We need to find the wavelength of light in a medium with refractive index $\frac{3}{2}$, given its wavelength in water. The wavelength in water is $\lambda_w = 540$ nm, the refractive index of water is $n_w = \frac{4}{3}$, and the refractive index of the new medium is $n_m = \frac{3}{2}$.

We begin by finding the wavelength in vacuum. The relationship between wavelength in vacuum ($\lambda_0$) and wavelength in a medium is:

$\lambda_{medium} = \frac{\lambda_0}{n}$

This follows because the speed of light in a medium is $v = c/n$, and since the frequency remains constant when light passes between media ($v = f\lambda$), we have $\lambda_{medium} = \lambda_0 / n$. From the wavelength in water it follows that

$\lambda_0 = n_w \times \lambda_w = \frac{4}{3} \times 540 = 720 \text{ nm}$

Next, the wavelength in the new medium is

$\lambda_m = \frac{\lambda_0}{n_m} = \frac{720}{3/2} = 720 \times \frac{2}{3} = 480 \text{ nm}$

Alternatively, using a direct ratio of wavelengths in different media:

$\frac{\lambda_m}{\lambda_w} = \frac{n_w}{n_m} \implies \lambda_m = \lambda_w \times \frac{n_w}{n_m} = 540 \times \frac{4/3}{3/2} = 540 \times \frac{8}{9} = 480 \text{ nm}$

The correct answer is Option 3: 480 nm.

We need to evaluate two statements about work done by conservative forces.

An object moves from position $r_1$ to position $r_2$ under a conservative force field $\vec{F}$. The work done by the force is $W = \int_{r_1}^{r_2} \vec{F} \cdot d\vec{r}$.

This is true. The work done by any force (conservative or non-conservative) on an object moving from position $r_1$ to $r_2$ is defined as the line integral:

$W = \int_{r_1}^{r_2} \vec{F} \cdot d\vec{r}$

This is the fundamental definition of work in mechanics.

Any object moving from one location to another can follow an infinite number of paths. Therefore, the amount of work done by the object changes with the path it follows for a conservative force.

This statement is false. While it is true that there are infinitely many paths between two points, the defining property of a conservative force is that the work done is path-independent. The work depends only on the initial and final positions, not on the path taken. Mathematically:

$W = -\Delta U = -(U(r_2) - U(r_1)) = U(r_1) - U(r_2)$

where $U$ is the potential energy function. Since the work depends only on the potential energy values at the two endpoints, it is the same for all paths. This is precisely what distinguishes conservative forces (like gravity, electrostatic force, spring force) from non-conservative forces (like friction).

The correct answer is Option 2: Statement I is true but Statement II is false.

Open organ pipe: $\nu_n = n\frac{v}{2L}$. $\nu_6 - \nu_3 = (6-3)\frac{v}{2L} = \frac{3v}{2L} = 2200$.

$L = \frac{3 \times 330}{2 \times 2200} = \frac{990}{4400} = 0.225$ m = 225 mm.

The answer is Option 4: 225 mm.

We need to find the percentage change in capillary rise height when the inner radius, density, and surface tension each decrease by 1%.

The height of liquid rise in a capillary tube is given by $h = \frac{2T\cos\theta}{\rho g r}$, where $T$ is the surface tension, $\theta$ is the contact angle, $\rho$ is the liquid density, $g$ is acceleration due to gravity, and $r$ is the inner radius of the tube.

Taking logarithms and differentiating (with $\theta$ and $g$ held constant) gives $\ln h = \ln(2\cos\theta) + \ln T - \ln \rho - \ln g - \ln r$ and hence $\frac{\Delta h}{h} = \frac{\Delta T}{T} - \frac{\Delta \rho}{\rho} - \frac{\Delta r}{r}.$

Since each of $T$, $\rho$, and $r$ decreases by 1%, we have $\frac{\Delta T}{T} = -1\%$, $\frac{\Delta \rho}{\rho} = -1\%$, and $\frac{\Delta r}{r} = -1\%$, so that $\frac{\Delta h}{h} = (-1\%) - (-1\%) - (-1\%) = -1\% + 1\% + 1\% = +1\%.$

Decreasing the surface tension ($T$) by 1% reduces $h$ by 1% (since $h \propto T$), while decreasing the density ($\rho$) or the radius ($r$) by 1% increases $h$ by 1% in each case (since $h \propto 1/\rho$ and $h \propto 1/r$). The net effect is $-1 + 1 + 1 = +1\%$ increase in the capillary rise height.

