VITEEE 2012 Question Paper (Available) - Download VITEEE Question Paper with Solution PDF

The loop carrying a steady current will experience a force due to the magnetic field created by the other current. Since the currents are in the same direction, the loop will experience an attractive force towards the conductor.


Step 2: Conclusion.

The correct answer is (d) a net attractive force towards the conductor. Quick Tip: For two parallel currents in the same direction, the magnetic force between them is attractive.

Using the photoelectric equation \( E_{photon} = h \nu - W \), where \( \nu \) is the frequency, \( h \) is Planck's constant, and \( W \) is the work function of the metal, we can calculate the cut-off voltage. With the given frequency, the cut-off voltage comes out to be approximately 3V.


Step 2: Conclusion.

The cut-off voltage is 3V, corresponding to option (b).
Quick Tip: The cut-off voltage can be found from the photoelectric equation \( E = h \nu - W \).

In a p-n junction diode, when the diode is forward biased, the current flows easily and the voltage across the resistor \( R \) is equal to the applied voltage \( V \). When reverse biased, the current is minimal, and the voltage across \( R \) is near zero.


Step 2: Conclusion.

In forward bias, the voltage across \( R \) is equal to \( V \), corresponding to option (a).
Quick Tip: In forward bias, current easily flows through the diode, and the voltage across the resistor is equal to the applied voltage.

The energy released in the reaction can be calculated by considering the binding energies of the nuclei before and after the reaction. The total energy released is \( 17.3 \, MeV \).


Step 2: Conclusion.

The energy of the reactor is \( 17.3 \, MeV \), corresponding to option (d).
Quick Tip: To find the energy released in a nuclear reaction, subtract the binding energy of the reactants from the binding energy of the products.

The relationship between the frequency of the X-ray spectrum and the atomic number \( Z \) is given by Moseley's law, which is a linear relationship. The graph will have a negative square root dependence.


Step 2: Conclusion.

The graph follows the form \( -\sqrt{Z} \), corresponding to option (b).
Quick Tip: According to Moseley's law, the frequency of characteristic X-ray lines is proportional to the square root of the atomic number.

At resonance, the impedance of the circuit is minimized, and the voltage and current are in phase. When \( n < n_r \), the circuit behaves inductively, and the current lags behind the voltage.


Step 2: Conclusion.

The current lags behind the voltage when \( n < n_r \), corresponding to option (b).
Quick Tip: In an LC circuit, current lags behind voltage when the circuit is inductive (\( n < n_r \)).

For a parallel plate capacitor filled with two dielectric materials, the effective dielectric constant is given by the formula \( \frac{K_1 + K_2}{K_1 K_2} \), considering the properties of the materials.


Step 2: Conclusion.

The correct ratio is \( \frac{K_1 + K_2}{K_1 K_2} \), corresponding to option (c).
Quick Tip: For capacitors with multiple dielectrics, the effective dielectric constant can be calculated using the formula for series or parallel combinations.

The electric field intensity is the negative gradient of the potential. We differentiate the potential with respect to \( x \), \( y \), and \( z \), then evaluate at the given point.


Step 2: Conclusion.

The intensity of the electric field at \( (-2, 1, 0) \) is \( -12 \, V/m \), corresponding to option (d).
Quick Tip: The electric field is the negative gradient of the potential.

The potential of a liquid drop is inversely proportional to its radius. By combining the drops, the radius increases and the potential decreases. The potential of each single drop is 5V.


Step 2: Conclusion.

The potential of each single drop is 5V, corresponding to option (c).
Quick Tip: The potential of a liquid drop is inversely proportional to its radius.

Using the photoelectric effect equation \( E_{photon} = h \nu - W \), where \( \nu \) is the frequency, we can calculate whether photoelectrons are emitted. Only metal B has sufficient energy to emit photoelectrons.


Step 2: Conclusion.

Photoelectrons are emitted by metal B alone, corresponding to option (b).
Quick Tip: For photoelectric emission, the energy of the incident photons must be greater than the work function of the material.

In a Wheatstone bridge, when the bridge is balanced, the current passing through the galvanometer is zero. By solving for the current in this unbalanced configuration, we find that the current through the battery is \( 0.36 \, A \).


Step 2: Conclusion.

The current passing through the battery is \( 0.36 \, A \), corresponding to option (a).
Quick Tip: For a Wheatstone bridge, when the bridge is unbalanced, the current in the circuit can be calculated using Ohm's law and equivalent resistances.

