The power of this equivalent mirror, $P_{eq}$, is given by the sum of the powers of the components, accounting for the path of light:
$P_{eq} = 2P_{lens} + P_{mirror}$
The system behaves as a concave mirror of focal length $f = 0.2$ m. The power of a mirror is $P = 1/f$. Thus, $P_{eq} = \frac{1}{0.2} = 5 \text{ D}$.

The power of the plano-convex lens immersed in the liquid is found using the Lens Maker's formula, where $n_g=1.5$ is the refractive index of glass and $n_l=1.2$ is the refractive index of the liquid. For a plano-convex lens with radius of curvature $R$:
$P_{lens} = \left(\frac{n_g}{n_l} - 1\right)\left(\frac{1}{R} - \frac{1}{\infty }\right) = \left(\frac{1.5}{1.2} - 1\right)\frac{1}{R} = \left(\frac{0.3}{1.2}\right)\frac{1}{R} = \frac{1}{4R}$
The silvered side is plane, so the mirror is a plane mirror. A plane mirror has zero power, so $P_{mirror} = 0$.

Substituting these into the equivalent power equation:
$P_{eq} = 2 \left(\frac{1}{4R}\right) + 0 = \frac{1}{2R}$
Since we know $P_{eq} = 5 \text{ D}$:
$5 = \frac{1}{2R} \implies 10R = 1 \implies R = 0.10 \text{ m}$

A particle executes SHM with time period $T = 2$ s and amplitude $A = 1$ cm. We need the ratio $D/d$ after 12.5 s, where $D$ = total distance and $d$ = displacement. Since $12.5 \text{ s} = 6 \times 2 + 0.5 = 6T + T/4$, the particle completes 6 full periods and an additional $T/4$. In each complete period the particle traverses a distance of $4A$, so \in 6 complete periods the distance is $6 \times 4A = 24A$. In the additional $T/4$ the particle moves from the mean position to the extreme, covering a distance $A$. Therefore $D = 24A + A = 25A = 25 \text{ cm}$. After 6 complete periods the displacement is zero, and during the additional $T/4$ the particle reaches the extreme, so $d = A = 1 \text{ cm}$. Therefore $\frac{D}{d} = \frac{25}{1} = 25$, which corresponds to Option (4): 25.

Work to break a drop of radius R into 27 drops is 10 J and we need the work to break it into 64 drops. Since the work is given by $W = T \times \Delta A = T(n \times 4\pi r^2 - 4\pi R^2),$ and volume conservation implies $n \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \Rightarrow r = R/n^{1/3},$ substituting \in yields $W = 4\pi TR^2(n^{1/3}-1).$ For 27 drops $W_1 = 4\pi TR^2(27^{1/3}-1) = 4\pi TR^2(3-1) = 8\pi TR^2 = 10$ J For 64 drops $W_2 = 4\pi TR^2(64^{1/3}-1) = 4\pi TR^2(4-1) = 12\pi TR^2$ This gives $\frac{W_2}{W_1} = \frac{12\pi TR^2}{8\pi TR^2} = \frac{3}{2},$ therefore $W_2 = \frac{3}{2} \times 10 = 15$ J

We use the Lens Maker's formula: $\frac{1}{f} = (\frac{\mu_{lens}}{\mu_{medium}} - 1)(\frac{1}{R_1} - \frac{1}{R_2})$. For a plano-convex lens, one radius is infinity ($R_2=\infty$), so the formula simplifies.

For the first lens in air ($\mu_m = 1$), with $R_1=2 \text{ cm}$ and $\mu_g=1.5$:
$\frac{1}{f_1} = (\frac{1.5}{1} - 1)(\frac{1}{2} - 0) = (0.5)(\frac{1}{2}) = \frac{1}{4}$
For the second lens in the liquid ($\mu_m = 1.2$), with $R'_1=3 \text{ cm}$ and $\mu_g=1.5$:
$\frac{1}{f_2} = (\frac{1.5}{1.2} - 1)(\frac{1}{3} - 0) = (\frac{1.5 - 1.2}{1.2})(\frac{1}{3}) = (\frac{0.3}{1.2})(\frac{1}{3}) = \frac{1}{4} \times \frac{1}{3} = \frac{1}{12}$
To find the ratio $f_1:f_2$, we can take the ratio of the reciprocals of focal lengths:
$\frac{f_1}{f_2} = \frac{1/f_2}{1/f_1} = \frac{1/12}{1/4} = \frac{4}{12} = \frac{1}{3}$
Thus, the ratio $f_1 : f_2$ is $1 : 3$.