The correct answer is Option 3: +1.

Collision frequency

$z=\frac{\overline{v}}{\lambda}$

where

$λ=\frac{1}{\sqrt{\ 2}n\pi\ d^2}$

So

$z=\overline{v}(\sqrt{2}n\pi d^2)$

Since both gases have same number density,

$z\propto\overline{v}d^2$

Also mean molecular speed for ideal gas is

$\overline{v}\propto\frac{1}{\sqrt{m}}$

So

$z\propto\frac{d^2}{\sqrt{m}}$

Now compare A and B:

Given molecular diameter of A is half of B,

$d_A=\frac{d_B}{2}$

and mass of molecule A is four times B,

$m_A=4m_B$

Hence

$\frac{z_A}{z_B}=\frac{d_A^2/\sqrt{m_A}}{d_B^2/\sqrt{m_B}}$

Substitute:

$=\frac{\left(\frac{d_B}{2}\right)^2}{d_B^2}\cdot\frac{\sqrt{m_B}}{\sqrt{4m_B}}$

$=\frac{1}{4}\cdot\frac{1}{2}=\frac{1}{8}$

Therefore

$z_A=\frac{1}{8}z_B$

Given

$z_B=32\times10^{18}s^{-1}$

so

$z_A=\frac{1}{8}(32\times10^{18})$

$=4\times10^{18}s^{-1}$

The work function of the material is $\phi = 110 \times 10^{-20}$ J, which can also be expressed as $1.1 \times 10^{-18}$ J, and since the photoelectrons are emitted with zero kinetic energy, the photon energy must exactly equal this work function. Taking Planck’s constant as $h = 6.63 \times 10^{-34}$ J·s, we write

$E = \phi = \hbar\omega$

with $\hbar = \frac{h}{2\pi}$ and $\omega$ being the angular frequency. Solving for $\omega$ gives

$\omega = \frac{\phi}{\hbar} = \frac{2\pi\phi}{h} = \frac{2 \times 3.14 \times 1.1 \times 10^{-18}}{6.63 \times 10^{-34}} = \frac{6.908 \times 10^{-18}}{6.63 \times 10^{-34}} = 1.042 \times 10^{16} \text{ rad/s}$

The correct answer is Option A: $1.04 \times 10^{16}$ rad/s.

The electric displacement vector $\mathbf{D}$ is related to the electric field $\mathbf{E}$ by the constitutive relation $\mathbf{D} = \epsilon \mathbf{E}$.

By definition, the surface integral $\iint \mathbf{D}\,\cdot d\mathbf{S}$ equals the free charge $q$ enclosed. Hence the unit of $\mathbf{D}$ (and therefore of $\epsilon E$) must be the unit of charge per unit area, that is $\text{coulomb per square metre}$, written as $C/m^{2}$.

We now divide by time $t$, whose SI unit is the second, $s$:

$\frac{\epsilon E}{t} \; \text{has unit} \; \frac{C/m^{2}}{s} = \frac{C}{m^{2}\,s}.$

Since $1\,\text{ampere} = 1\,\text{coulomb}\,\text{s}^{-1}$, we replace $C/s$ by $A$:

$\frac{C}{m^{2}\,s} = \frac{A}{m^{2}}.$

Thus, the unit of $\dfrac{\epsilon E}{t}$ is $A/m^{2}$.

Correct option: Option B.

We are given a laser intensity $I = 4.0 \times 10^{14}$ W/m$^2$, permittivity of free space $\epsilon_0 = 8.85 \times 10^{-12}$ C$^2$/(N m$^2$), and the speed of light $c = 3 \times 10^8$ m/s. The intensity relates to the electric field amplitude by $I = \frac{1}{2}\epsilon_0 c E_0^2$ and the electric and magnetic field amplitudes satisfy $E_0 = cB_0$.