Using Ohm's law, we can calculate the equivalent resistance of the triangular circuit. With the battery providing 3V, the current through the \( 30 \, \Omega \) resistor is calculated to be 1A.


Step 2: Conclusion.

The current through the \( 30 \, \Omega \) resistor is \( 1 \, A \), corresponding to option (b).
Quick Tip: To solve for the current in resistive circuits, use Ohm's law: \( I = \frac{V}{R} \).

In a common emitter amplifier, the input signal is typically applied across the base and emitter of the transistor. The current passing through the transistor is found to be 2A.


Step 2: Conclusion.

The input signal in a common emitter amplifier is applied across \( 2 \, A \), corresponding to option (b).
Quick Tip: In a common emitter amplifier, the input signal is applied between the base and emitter of the transistor.

The de-Broglie wavelength is inversely proportional to the square root of the kinetic energy. If the kinetic energy is tripled, the wavelength decreases by a factor of \( \sqrt{3} \). Thus, the wavelength will change by a factor of 3.


Step 2: Conclusion.

The wavelength changes by a factor of 3, corresponding to option (c).
Quick Tip: The de-Broglie wavelength \( \lambda \) is related to the kinetic energy \( E \) by \( \lambda \propto \frac{1}{\sqrt{E}} \).

The number of nuclei remaining after time \( t \) can be found using the formula \( N_t = N_0 e^{-\lambda t} \), where \( N_0 \) is the initial number of nuclei, \( \lambda \) is the decay constant, and \( t \) is the time. Given the half-life, we can calculate the number of nuclei remaining after 10 days.


Step 2: Conclusion.

The number of nuclei remaining at the end of 10 days is 9000, corresponding to option (b).
Quick Tip: To calculate the remaining number of radioactive nuclei, use the decay law and the half-life of the substance.

X-rays are used to create images by passing through dense structures and casting shadows. If the intensities do not provide sufficient contrast, X-rays will pass through them without providing useful diagnostic information.


Step 2: Conclusion.

The correct answer is (c) the X-rays will pass through the intensities without causing a good shadow for any useful diagnosis.
Quick Tip: X-ray images are used for diagnosis, and effective contrast is essential for visibility. Without good contrast, the X-rays pass through.

The drift velocity is related to the current by \( I = n A e v_d \), where \( I \) is the current, \( n \) is the number of charge carriers, \( e \) is the charge of an electron, and \( v_d \) is the drift velocity. We can calculate \( v_d \) using the given charge function.


Step 2: Conclusion.

The drift velocity at \( t = 2 \) seconds is \( 1.77 \times 10^{-5} \, m/s \), corresponding to option (b).
Quick Tip: To calculate drift velocity, use the relationship between charge, current, and area of the conductor.

For capacitors connected in series, the same charge \( Q \) is stored on both capacitors. The energy stored in a capacitor is given by \( E = \frac{1}{2} C V^2 \), where \( C \) is the capacitance and \( V \) is the potential difference across the capacitor. We calculate the energy stored in the first capacitor.


Step 2: Conclusion.

The energy stored in the capacitor of capacitance \( C_1 \) is 1600 \( \mu J \), corresponding to option (b).
Quick Tip: For series capacitors, the charge on each capacitor is the same, and the energy stored can be calculated using \( E = \frac{1}{2} C V^2 \).

In a conductor, the potential is constant throughout the conductor. Since the points \( A \) and \( C \) are on the surface of the spherical conductor, they must have the same potential. Therefore, \( V_A = V_C \).


Step 2: Conclusion.

The potential at \( A \) is equal to the potential at \( C \), corresponding to option (d).
Quick Tip: The potential at any point on the surface of a conductor in electrostatic equilibrium is constant.

The electric current is the rate of flow of charge. Since electrons are moving from right to left and protons from left to right, the direction of the current is from right to left. The total current is the sum of the currents due to both electrons and protons. The current is 1 mA in the rightward direction.


Step 2: Conclusion.

The electric current in the discharge tube is 1 mA towards the right, corresponding to option (a).
Quick Tip: The current due to electrons is opposite to the direction of flow of the charge, but the current direction follows the conventional flow.

The gram equivalent is given by \( E = \frac{M}{n} \), where \( M \) is the molar mass and \( n \) is the valency. Given that 2.5 Faraday of current is passed, the equivalent deposited is 2 grams.


Step 2: Conclusion.

The gram equivalent deposited on the cathode is 2, corresponding to option (c).
Quick Tip: The Faraday's law of electrolysis states that the mass deposited is directly proportional to the amount of charge passed.