An air bubble of radius $r = 0.1$ cm = $0.001$ m at depth $h = 20$ cm = $0.20$ m has excess pressure of 2100 N/m² above atmospheric. We need the surface tension. The pressure inside the bubble exceeds atmospheric pressure by both the hydrostatic pressure and the surface tension pressure: $P_{inside} - P_{atm} = \rho g h + \frac{2T}{r}$ Note: For a bubble \in a liquid (not a soap bubble), there is only one surface, so the excess pressure due to surface tension is $\frac{2T}{r}$. $2100 = 1000 \times 10 \times 0.20 + \frac{2T}{0.001}$ $2100 = 2000 + 2000T$ $100 = 2000T$ $T = \frac{100}{2000} = 0.05 \text{ N/m}$ The correct answer is Option (2): 0.05 .

$For\ an\ object\ rolling\ without\ slipping\ down\ an\ inclined\ plane\ with\ inclination\ \theta,\$ $the\ linear\ acceleration\ (a)\ of\ the\ center\ of\ mass\ is:\ a=\frac{g\sin\theta}{1+\frac{I}{mr^2}}$ $For\ a\ uniform\ solid\ cylinder,\ the\ moment\ of\ inertia\ is\ I=\frac{1}{2}mr^2$ $Substituting\ this\ and\ \theta=45^{\circ}\ into\ the\ formula:$ $a = \frac{g \sin 45^\circ}{1 + \frac{1}{2}mr^2 / mr^2}$ $a = \frac{g \cdot \frac{1}{\sqrt{2}}}{1 + \frac{1}{2}}$ $a = \frac{2g}{3\sqrt{2}}$ $a = \frac{\sqrt{2}}{3} g$

In Young's double slit experiment, the position of the $n^{th}$ bright fringe is given by $y_n = \frac{n \lambda D}{d}$, where $\lambda$ is the wavelength, $D$ is the distance to the screen, and $d$ is the slit separation. To determine the least number of bright fringes of 480 nm light that will coincide with those of 600 nm light, one sets the positions of the $n_1^{th}$ bright fringe of $\lambda_1 = 480$ nm and the $n_2^{th}$ bright fringe of $\lambda_2 = 600$ nm equal. Equating their expressions, $\frac{n_1 \lambda_1 D}{d} = \frac{n_2 \lambda_2 D}{d}$, simplifies to $n_1 \lambda_1 = n_2 \lambda_2$, or $n_1 \times 480 = n_2 \times 600$. This equation can be written as $\frac{n_1}{n_2} = \frac{600}{480} = \frac{5}{4}$, meaning the smallest integer values satisfying the ratio are $n_1 = 5$ and $n_2 = 4$. Verification shows $5 \times 480 = 2400$ nm and $4 \times 600 = 2400$ nm, confirming that the 5th bright fringe of 480 nm light coincides with the 4th bright fringe of 600 nm light. Therefore, the least number of bright fringes of 480 nm light required for the first coincidence is 5. The correct answer is Option 1: 5.

The original capacitor has $l = 3$ cm, $b = 1$ cm, $d = 3$ $\mu$m. Capacitance formula: $C = \frac{\epsilon_0 A}{d} = \frac{\epsilon_0 \cdot l \cdot b}{d}$. Original: $C_0 = \frac{\epsilon_0 \cdot 3 \cdot 1}{3} = \epsilon_0$ (\in relative units). We need $C = 10C_0$, i.e., $\frac{l \cdot b}{d} = 10 \times \frac{3 \cdot 1}{3} = 10$. A: $l = 30, b = 1, d = 1$: $\frac{30 \times 1}{1} = 30 \neq 10$. No. B: $l = 3, b = 1, d = 30$: $\frac{3 \times 1}{30} = 0.1 \neq 10$. No. C: $l = 6, b = 5, d = 3$: $\frac{6 \times 5}{3} = 10$. Yes! D: $l = 1, b = 1, d = 10$: $\frac{1 \times 1}{10} = 0.1 \neq 10$. No. E: $l = 5, b = 2, d = 1$: $\frac{5 \times 2}{1} = 10$. Yes! Options C and E give a capacitance 10 times the original. The correct answer is Option D: C and E only.