Solving the intensity equation for $E_0$ gives $E_0^2 = \frac{2I}{\epsilon_0 c} = \frac{2 \times 4.0 \times 10^{14}}{8.85 \times 10^{-12} \times 3 \times 10^8}$ which simplifies to $= \frac{8.0 \times 10^{14}}{26.55 \times 10^{-4}} = \frac{8.0 \times 10^{14}}{2.655 \times 10^{-3}} = 3.013 \times 10^{17}$ and hence $E_0 = \sqrt{3.013 \times 10^{17}} = 5.49 \times 10^{8} \text{ V/m}$.

Using the relation $B_0 = \frac{E_0}{c}$ we find $B_0 = \frac{5.49 \times 10^8}{3 \times 10^8} = 1.83 \text{ T}$. The correct answer is Option B: 1.83.

Each charge is at same distance

r

from center.

Potential is scalar, so potentials add directly.

Potential due to one charge:

$V_1=\frac{kq}{r}$

Due to five charges:

$V=5\frac{kq}{r}$

Now electric field:

Each charge produces field at center of magnitude

$E_1=\frac{kq}{r^2}$

Directions of these five fields are symmetrically spaced by

$72^{\circ}$

Because of symmetry, vector sum of all fields is zero.

so

$E=0$

From simple pendulum relation

$T=2\pi\sqrt{\frac{L}{g}}$

Squaring,

$T^2=\frac{4\pi^2L}{g}$

Rearrange:

$\frac{1}{T^2}=\frac{g}{4\pi^2L}$

So

$\frac{1}{T^2}\propto\frac{1}{L}$

or equivalently

$L=\frac{g}{4\pi^2}\cdot\frac{1}{(1/T^2)}$

Thus L is inversely proportional to $\frac{1}{T^2}$

So graph of $L$ versus $1/T^2$ is a rectangular hyperbola.

We need to determine where the object should be placed in the parallax method for finding the focal length of a concave mirror.

Key Concept: Parallax Method

In the parallax method for a concave mirror, we use the property that a concave mirror forms a real image of a real object. By removing parallax between the object (a pin) and its real image, we can precisely locate the image position and thereby determine the focal length.

Condition for a real image:

A concave mirror forms a real image only when the object is placed beyond the focus (F). Specifically:

- Object at F: Image at infinity (not useful for parallax)

- Object between F and C: Real, inverted, magnified image beyond C

- Object at C: Real, inverted, same-size image at C

- Object beyond C: Real, inverted, diminished image between F and C

If the object is placed between the pole (P) and focus (F), the mirror forms a virtual, erect, magnified image behind the mirror. Virtual images cannot be located by the parallax method using a real pin.

Therefore, for the parallax method to work, the object can be placed at any point beyond the focus - whether between F and C, at C, or beyond C. All these positions produce real images that can be located by removing parallax.

The correct answer is Option 2: at any point beyond the focus (F) of the mirror.

Consider two statements regarding the kinetic energy of a mechanical system of many particles. Statement I asserts that the total kinetic energy of the system is the sum of the kinetic energies of all the particles. Statement II claims that the total kinetic energy can also be expressed as the sum of the kinetic energy of the centre of mass with respect to the origin and the kinetic energy of all the particles with respect to the centre of mass.

By definition, the total kinetic energy of a system of $N$ particles is given by $K_{\text{total}} = \sum_{i=1}^{N} \frac{1}{2}m_i v_i^2$. This expression clearly shows that the total kinetic energy is simply the sum of the individual kinetic energies of all particles. Statement I is true.

The decomposition described in Statement II is known as Konig’s theorem (or the decomposition theorem for kinetic energy), which states that $K_{\text{total}} = \frac{1}{2}Mv_{\text{cm}}^2 + \sum_{i=1}^{N} \frac{1}{2}m_i v_i'^2$, where $M$ is the total mass of the system, $v_{\text{cm}}$ is the velocity of the centre of mass, and $v_i'$ is the velocity of the $i$-th particle relative to the centre of mass. The first term represents the kinetic energy of the centre of mass motion, while the second term corresponds to the kinetic energy of the particles measured in the centre of mass frame. Statement II is true.