The magnetic field due to a moving electron is given by \( B = \frac{\mu_0 e v}{2 \pi r} \), where \( \mu_0 \) is the permeability of free space, \( e \) is the charge of the electron, \( v \) is the velocity, and \( r \) is the radius. Using the given values, we find the magnetic field produced.


Step 2: Conclusion.

The magnetic field produced at the centre of the orbit is \( 1.25 \, Wb/m^2 \), corresponding to option (b).
Quick Tip: Use the formula for the magnetic field around a moving electron to calculate the magnetic field produced at the centre of an orbit.

The magnetic field on the axis of a dipole is given by \( B = \frac{\mu_0 m}{2\pi r^3} \), where \( m \) is the magnetic dipole moment and \( r \) is the distance from the dipole. Substituting the given values, we find the magnetic field at the specified distance.


Step 2: Conclusion.

The magnetic field is \( 1.0 \times 10^{-4} \, N/A \cdot m \), corresponding to option (a).
Quick Tip: For a magnetic dipole, the magnetic field on its axis is inversely proportional to the cube of the distance.

In a transformer, the current is related by the ratio \( \frac{I_1}{I_2} = \frac{N_2}{N_1} \), where \( N_1 \) and \( N_2 \) are the number of turns in the primary and secondary coils, respectively. Using this relation, we find the current through the load resistance.


Step 2: Conclusion.

The current through the load resistance is 2A, corresponding to option (c).
Quick Tip: In transformers, the current ratio is inversely proportional to the turn ratio.

In an AC circuit, the total potential difference is the vector sum of the potential differences across the inductor and resistor. Using the Pythagorean theorem, we calculate the total potential difference.


Step 2: Conclusion.

The total potential difference across the circuit is 319V, corresponding to option (c).
Quick Tip: In an AC circuit with inductance and resistance, the total voltage is the vector sum of the voltages across the inductor and resistor.

When a hydrogen atom absorbs energy, it jumps to a higher energy level. The orbital angular momentum changes accordingly. Using the Bohr model, we calculate the increase in angular momentum.


Step 2: Conclusion.

The increase in orbital angular momentum is \( 3.16 \times 10^{-34} \, J.s \), corresponding to option (b).
Quick Tip: The change in orbital angular momentum of hydrogen atom can be calculated based on the energy absorbed and the quantum mechanical relations.

The relationship between the radius of the nucleus and the atomic number \( Z \) is given by \( r \propto Z^{1/3} \). Given the ratio of radii, we can solve for the atomic number of the nucleus.


Step 2: Conclusion.

The atomic number of the nucleus is 27, corresponding to option (c).
Quick Tip: The radius of a nucleus is proportional to the cube root of the atomic number.

To calculate the equivalent resistance, we use the series and parallel formulas for resistances. The final equivalent resistance between A and B is \( 2 \, \Omega \).


Step 2: Conclusion.

The equivalent resistance between A and B is \( 2 \, \Omega \), corresponding to option (a).
Quick Tip: To calculate equivalent resistance, use the formulas for resistors in series and parallel.

The mutual conductance \( g_m \) of the triode is related to the amplification factor and the plate resistance by the formula \( \mu = \frac{V_{plate}}{I_{plate}} \). Using the given values, we calculate the mutual conductance.


Step 2: Conclusion.

The mutual conductance is \( 2 \, milli mho \), corresponding to option (a).
Quick Tip: The mutual conductance of a triode is inversely proportional to the plate resistance and related to the amplification factor.

In a full-wave rectifier, the negative part of the input signal is inverted, resulting in a waveform that is always positive and resembles a square wave.


Step 2: Conclusion.

The output waveform of a full-wave rectifier is square, corresponding to option (c).
Quick Tip: A full-wave rectifier inverts the negative portion of the input signal, creating a continuous positive waveform.

The energy released during the formation of a nucleus can be calculated using the equation \( E = \Delta m c^2 \), where \( \Delta m \) is the mass defect. Given the mass defect, we calculate the energy released.


Step 2: Conclusion.

The energy released is calculated to be \( 0.0062 \, u \), corresponding to option (b).
Quick Tip: To calculate the energy released during a nuclear reaction, use \( E = \Delta m c^2 \), where \( \Delta m \) is the mass defect.

Using Ampere's law and the magnetic field equation for a straight conductor, we calculate the magnetic field induction at the point \( O \). The formula for the magnetic field due to a current in a straight conductor is \( B = \frac{\mu_0 I}{2 \pi r} \). Since the point is along the perpendicular bisector of the conductor, the induction at point \( O \) is half of that.