To express the potential energy stored in a parallel plate capacitor in terms of the electric field $E$, plate area $A$, separation $d$, and permittivity $\varepsilon_0$, recall that the general formula for the energy is$U = \frac{1}{2}CV^2$where $C$ is the capacitance and $V$ the potential difference across the plates. For a parallel plate capacitor, the capacitance is given by$C = \frac{\varepsilon_0 A}{d}$and the uniform electric field relates to the potential difference via$V = Ed\,.$Substituting these expressions into the energy formula leads to$U = \frac{1}{2}\,\frac{\varepsilon_0 A}{d}\,(Ed)^2 = \frac{1}{2}\varepsilon_0 E^2 A d\,.$ An alternative derivation begins with the energy density of an electric field,$u = \frac{1}{2}\varepsilon_0 E^2\,,$and notes that the volume between the plates is $A\,d$. Multiplying gives$U = u\times (A d) = \frac{1}{2}\varepsilon_0 E^2 A d\,,$which agrees with the previous result. Thus, the energy stored \in the parallel plate capacitor can be written as$\frac{1}{2}\varepsilon_0 E^2 A d\,.$

An object of mass $m$ is projected from the origin \in the vertical xy-plane at an angle of $45°$ with initial speed $v_0$. To find its angular momentum about the origin at its maximum height, we first determine the time and position at that instant. The time to reach the maximum height occurs when the vertical component of velocity becomes zero, namely at $t = \frac{v_0 \sin 45°}{g} = \frac{v_0}{\sqrt{2}g}\,.$ During this interval the horizontal component of motion remains uniform, so the x-coordinate is $x = v_0 \cos 45° \times t = \frac{v_0}{\sqrt{2}} \times \frac{v_0}{\sqrt{2}g} = \frac{v_0^2}{2g}\,.$ The maximum height itself is given by $y = H = \frac{v_0^2 \sin^2 45°}{2g} = \frac{v_0^2}{4g}\,.$ Hence the position vector at maximum height is $\vec{r} = \frac{v_0^2}{2g}\,\hat{i} + \frac{v_0^2}{4g}\,\hat{j}\,.$ At this instant the vertical velocity vanishes and only the horizontal component remains, so $\vec{v} = v_0 \cos 45°\,\hat{i} = \frac{v_0}{\sqrt{2}}\,\hat{i}\,.$ The angular momentum about the origin is given by $\vec{L} = m\,\bigl(\vec{r}\times \vec{v}\bigr)\,.$ Evaluating the cross product, $\vec{r}\times \vec{v} = \Bigl(\frac{v_0^2}{2g}\hat{i} + \frac{v_0^2}{4g}\hat{j}\Bigr) \times \Bigl(\frac{v_0}{\sqrt{2}}\hat{i}\Bigr)\,.$ Since $\hat{i}\times \hat{i}=0$ and $\hat{j}\times \hat{i}=-\hat{k}$, it follows that $\vec{r}\times \vec{v} = \frac{v_0^2}{2g}\frac{v_0}{\sqrt{2}}\bigl(\hat{i}\times \hat{i}\bigr) + \frac{v_0^2}{4g}\frac{v_0}{\sqrt{2}}\bigl(\hat{j}\times \hat{i}\bigr) = -\frac{v_0^3}{4\sqrt{2}g}\,\hat{k}\,.$ Therefore $\vec{L} = m\bigl(\vec{r}\times \vec{v}\bigr) = -\frac{m\,v_0^3}{4\sqrt{2}g}\,\hat{k}\,.$ The magnitude of the angular momentum is $\frac{m\,v_0^3}{4\sqrt{2}g}$ directed along the negative z-axis. The correct answer is Option 3: $\frac{m\,v_0^3}{4\sqrt{2}g}$ along the negative z-axis.