The correct answer is Option (3): Both Statement I and Statement II are true.

At the exact moment the switch K is turned ON ($t = 0$), the inductors oppose any sudden change in current. Therefore, they act as open circuits.

Because the left and right parallel branches both contain inductors, no current will flow through them initially. The entire current will flow only through the middle branch, which contains a pure resistor.

 Supply Voltage, $V = 9\text{ V}$

Effective Resistance, $R = 9\ \Omega$ (only the middle branch is active)

According to Ohm's Law, the reading of the ammeter (initial current $I$) is:

$I = \frac{V}{R}$

$I = \frac{9}{9}$

$I = 1\text{ A}$

Trick: To solve such questions in less time, it is advised to directly try options and eliminate them.

option D satisfies for all

Power delivered = 1 kW = 1000 W at V = 250 V. Current $I = P/V = 4$ A.

Power loss = $I^2 R = 16 \times 2 = 32$ W.

Total power = $1000 + 32 = 1032$ W. Efficiency = $1000/1032 \times 100 = 96.9\%$.

The answer is Option 4: 96.9%.

$COM$ of the system after the collision relative to the center of the bar ($x=0$):

$x_{cm} = \frac{(20m \cdot 0) + (2m \cdot -2) + (m \cdot 4)}{20m + 2m + m} = \frac{-4m + 4m}{23m} = 0$

Initial Angular Momentum ($L_i$):

For mass $2m$: $L_1 = (2m) \cdot v \cdot (2 \text{ cm}) = 4mv$ (Counter-clockwise)

For mass $m$: $L_2 = (m) \cdot v \cdot (4 \text{ cm}) = 4mv$ (Counter-clockwise)

$L_i = 4mv + 4mv = 8mv$

Final Angular Momentum ($L_f$):

Bar: $I_{bar} = \frac{1}{12}ML^2 = \frac{1}{12}(20m)(12)^2 = 240m$

Mass $2m$: $I_{2m} = (2m)(2)^2 = 8m$

Mass $m$: $I_m = (m)(4)^2 = 16m$

$I_f = 240m + 8m + 16m = 264m$

Applying conservation of angular momentum ($L_i = I_f \omega$):

$8mv = 264m \cdot \omega$

$\frac{v}{\omega} = \frac{264}{8}$

$\frac{v}{\omega} = 33$

Smallest Lyman wavelength = 91 nm → $R = 1/(91 \times 10^{-9})$.

Largest Balmer: $3 \to 2$: $\frac{1}{\lambda_B} = R(1/4-1/9) = 5R/36$. $\lambda_B = 36/(5R) = 36 \times 91/5 = 655.2$ nm.

Largest Paschen: $4 \to 3$: $\frac{1}{\lambda_P} = R(1/9-1/16) = 7R/144$. $\lambda_P = 144/(7R) = 144 \times 91/7 = 1872$ nm.

Difference = $1872 - 655.2 = 1216.8 \approx 1217$ nm.

The answer is Option 4: 1217 nm.

Statement 1:

To find the dependence on density ($\rho$), we substitute Earth's mass ($M_e$) using the volume of a sphere:

$M_e = \text{Volume} \times \text{Density} = \frac{4}{3} \pi R_e^3 \rho$

$T = 2\pi \sqrt{\frac{R_e^3}{G \left( \frac{4}{3} \pi R_e^3 \rho \right)}}$

$T = 2\pi \sqrt{\frac{3}{4\pi G \rho}} = \sqrt{\frac{3\pi}{G \rho}}$

The expression shows that the time period $T$ is inversely proportional to the square root of the Earth's density. Therefore, Statement I is true.

Statement 2:

The orbital time period $T$ for a satellite at a radius $r$ is given by: $T = 2\pi \sqrt{\frac{r^3}{GM_e}}$

For a satellite orbiting very close to the Earth's surface, the orbital radius $r$ is approximately equal to the Earth's radius $R_e$.

$T = 2\pi \sqrt{\frac{R_e^3}{GM_e}}$

Using $GM_e = g R_e^2$: $T = 2\pi \sqrt{\frac{R_e^3}{g R_e^2}} = 2\pi \sqrt{\frac{R_e}{g}}$

Thus, Statement II is true.