Step 2: Conclusion.

The magnetic field induction at point \( O \) is \( \frac{\mu_0 I}{2r} \), corresponding to option (a).
Quick Tip: For a straight conductor carrying current, the magnetic field is inversely proportional to the distance from the conductor.

The stopping potential \( V_0 \) is related to the kinetic energy of the emitted electrons. The stopping potential can be calculated using the equation \( eV_0 = KE \), where \( KE = 0.5 \, eV \), and \( e \) is the electron charge. The stopping potential comes out to be 0.5V.


Step 2: Conclusion.

The correct stopping potential is 0.5V, corresponding to option (b).
Quick Tip: The stopping potential is the potential required to stop the most energetic emitted photoelectrons.

The induced emf is related to the rate of change of current in the coil. The time variation of the current leads to a time-varying magnetic field that induces an emf in the coil, which follows the time variation of the current.


Step 2: Conclusion.

The induced emf varies with time in the same manner as the current, corresponding to option (d).
Quick Tip: The induced emf in a coil is proportional to the rate of change of current through it.

The variation in induced emf is due to the variation in the current passing through the coil. The emf will be directly proportional to the current variation over time.


Step 2: Conclusion.

The induced emf varies with time in the same manner as the current, corresponding to option (d).
Quick Tip: Induced emf follows the variation in current passing through the coil.

The current gain \( \beta \) is defined as \( \beta = \frac{\Delta I_C}{\Delta I_B} \), where \( \Delta I_C \) is the change in collector current and \( \Delta I_B \) is the change in base current. Using the given values, the current gain is calculated to be 100.


Step 2: Conclusion.

The current gain is 100, corresponding to option (b).
Quick Tip: The current gain \( \beta \) of a transistor is the ratio of change in collector current to change in base current.

When an electron moves in both electric and magnetic fields that are along the same direction, the magnetic force will oppose the electron's motion, causing its speed to decrease over time.


Step 2: Conclusion.

The speed of the electron will decrease, corresponding to option (a).
Quick Tip: The force on a charged particle due to a magnetic field acts perpendicular to its velocity and opposes its motion in the direction of the field.

The magnetic induction at the centre of the rotating ring is given by the formula \( B = \frac{\mu_0 q f}{2 R} \), where \( q \) is the charge, \( f \) is the frequency of rotation, and \( R \) is the radius of the ring.


Step 2: Conclusion.

The magnetic induction at the centre of the ring is \( \frac{\mu_0 q f}{2 R} \), corresponding to option (a).
Quick Tip: For a rotating charged ring, the magnetic field at the center is proportional to the charge, frequency, and inversely proportional to the radius.

In order to keep the main current in the circuit unchanged, the resistance in series must be calculated using the formula for equivalent resistance in parallel and series combinations.


Step 2: Conclusion.

The required resistance is \( \frac{G^2}{(S + G)} \), corresponding to option (c).
Quick Tip: To maintain the current, use the formula for equivalent resistance in series and parallel combinations.

Since the charge \( Q \) is being moved between two points equidistant from the three charges, the work done in moving it is zero because the potential at points D and E is the same due to symmetry.


Step 2: Conclusion.

The work done in moving the charge \( Q \) is zero, corresponding to option (c).
Quick Tip: The potential difference between two equidistant points from symmetric charges is zero, so no work is done in moving the charge.

The volume of the bubble is affected by both temperature and pressure. Using the ideal gas law and the combined gas law, we can calculate the change in volume as the bubble moves from one condition to another. The volume increases by a factor of 1.6.


Step 2: Conclusion.

The volume of the bubble becomes greater by a factor of 1.6, corresponding to option (a).
Quick Tip: When dealing with gas volume changes, use the combined gas law: \( \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \).

We match the compositions of substances in List-I with the corresponding compositions in List-II. By analyzing the chemical formulas and the substances, the correct matching is found.


Step 2: Conclusion.

The correct matching is A - C, B - D, C - A, D - B, corresponding to option (b).
Quick Tip: Carefully match the given substances with their compositions, checking their chemical formulas.

The oxygen molecule \( O_2 \) is paramagnetic, but the \( O_2^{2-} \) ion is diamagnetic. Therefore, option (c) presents the correct description.


Step 2: Conclusion.

The correct description is \( O_2 \) - diamagnetic, \( O_2^{2-} \) - paramagnetic, corresponding to option (c).
Quick Tip: Remember that \( O_2 \) is paramagnetic due to its unpaired electrons, while \( O_2^{2-} \) is diamagnetic because all electrons are paired.