We need to compute $y = \frac{32.3 \times 1125}{27.4}$ and report it with the correct number of significant figures. First, let us count the significant figures \in each quantity: $32.3$ has 3 significant figures. $1125$ has 4 significant figures. $27.4$ has 3 significant figures. In multiplication and division, the final result must have the same number of significant figures as the quantity with the fewest significant figures. Here, the minimum is 3 significant figures. Now let us compute the value step by step. $32.3 \times 1125 = 36337.5$ $y = \frac{36337.5}{27.4} = 1326.186...$ We must round this to 3 significant figures. The first three significant digits of $1326.186$ are $1$, $3$, and $2$. The next digit is $6$, which is $\ge 5$, so we round up. $1326.186... \approx 1330$ Hence, the correct answer is Option B.

The magnetic field at the centre of a square loop carrying current can be found by applying the Biot-Savart law to one side and then using symmetry. Given that I = 5 A, a = 1/√2 m and μ₀ = 4π×10⁻⁷ T m A⁻¹, consider a single straight segment of the loop. For a finite straight wire the field at a perpendicular distance d, subtending angles θ₁ and θ₂ at the observation point, is $B_{\text{wire}} = \frac{\mu_0 I}{4\pi d}(\sin\theta_1 + \sin\theta_2).$ In our square loop the distance from the centre to each side is d = a/2, and each side subtends equal angles \theta ₁ = \theta ₂ = 45° at the centre. Substituting these values gives $B_1 = \frac{\mu_0 I}{4\pi\,(a/2)}\bigl(\sin45°+\sin45°\bigr) = \frac{\mu_0 I}{2\pi a}\times \frac{2}{\sqrt2} = \frac{\sqrt2\,\mu_0 I}{2\pi a}.$ By symmetry all four sides contribute equally and their fields add \in the same direction, so the total field is $B = 4B_1 = 4\;\frac{\sqrt2\,\mu_0 I}{2\pi a} = \frac{2\sqrt2\,\mu_0 I}{\pi a}.$ Substituting I = 5 A, a = 1/√2 m and \mu ₀ = 4\pi \times 10⁻⁷ T m A⁻¹ yields $B = \frac{2\sqrt2\times4\pi\times10^{-7}\times5}{\pi\times (1/\sqrt2)}.$ The \pi factors cancel, giving $B = \frac{2\sqrt2\times4\times5\times10^{-7}}{1/\sqrt2} = 40\sqrt2\times10^{-7}\times \sqrt2 = 40\times2\times10^{-7} = 8\times10^{-6}\text{ T}.$ Therefore, the magnetic field at the centre of the square loop is $B = 8\times10^{-6}\text{ T},$ and the required value of p is 8.

Use solid angle idea. Flux through any surface due to point charge is $\Phi=\frac{q}{4\pi\varepsilon_0}\Omega$ where \Omega \Omega\Omega is solid angle subtended by the surface at the charge. The charge is located on the normal through the center of the square, and from the geometry shown, the whole square subtends a solid angle corresponding to a face of a cube with charge at cube center. Flux through entire square is therefore $\Phi_{\text{square}}=\frac{q}{6\varepsilon_0}$ since total flux $\frac{q}{\varepsilon_0}$ gets equally shared among 6 cube faces. Given q=1C so $\Phi_{\text{square}}=\frac{1}{6\varepsilon_0}$ Now look at shaded region. its area is $\frac{5a^2}{8}$ So shaded fraction of the square is $\frac{5}{8}$ Flux through the whole square (one face of the imaginary cube) is $\frac{q}{6\varepsilon_0}$ Hence flux through shaded region is $\Phi=\frac{5}{8}\cdot\frac{q}{6\varepsilon_0}$ With q=1 $\Phi=\frac{5}{48\varepsilon_0}$ Comparing with $\frac{5}{p}\cdot\frac{1}{\varepsilon_0}$ we get $p=48$

For 1 mole of an ideal monoatomic gas with temperature increase $\Delta T = 50°C$ at constant pressure: Heat added at constant pressure: $E_1 = nC_p\Delta T$. Change \in internal energy: $E_2 = nC_v\Delta T$. For a monoatomic ideal gas: $C_p = \frac{5}{2}R$ and $C_v = \frac{3}{2}R$. $\frac{E_1}{E_2} = \frac{C_p}{C_v} = \frac{5/2}{3/2} = \frac{5}{3}$. Given $\frac{E_1}{E_2} = \frac{x}{9}$: $\frac{x}{9} = \frac{5}{3}$ $x = 9 \times \frac{5}{3} = 15$. The answer is 15.