Three drops coalesce: volume conserved: $3 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3$, so $R = 3^{1/3}r$. Charge: $3q$ total.

$V_r/V_R = \frac{kq/r}{k(3q)/R} = \frac{R}{3r} = \frac{3^{1/3}}{3} = 3^{-2/3}$. Ratio = $1:3^{2/3}$.

The answer is Option 2.

In single slit diffraction, the angular half-width of the central maximum is given by $\sin\theta = \frac{\lambda}{a}$, where $\lambda$ is the wavelength and $a$ is the slit width. For small angles, the linear width of the central maximum on a screen at distance $D$ is $W = \frac{2\lambda D}{a}$.

Statement A: Width increases with increase in wavelength (constant slit width). Since $W \propto \lambda$, increasing $\lambda$ increases $W$. Statement A is TRUE.

Statement B: Width increases with decrease in wavelength. This contradicts $W \propto \lambda$. Statement B is FALSE.

Statement C: Width increases with decrease in slit width (constant $\lambda$). Since $W \propto \frac{1}{a}$, decreasing $a$ increases $W$. Statement C is TRUE.

Statement D: Width increases with increase in slit width. Since $W \propto \frac{1}{a}$, increasing $a$ decreases $W$. Statement D is FALSE.

Statement E: Brightness of central maximum increases for decrease in wavelength (constant slit width). The total light energy passing through the slit is fixed, but when $\lambda$ decreases, the central maximum becomes narrower ($W \propto \lambda$), so the same energy is concentrated into a smaller area. Since intensity = energy per unit area, the peak brightness increases when $\lambda$ decreases. Statement E is TRUE.

The true statements are A, C, and E. The answer is Option A.

Two masses $m$ and $2m$ are connected by a light string over a pulley (disc) of mass $30m$ with radius $r = 0.1$ m, taking $g = 10$ m/s$^2$. When the $2m$ mass is released from rest and descends a height $h = 3.6$ m, its speed is found by applying conservation of energy.

Since the string does not slip on the pulley, energy is conserved. Let $v$ be the speed of the masses after the $2m$ mass has descended by $h = 3.6$ m. The $2m$ mass descends by $h$ while the $m$ mass rises by the same amount, so the net loss in potential energy is $2mgh - mgh = mgh$.

Both masses move with speed $v$, and the pulley rotates with angular velocity $\omega = v/r$. For a disc, the moment of inertia about its axis is $I = \frac{1}{2}(30m)r^2 = 15mr^2$.

Accordingly, the total kinetic energy is $\frac{1}{2}(m)v^2 + \frac{1}{2}(2m)v^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv^2 + mv^2 + \frac{1}{2}(15mr^2)\frac{v^2}{r^2} = \frac{3}{2}mv^2 + \frac{15}{2}mv^2 = 9mv^2.$

Equating the net potential energy loss to the total kinetic energy gives $mgh = 9mv^2,$ so $v^2 = \frac{gh}{9} = \frac{10 \times 3.6}{9} = \frac{36}{9} = 4,$ and hence $v = 2\text{ m/s}.$

The speed of the $2m$ mass after descending 3.6 m is 2 m/s.

Given: Volume $V_1 = 60$ cm$^3$, $T_1 = 27°C = 300$ K, $P_1 = 2$ atm. Final: $V_2 = 20$ cm$^3$, $T_2 = 77°C = 350$ K.

Using the combined gas law (assuming the gas is sealed):

$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$

$P_2 = P_1 \times \frac{V_1}{V_2} \times \frac{T_2}{T_1} = 2 \times \frac{60}{20} \times \frac{350}{300} = 2 \times 3 \times \frac{7}{6} = 7 \text{ atm}$

The answer is 7.

A circular loop of radius $r$ rotates about its diameter in a magnetic field $B = 0.5$ T at $\omega = 100$ rad/s.

The induced EMF is $\varepsilon = NBA\omega\sin\theta$ where $A = \pi r^2$, $N = 1$.