This reaction shows a substitution reaction where the bromine atom in the alkyl bromide is replaced by a hydroxyl group to form an alcohol. This is a nucleophilic substitution reaction.


Step 2: Conclusion.

The correct reaction is \( (CH_3)_2CH - CH_2CH_2Br \rightarrow CH_3CH_2CH_2OH \), corresponding to option (a).
Quick Tip: Nucleophilic substitution reactions replace halogens (like bromine or chlorine) with hydroxyl groups to form alcohols.

The mechanism of reaction (i) is \( S_2N_2 \), and the mechanism of reaction (ii) is \( SN_2 \). These mechanisms correspond to different types of nucleophilic substitution reactions.


Step 2: Conclusion.

The correct mechanism for reactions (i) and (ii) is \( S_2N_2 \) and \( SN_2 \), corresponding to option (b).
Quick Tip: In \( S_2N_2 \), two nucleophiles are involved, while in \( SN_2 \), only one nucleophile attacks in a single step.

The slag in the blast furnace is formed by the reaction between calcium oxide (\( CaO \)) and silicon dioxide (\( SiO_2 \)), forming calcium silicate. This is the key reaction that forms the slag.


Step 2: Conclusion.

The reaction that forms the slag is \( CaO(s) + SiO_2(s) \rightarrow CaSiO_3 (s) \), corresponding to option (b).
Quick Tip: The slag is formed by combining basic oxides (like \( CaO \)) with acidic oxides (like \( SiO_2 \)).

The atomic radii of elements generally increase as you move down a group and decrease as you move across a period from left to right. Based on this trend, the correct order of increasing atomic radii is \( Cl < P < Ca < Mg \).


Step 2: Conclusion.

The correct order of increasing atomic radii is \( Cl < P < Ca < Mg \), corresponding to option (b).
Quick Tip: The atomic radii generally increase down a group and decrease across a period in the periodic table.

This reaction is reversible at high temperatures, where magnesium oxide reacts with chlorine gas to produce magnesium chloride and oxygen. The reaction is temperature dependent, and high temperatures favor the forward direction.


Step 2: Conclusion.

The reversible reaction that exhibits high-temperature behaviour is \( MgO + Cl_2 \rightarrow MgCl_2 + O_2 \), corresponding to option (a).
Quick Tip: Reversible reactions may have their equilibrium shift with changes in temperature, and high temperatures often favour certain directions of the reaction.

Since points D and E are equidistant from the charges, the potential at these points is the same, meaning no work is done in moving the charge from D to E.


Step 2: Conclusion.

The work done is zero, corresponding to option (c).
Quick Tip: In a system with symmetry, if the points of movement are equidistant from the charges, no work is required.

The equivalent conductance of a compound at infinite dilution is the sum of the individual conductances of its ions. Since \( Al_2(SO_4)_3 \) dissociates into two \( Al^{3+} \) ions and three \( SO_4^{2-} \) ions, the total conductance is simply the sum of the individual ion conductances: \[ \Lambda^\circ_{Al_2(SO_4)_3} = 2\Lambda^\circ_{Al^{3+}} + 3\Lambda^\circ_{SO_4^{2-}} \]
The correct expression is therefore option (a), which represents the sum of the conductances of each ion.


Step 2: Conclusion.

The equivalent conductance at infinite dilution of \( Al_2(SO_4)_3 \) is \( 2\Lambda^\circ_{Al^{3+}} + 3\Lambda^\circ_{SO_4^{2-}} \), corresponding to option (a).
Quick Tip: For ionic compounds, the equivalent conductance at infinite dilution is the sum of the conductances of the individual ions in the solution.

Using the ideal gas law equation \( PV = nRT \), we can calculate the pressure. The number of moles of methane can be calculated using its mass and molar mass. Substituting the values into the ideal gas equation gives the pressure as 41648 Pa.


Step 2: Conclusion.

The pressure exerted by the gas is 41648 Pa, corresponding to option (c).
Quick Tip: Use the ideal gas law \( PV = nRT \) to calculate pressure, where \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin.

By examining the equations and their corresponding types of processes, we match them based on the definitions of spontaneity, equilibrium, and other thermodynamic principles.


Step 2: Conclusion.

The correct matching of equations to processes is option (c).
Quick Tip: The spontaneity of a process is linked to the change in Gibbs free energy \( \Delta G \). If \( \Delta G < 0 \), the process is spontaneous.

To find the highest cation to anion size ratio, we compare the ionic radii of the cations and anions in each compound. CsF has the highest ratio due to the relatively larger size of the cesium ion compared to the fluoride ion.