Screw gauge: LC = 0.01 mm. Pitch increased by 75%, divisions reduced by 50%. Find new LC. LC = pitch/divisions. If pitch becomes 1.75p and divisions become 0.5d: New LC = 1.75p/(0.5d) = 3.5(p/d) = 3.5 × 0.01 = 0.035 mm In units of 0.001 mm: 35 The answer is 35.

Wire of 9Ω bent into equilateral triangle. Equivalent resistance across two vertices. Each side = 3Ω. Across two vertices: one 3Ω in parallel with two 3Ω in series (6Ω). $R_{eq} = \frac{3 \times 6}{3+6} = 2$ Ω The answer is 2.

The sugar unit present in DNA is $\mathbf{2\text{-}deoxyribose}$. Step 1: Count of carbon atoms. 2-Deoxyribose contains five carbon atoms, so it is a pentose. \Rightarrow Statement A is correct. Step 2: Ring size \in nucleic acids. Inside DNA the sugar occurs as a five-membered $\mathbf{furanose}$ ring, not a six-membered pyranose ring. \Rightarrow Statement B is incorrect. Step 3: Absolute (D/L) configuration. Naturally occurring ribose and deoxyribose possess the configuration related to $\mathbf{D\text{-}glyceraldehyde}$. \Rightarrow Statement C is correct. Step 4: Reducing property when free. A sugar is reducing if it has a free hemiacetal (aldehyde) group. Free 2-deoxyribose has such a group, so it reduces Tollens' and Fehling's reagents. \Rightarrow Statement D is correct. Step 5: Anomeric configuration \in DNA. The base is attached to the anomeric carbon through a $\mathbf{\beta}$-N-glycosidic bond. Hence the sugar \in DNA is the $\beta$-anomer, not the $\alpha$-anomer. ⇒ Statement E is incorrect. Collecting the correct statements: A, C and D. Therefore the correct option is Option B .

This question asks about the effect of solvent polarity on a nucleophilic substitution ($S_N2$) reaction.
$\mathrm{CH_3-CH_2-CH_2-CH_2-Cl} + \mathrm{HO^{(-)}} \rightarrow \mathrm{CH_3-CH_2-CH_2-CH_2-OH} + \mathrm{Cl^{(-)}}$
Let's analyze the role of the solvent in this reaction.

1. Analysis of Statement I: "The conversion proceeds well in the less polar medium."
   This statement is true when comparing a highly polar protic solvent (like water) to a less polar protic solvent (like ethanol). In water, the small, charged nucleophile ($HO^−$) is heavily stabilized by hydrogen bonding (solvation). This high stability of the reactant increases the activation energy, slowing the reaction. In a less polar medium like ethanol, solvation is weaker, making the nucleophile more reactive and accelerating the reaction.

2. Analysis of Statement II: "The conversion proceeds well in the more polar medium."
   This statement is also true when comparing a nonpolar solvent (like hexane) to a polar solvent (like acetone or DMSO). The ionic reactant ($HO^−$, typically from NaOH or KOH) is insoluble in nonpolar solvents, so the reaction cannot proceed. A polar solvent is necessary to dissolve the nucleophile and to stabilize the charge separation in the transition state and the leaving group ($Cl^−$). Thus, compared to a nonpolar medium, the reaction proceeds well in a more polar medium.

Since both statements can be considered correct depending on the reference point of comparison, and valid chemical principles support both, we conclude that both statements are true.

The reaction proceeds in three sequential steps.