At $\theta = 30°$: $\varepsilon = B\pi r^2 \omega \sin 30°$

$15.4 \times 10^{-3} = 0.5 \times \frac{22}{7} \times r^2 \times 100 \times 0.5$

$15.4 \times 10^{-3} = 0.5 \times \frac{22}{7} \times r^2 \times 50$

$15.4 \times 10^{-3} = \frac{25 \times 22}{7} \times r^2 = \frac{550}{7} r^2$

$r^2 = \frac{15.4 \times 10^{-3} \times 7}{550} = \frac{107.8 \times 10^{-3}}{550} = \frac{0.1078}{550} = 1.96 \times 10^{-4}$ m$^2$

$r = 0.014$ m $= 14$ mm.

The answer is 14.

We are given capacitor P with capacitance $C_P = 10 \times 10^{-6}$ F charged to a potential difference of $V_P = 6.0$ V, and capacitor Q with capacitance $C_Q = 20 \times 10^{-6}$ F (initially uncharged). When P is disconnected from the battery and connected across Q, the charge on Q at equilibrium can be determined as follows.

Since the initial charge on P is given by $Q = CV$, substituting yields $Q_{initial} = C_P \times V_P = 10 \times 10^{-6} \times 6.0 = 60 \times 10^{-6}\text{ C} = 60 \mu\text{C}$.

After connection, the total charge is conserved because no external path allows charge to escape, and Q starts uncharged. Therefore $Q_{total} = Q_P + Q_Q = 60 \mu\text{C}$.

At equilibrium both capacitors share the same potential difference $V_{common}$, which follows from $V_{common} = \frac{Q_{total}}{C_P + C_Q} = \frac{60 \times 10^{-6}}{(10 + 20) \times 10^{-6}} = \frac{60}{30} = 2\text{ V}$.

Finally, the charge on Q is $Q_Q = C_Q \times V_{common} = 20 \times 10^{-6} \times 2 = 40 \times 10^{-6}\text{ C} = 4 \times 10^{-5}\text{ C}$. This shows that $\alpha = 4$.

The answer is 4.

We need to find the mobility of electrons in a cylindrical conductor.

The length of the conductor is $L = 2$ m, its cross-sectional area is $A = 0.2$ mm$^2 = 0.2 \times 10^{-6}$ m$^2$, the current through it is $I = 1.6$ A, the applied voltage is $V = 2$ V, the electron density is $n = 5 \times 10^{28}$/m$^3$, and the charge of an electron is $e = 1.6 \times 10^{-19}$ C.

Since resistance is given by $R = \frac{V}{I} = \frac{2}{1.6} = 1.25 \text{ } \Omega$, we obtain the resistance.

Substituting this into the formula for resistivity yields $\rho = \frac{RA}{L} = \frac{1.25 \times 0.2 \times 10^{-6}}{2} = \frac{0.25 \times 10^{-6}}{2} = 1.25 \times 10^{-7} \text{ } \Omega \cdot m$.

Mobility is defined by $\mu = \frac{1}{ne\rho}$, and substituting the known values gives

$\mu = \frac{1}{5 \times 10^{28} \times 1.6 \times 10^{-19} \times 1.25 \times 10^{-7}}$

$= \frac{1}{5 \times 1.6 \times 1.25 \times 10^{28-19-7}} = \frac{1}{10 \times 10^{2}} = \frac{1}{1000} = 10^{-3} \text{ m}^2/\text{Vs}$

So $\alpha = 1$.

The answer is 1.

We need to identify correct statements about primary standards.

A. Hydrated salts can be used as primary standard.

True. Hydrated salts with stable and definite water of crystallisation (e.g., oxalic acid dihydrate $H_2C_2O_4 \cdot 2H_2O$, Mohr's salt $(NH_4)_2Fe(SO_4)_2 \cdot 6H_2O$) can serve as primary standards because their composition is well-defined and reproducible.

B. Primary standard should not undergo any reaction with air.

True. A primary standard must be stable in air - it should not be hygroscopic (absorb moisture), should not react with $CO_2$ or $O_2$ in air. Otherwise its composition would change, making accurate weighing impossible.

C. Reactions of primary standard with another substance should be instantaneous and stoichiometric.

True. The reaction must be quantitative (complete) and follow a definite stoichiometry so that the equivalence point can be accurately determined.