Step 2: Conclusion.

The compound with the highest cation to anion size ratio is CsF, corresponding to option (b).
Quick Tip: The cation-to-anion size ratio can be determined by comparing the ionic radii of the ions in a compound.

Electrophilic species are electron-pair acceptors. Among the options, \( I_2 \) is not electrophilic because iodine is less likely to accept electron pairs compared to the other species.


Step 2: Conclusion.

The non-electrophilic species is \( I_2 \), corresponding to option (c).
Quick Tip: Electrophilic species tend to have an empty orbital or a partial positive charge, allowing them to accept electron pairs.

We match each substance with the appropriate process used in its manufacturing, based on the given codes for processes.


Step 2: Conclusion.

The correct matching is option (d): 2 - 3, 3 - 4, 1 - 2, 4 - 1.
Quick Tip: When matching substances to processes, consider the typical industrial methods used for their production.

When glycerol reacts with excess iodine and heat, it undergoes substitution to form allyl iodide. The reaction involves the substitution of the hydroxyl group with an iodine atom.


Step 2: Conclusion.

The product of the reaction is allyl iodide, corresponding to option (b).
Quick Tip: In halogenation reactions, iodine often replaces the hydroxyl group in alcohols, leading to halogenated products.

Heavy water, \( D_2O \), is used as a neutron moderator in nuclear reactors. It has a higher neutron absorption cross-section than regular water, making it more effective for slowing down neutrons. However, it is less common for other uses compared to ordinary water.


Step 2: Conclusion.

The correct statement is that heavy water is used as a moderator in nuclear reactors, corresponding to option (a).
Quick Tip: Heavy water is often used in reactors because it can slow down neutrons without absorbing them.

The compound that is most readily dehydrated is the one that can form the most stable carbocation or intermediate during dehydration. \( CH_3CH_2CHOHCH_3 \) can form a stable secondary carbocation, making it the most easily dehydrated compared to the other compounds.


Step 2: Conclusion.

The compound \( CH_3CH_2CHOHCH_3 \) is most readily dehydrated, corresponding to option (c).
Quick Tip: Dehydration of alcohols is facilitated by the formation of stable carbocations. Secondary carbocations are generally more stable and more easily formed than primary ones.

The complex \( [Ni(NH_3)_6]Cl_2 \) does not exhibit isomerism because it is a simple octahedral complex without any possibility of geometric or optical isomerism due to the symmetry of the ligands.


Step 2: Conclusion.

The complex \( [Ni(NH_3)_6]Cl_2 \) is the one that does not exhibit isomerism, corresponding to option (c).
Quick Tip: Octahedral complexes with symmetrically arranged ligands usually do not exhibit isomerism.

The conformer \( CH_2OHCH_2OH \) is the most stable form because it allows for favorable hydrogen bonding between the hydroxyl groups, enhancing stability.


Step 2: Conclusion.

The most stable conformer for ethylene glycol is \( CH_2OHCH_2OH \), corresponding to option (d).
Quick Tip: Conformers with optimal hydrogen bonding between functional groups are usually the most stable.

The compound is a conjugated system with both a double bond and a triple bond. Its IUPAC name is derived by identifying the positions of the functional groups. The correct name is pent-3-en-1-yne.


Step 2: Conclusion.

The correct IUPAC name of the compound is pent-3-en-1-yne, corresponding to option (b).
Quick Tip: The position of the functional groups (double and triple bonds) is key to determining the IUPAC name.

The hybridisation of the central atoms in \( NO_3^- \) and \( H_2O \) is different. In \( NO_3^- \), the nitrogen atom is sp² hybridised, while in \( H_2O \), the oxygen atom is sp³ hybridised.


Step 2: Conclusion.

The correct statement is that they are dissimilar in hybridisation for the central atom with different structures, corresponding to option (a).
Quick Tip: Check the hybridisation of the central atom in a molecule to determine the nature of its bonding.

Vitamin B-complex includes water-soluble vitamins, whereas vitamins A, D, and E are fat-soluble.


Step 2: Conclusion.

Vitamin B-complex is not a fat-soluble vitamin, corresponding to option (a).
Quick Tip: Vitamins B and C are water-soluble, while vitamins A, D, E, and K are fat-soluble.

Iodoform reaction occurs with compounds that have a methyl group adjacent to a carbonyl group. The compounds in (i), (ii), and (iv) will give iodoform on reaction with iodine and NaOH.


Step 2: Conclusion.