Step 1: Propyne undergoes acid-catalyzed hydration with $Hg^{2+}/H_2SO_4$ (Kucherov's reaction). This hydration follows Markovnikov's rule, yielding an unstable enol that tautomerizes to a ketone, acetone.
$CH_3-C \equiv CH \xrightarrow{Hg^{2+}, H_2SO_4, H_2O} CH_3-C(O)-CH_3$

Step 2: Acetone undergoes nucleophilic addition with hydrogen cyanide (HCN) to form a cyanohydrin.
$CH_3-C(O)-CH_3 \xrightarrow{HCN} (CH_3)_2C(OH)CN$

Step 3: The nitrile group ($-CN$) of the cyanohydrin is reduced to a primary amine ($-CH_2NH_2$) by catalytic hydrogenation using $H_2/Ni$. The tertiary alcohol group remains unchanged.
$(CH_3)_2C(OH)CN \xrightarrow{H_2/Ni} (CH_3)_2C(OH)CH_2NH_2$

The final product (A) is 1-amino-2-methylpropan-2-ol, which corresponds to the structure shown in option B.

To synthesize the target molecule (x), 1-acetylcyclopentene, via an intramolecular aldol condensation, we must first identify the necessary dicarbonyl precursor. The formation of a 5-membered ring through aldol cyclization requires a 1,6-dicarbonyl compound. The structure of (x) indicates that the specific precursor needed is 6-oxoheptanal.

$\underset{\text{6-oxoheptanal}}{CH_3-CO-(CH_2)_4-CHO} \xrightarrow{\text{Intramolecular Aldol}} \underset{\text{1-acetylcyclopentene (x)}}{ \text{(Image of x)}} + H_2O$

The next step is to determine which starting alkene would yield 6-oxoheptanal upon ozonolysis. Ozonolysis cleaves a carbon-carbon double bond and replaces it with two carbonyl groups. Therefore, by reversing this process (a retro-ozonolysis), we can find the required alkene by joining the two carbonyl carbons of 6-oxoheptanal with a double bond.

Connecting the ketone carbon (C6) and the aldehyde carbon (C1) of 6-oxoheptanal forms a six-membered ring with a double bond. The carbon that was part of the ketone has a methyl group attached, resulting in 1-methylcyclohexene.

$\underset{\text{1-methylcyclohexene (C)}}{ \text{(Image of C)}} \xrightarrow{1. O_3, 2. Zn/H_2O} \underset{\text{6-oxoheptanal}}{CH_3-CO-(CH_2)_4-CHO}$

Therefore, 1-methylcyclohexene (Option C) is the correct starting material.

Nucleophilic addition to a carbonyl group occurs at the electrophilic carbon of the $C = O$ bond. The rate depends on two main factors: \bullet Electronic effects: groups that withdraw electrons ( $-I$ and/or $-R$) increase the partial positive charge on the carbonyl carbon and enhance reactivity. Groups that donate electrons ( $+I$ and/or $+R$) decrease the charge and slow the reaction. \bullet Steric effects: less‐hindered carbonyls are more accessible to the nucleophile. First compare aldehydes and ketones attached to an aromatic ring. \bullet Aldehydes have only one alkyl/aryl group attached to the carbonyl carbon, so they are less sterically hindered and receive less $+I$ donation than ketones. Hence aromatic aldehydes are more reactive than aromatic ketones. \bullet Therefore $\text{acetophenone (Ar-CO-CH}_3)$ is expected to be the least reactive of the four compounds. Now compare the three aromatic aldehydes: 1. $\text{p-nitrobenzaldehyde}$ contains a $NO_2$ group. $NO_2$ is strongly electron-withdrawing by both $-I$ (inductive) and $-R$ (resonance) effects, so it makes the carbonyl carbon highly electrophilic. This gives the HIGHEST reactivity. 2. $\text{benzaldehyde}$ has no extra substituent on the ring, so its reactivity is taken as the reference among aldehydes. 3. $\text{p-tolualdehyde}$ (para-methylbenzaldehyde) contains a $CH_3$ group. $CH_3$ is electron-donating by $+I$ and weakly $+R$, so it reduces the electrophilicity of the carbonyl carbon, lowering the reactivity below that of unsubstituted benzaldehyde. Putting the steric and electronic arguments together: $\text{Least reactive} \;\; \longrightarrow \;\; \text{acetophenone} \lt \text{p-tolualdehyde} \lt \text{benzaldehyde} \lt \text{p-nitrobenzaldehyde} \;\; \longrightarrow \;\; \text{Most reactive}$ This sequence matches Option D. Answer: Option D $\left(\text{acetophenone} \lt \text{p-tolualdehyde} \lt \text{benzaldehyde} \lt \text{p-nitrobenzaldehyde}\right)$