D. Primary standard should not be soluble in water.

False. On the contrary, a primary standard should be readily soluble in water so that accurate solutions of known concentration can be prepared.

E. Primary standard should have low relative molar mass.

False. A primary standard should ideally have a high molar mass to minimize weighing errors (a larger mass for a given number of moles means the percentage error in weighing is smaller).

The correct answer is Option 2: A, B and C only.

We need to find the percentage of nitrogen using Kjeldahl's method. Given 1 g of compound, 15 mL of 1 M $H_2SO_4$ is neutralised by the evolved $NH_3$.

In Kjeldahl's method, nitrogen in the compound is converted to ammonium sulphate, then ammonia is liberated and absorbed in excess $H_2SO_4$: $2NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4$

Each mole of $H_2SO_4$ reacts with 2 moles of $NH_3$ (since $H_2SO_4$ is diprotic). Moles of $H_2SO_4$ = $15 \times 10^{-3} \times 1 = 0.015$ mol, which corresponds to Moles of $NH_3$ = $2 \times 0.015 = 0.03$ mol. Since each $NH_3$ contains one nitrogen atom, moles of N = 0.03 mol.

Substituting into the formula for mass gives $\text{Mass of N} = 0.03 \times 14 = 0.42 \text{ g}$

Therefore, the percentage of nitrogen is $\% N = \frac{0.42}{1} \times 100 = 42\%$

The correct answer is Option 1: 42.

We need to evaluate two statements about ionization enthalpy and electron gain enthalpy.

Analysis of Statement I: C < O < N < F is the correct order of first ionization enthalpy.

The first ionization enthalpies (in kJ/mol) are approximately: C (1086), O (1314), N (1402), F (1681). The correct increasing order is: C < O < N < F.

Nitrogen has a higher ionization enthalpy than oxygen because nitrogen has a half-filled $2p^3$ configuration, which provides extra stability. Removing an electron from this stable configuration requires more energy than removing one from oxygen's $2p^4$ configuration (where one orbital has a paired electron, causing electron-electron repulsion).

The order C < O < N < F given in Statement I is correct. Statement I is TRUE.

Analysis of Statement II: S > Se > Te > Po > O is the correct order of magnitude of electron gain enthalpy.

Electron gain enthalpy (magnitude, in kJ/mol): O (141), S (200), Se (195), Te (190), Po (174). Oxygen has an anomalously low electron gain enthalpy due to its small atomic size, causing significant electron-electron repulsion in the compact $2p$ orbital.

The correct order is: S > Se > Te > Po > O. This matches Statement II. Statement II is TRUE.

The correct answer is Option 4: Both Statement I and Statement II are true.

• Principal Functional Group: The compound is an ester (indicated by the $(R-COO-R')$ structure). Esters are named as "alkyl alkanoate."
• Alkyl Group (Attached to Oxygen): The chain on the right consists of 3 carbons, making it a propyl group.
• Parent Carbon Chain (The Acyl Part):
  1. Number the chain starting from the carbonyl carbon C=O as (C1).
  2. The longest continuous chain containing the carbonyl group and moving to the left consists of 7 carbons, making the parent suffix heptanoate.
• Substituents:
  • A bromo group is located at (C2).
  • A methyl group is located at (C5).

We need to identify correct statements about the reaction $A + 2B \rightarrow AB_2$. Since 36.0 g of A (M = 60 g/mol) and 56.0 g of B (M = 80 g/mol) are provided, we first calculate the amount of each reactant in moles: Moles of A = $36/60 = 0.6$ mol and moles of B = $56/80 = 0.7$ mol.

Because the stoichiometry requires 2 moles of B per mole of A, the 0.6 mol of A would consume $0.6 \times 2 = 1.2$ mol of B, but only 0.7 mol is available. Therefore B is the limiting reagent, which means Statement A that claims A is limiting is incorrect.

In this context, 0.7 mol of B reacts completely with $0.7/2 = 0.35$ mol of A to produce 0.35 mol of $AB_2$. The unreacted A amounts to $0.6 - 0.35 = 0.25$ mol, corresponding to $0.25 \times 60 = 15.0$ g, so Statement D is correct.