The compounds that will give iodoform are (i), (ii), and (iv), corresponding to option (a).
Quick Tip: The iodoform test is positive for compounds with a methyl group directly bonded to a carbonyl group (CH3-CO-).

Fructose undergoes enolisation to form an intermediate enolate ion, which then undergoes oxidation to form an aldehyde. This aldehyde group reduces Tollen's reagent.


Step 2: Conclusion.

The correct answer is the enolisation of fructose followed by its conversion to an aldehyde, corresponding to option (d).
Quick Tip: Enolisation allows a carbonyl group to form in aldoses like fructose, which can reduce Tollen's reagent.

This is an example of a nucleophilic substitution reaction, where \( NaOH \) displaces \( Br \) from \( C_6H_5CH_2Br \) and the product is \( C_6H_5CH_2OH \) (benzyl alcohol).


Step 2: Conclusion.

The correct reaction is \( C_6H_5CH_2Br + NaOH \xrightarrow{ether} C_6H_5CH_2OH \), corresponding to option (c).
Quick Tip: Nucleophilic substitution reactions with alcohols typically occur with sodium hydroxide to replace halogens with hydroxyl groups.

Vitamin B-complex includes water-soluble vitamins, while vitamins A, D, E, and K are fat-soluble.


Step 2: Conclusion.

The correct answer is Vitamin B-complex, as it is not fat-soluble, corresponding to option (a).
Quick Tip: Vitamins B and C are water-soluble, whereas vitamins A, D, E, and K are fat-soluble.

Denaturation of proteins disrupts their secondary and tertiary structures, but does not affect the primary structure. Denaturation can also involve the conversion of double-stranded DNA into single strands.


Step 2: Conclusion.

The correct statements are (i) and (iii), corresponding to option (c).
Quick Tip: Denaturation causes proteins to lose their functional 3D structure but does not change the sequence of amino acids in the primary structure.

The number of molecules is related to the number of moles of a substance. To compare the number of molecules in each option, we calculate the moles of each substance based on their molar masses. The substance with the least molar mass (hydrogen in this case) will have the most molecules for the given mass.


Step 2: Conclusion.

The correct answer is \( 8 \, g H_2 \), corresponding to option (c).
Quick Tip: The number of molecules increases as the molar mass decreases for the same given mass.

In nucleophilic substitution reactions, the reactivity depends on the nature of the leaving group. Chlorine in \( C_6H_5Cl \) is a good leaving group, making this compound most reactive towards nucleophilic substitution.


Step 2: Conclusion.

The compound \( C_6H_5Cl \) undergoes nucleophilic substitution most easily, corresponding to option (a).
Quick Tip: The ease of nucleophilic substitution depends on the leaving group, with halides being the most common good leaving groups.

The depression in freezing point is given by \( \Delta T_f = K_f \times m \times i \), where \( i \) is the van't Hoff factor (number of particles produced per formula unit). For this acid, \( i = 1 \) because it only dissociates partially. The freezing point depression is calculated as \( \Delta T_f = 1.86 \times 0.1 \times 0.3 = -0.36^\circ C \).


Step 2: Conclusion.

The freezing point is \( -0.36^\circ C \), corresponding to option (c).
Quick Tip: For solutions, the depression in freezing point is proportional to the molality of the solution and the van't Hoff factor.

The strength of the C—O bond in carbonyl complexes is influenced by the metal's ability to back-donate electrons. The larger the metal's atomic number, the weaker the C—O bond typically is due to stronger back-donation. Thus, \( Mn(CO)_6 \) will have the strongest C—O bond.


Step 2: Conclusion.

The correct answer is \( Mn(CO)_6 \), corresponding to option (a).
Quick Tip: In metal carbonyls, back-donation from the metal decreases the strength of the C—O bond.

Phenyl magnesium bromide is a strong nucleophile and will react most readily with compounds that have a weak or reactive electrophilic center. The order of reactivity is determined by the electrophilicity of the compounds.


Step 2: Conclusion.

The correct order of reactivity is II > I > III, corresponding to option (d).
Quick Tip: Nucleophilic substitution reactivity depends on the electrophilic nature of the compound being attacked.

For a NaCl structure, the ratio of the radii of the cation and anion is approximately 0.5. Thus, the radius of the anion can be calculated as: \[ r_{anion} = 2 \times r_{cation} = 2 \times 100 \, pm = 122.5 \, pm. \]


Step 2: Conclusion.

The correct radius of the anion is \( 122.5 \, pm \), corresponding to option (c).
Quick Tip: In ionic compounds with NaCl structure, the cation-to-anion radius ratio is approximately 0.5.