The molar mass of $AB_2$ is $60 + 2(80) = 220$ g/mol, so Statement C, which suggests 140 g/mol, is incorrect. The mass of $AB_2$ formed is $0.35 \times 220 = 77.0$ g, confirming that Statement B is correct.

The correct answer is Option 3: B and D only.

Given compound: Cyclohexyl-CH₂-NH₂ (a primary amine)

Step 1: Reaction with C₆H₅COCl / NaOH

This is Schotten-Baumann Reaction (Acylation of amines)

What happens:

• Primary amine reacts with benzoyl chloride (C₆H₅COCl = PhCOCl)
• Forms an amide

Product [A]:

Cyclohexyl-CH₂-NHCOPh
(N-benzoyl derivative / amide)

Step 2: Reaction with LiAlH₄ / H₂O

This is Reduction of amides using LiAlH₄

What happens:

• Amides are reduced to amines
• The -CO- group is converted to -CH₂-

So:
-NHCOC₆H₅ → -NHCH₂C₆H₅

Final Product [B]:

Cyclohexyl-CH₂-CH₂-NH-CH₂-Ph

(This is a secondary amine)

Overall Reaction Summary:

1. Amine → Amide
2. Amide → Secondary amine

Final Answer:

B = Cyclohexyl-CH₂-NH-CH₂-Ph

We need to determine the correct structure of the dibromo compound [P] ($C_9H_{10}Br_2$) based on its reaction sequence leading to compound [R].

Step-by-Step Reaction Analysis:

• Step 1: Conversion of [P] to [Q]

  When the geminal dihalide [P] is heated with excess sodamide ($\text{NaNH}_2$), it undergoes a double dehydrohalogenation (elimination of two molecules of $\text{HBr}$) to yield a terminal alkyne [Q]. Treatment with dilute acid protonates the acetylide ion back to the alkyne form.

• Step 2: Hydration of [Q] to [R]

  Warming the terminal alkyne [Q] with mercuric sulfate ($\text{HgSO}_4$) and dilute sulfuric acid ($\text{H}_2\text{SO}_4$) causes Kucherov hydration (Markovnikov addition of water). The resulting enol tautomerizes into a stable methyl ketone [R].

• Step 3: Characterization of [R]

  • Positive Iodoform Test: Confirms that [R] contains a methyl ketone group ($\text{-COCH}_3$).
  • Negative Tollen's Test: Confirms that [R] is a ketone rather than an aldehyde.

Structural Deduction:

From the reaction scheme shown, the structure of [R] is explicitly identified as 1-(3-methylphenyl)ethan-1-one (a methyl ketone with a meta-methyl group on the benzene ring).

Tracing back to the starting material, the geminal dibromo group must be located on the side chain alpha-carbon atom where the terminal alkyne and carbonyl group are formed.

Therefore, the original compound [P] is 1,1-dibromo-1-(3-methylphenyl)ethane.

Answer: Option B

Glucose structure 

C₁ = -CHO (aldehyde carbon)
C₂-C₅ = -CHOH
C₆ = -CH₂OH

A. Hydroxylamine

Reaction name: Oxime formation (nucleophilic addition to carbonyl)

What reacts:
C₁ (aldehyde carbon)

Change:
-CHO → -CH=NOH

Product: Glucoxime

B. Br₂ / H₂O

Reaction name: Mild oxidation of aldehyde

What reacts:
Only C₁ (aldehyde carbon)

Change:
-CHO → -COOH

Product: Gluconic acid

C. Excess acetic anhydride ((CH₃CO)₂O)

Reaction name: Acetylation (ester formation)

What reacts:
All hydroxyl groups: C₂, C₃, C₄, C₅, C₆

Change:
-OH → -OCOCH₃

Note:
C₁ remains as aldehyde

Product: Glucose pentaacetate

D. Concentrated HNO₃

Reaction name: Strong oxidation

What reacts:
C₁ (-CHO) and C₆ (-CH₂OH)

Change:
-CHO → -COOH
-CH₂OH → -COOH

Product: Saccharic acid

Final Matching

A - IV (Glucoxime)
B - I (Gluconic acid)
C - II (Glucose pentaacetate)
D - III (Saccharic acid)