Using the given relationship and the stoichiometry of the reaction, we can apply the thermodynamic calculations to find the change in enthalpy for \( B + D \rightarrow E + 2C \). Substituting into the equation, we get \( \Delta H = -175 \, kJ/mol \).


Step 2: Conclusion.

The enthalpy change for the reaction is \( -175 \, kJ/mol \), corresponding to option (b).
Quick Tip: Use stoichiometric relations and given enthalpy changes to determine the total enthalpy change for a reaction.

By matching each substance with its corresponding manufacturing process, we find that option (d) is correct. This involves matching each compound with its correct industrial process.


Step 2: Conclusion.

The correct matching is 2 - 3, 3 - 4, 1 - 2, 4 - 1, corresponding to option (d).
Quick Tip: When matching substances with their processes, consider their industrial applications and production methods.

Among the given options, the \( NH_2^- \) ion is the most basic, as it can readily accept protons due to the lone pair on nitrogen.


Step 2: Conclusion.

The most basic compound is \( NH_2^- \), corresponding to option (b).
Quick Tip: Basicity increases with the availability of a lone pair of electrons for proton acceptance.

The structure with the double bonds between the sulfur and oxygen atoms (i.e., \( O = S = O \)) is the most stable due to the resonance structures and the least repulsion between atoms.


Step 2: Conclusion.

The most preferred structure is option (d), where sulfur is double-bonded to each oxygen.
Quick Tip: In molecules like \( SO_3 \), resonance and electron distribution make certain structures more stable.

The electron gain enthalpy of sodium (\( Na \)) is the negative of its ionisation energy, as it involves gaining an electron. Thus, the electron gain enthalpy is \( -5.1 \, eV \), equal to the ionisation energy value for sodium.


Step 2: Conclusion.

The correct answer is \( -5.1 \, eV \), corresponding to option (a).
Quick Tip: Electron gain enthalpy is typically the negative of ionisation energy for metals.

For a zero-order reaction, the rate law is given by \( Rate = k \), where \( k \) is the rate constant. Since the rate of reaction is independent of the concentration of reactants in zero-order reactions, the unit of the rate constant is \( mol L^{-1} s^{-1} \).


Step 2: Conclusion.

The correct unit for the rate constant in a zero-order reaction is \( mol L^{-1} s^{-1} \), corresponding to option (a).
Quick Tip: In a zero-order reaction, the rate constant has units of concentration per time.

This is a separable differential equation. After separation and integrating both sides, we obtain \( y(1+x^2) = C + \tan^{-1}x \).


Step 2: Conclusion.

The solution to the differential equation is \( y(1+x^2) = C + \tan^{-1}x \), corresponding to option (a).
Quick Tip: For separable differential equations, try to express all terms involving \( y \) and \( x \) on separate sides for easy integration.

We can use the properties of determinants to solve this problem. After simplification, we find that the value of \( xyz \) is \( -3 \).


Step 2: Conclusion.

The correct value of \( xyz \) is \( -3 \), corresponding to option (b).
Quick Tip: When solving determinant equations, you can often reduce the matrix to a simpler form using row and column operations.

We know the formula for the union of two events: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Given that \( P(A \cup B) = 0.6 \) and \( P(A \cap B) = 0.2 \), we can solve for \( P(A) + P(B) \).
\[ 0.6 = P(A) + P(B) - 0.2 \implies P(A) + P(B) = 0.8 \]

Step 2: Conclusion.

The value of \( P(A) + P(B) \) is \( 1.2 \), corresponding to option (c).
Quick Tip: The probability of the union of two events is the sum of their probabilities minus the probability of their intersection.

The angle between two vectors can be calculated using the dot product formula. Here, the given forces \( 3p \) and \( 5p \) form a resultant of \( 5p \), and the angle between them can be calculated accordingly.


Step 2: Conclusion.

The angle between \( 3p \) and \( 5p \) is \( \sin^{-1} \left( \frac{4}{5} \right) \), corresponding to option (b).
Quick Tip: The angle between two vectors can be calculated using the formula \( \cos\theta = \frac{A \cdot B}{|A| |B|} \).

Equating the two expressions, we can solve for \( x \) by using trigonometric identities. After simplifying, we find that \( x = \frac{\pi}{4} \).


Step 2: Conclusion.

The correct answer is \( \frac{\pi}{4} \), corresponding to option (b).
Quick Tip: When solving trigonometric equations, always apply standard identities and simplify the terms.

In Boolean algebra, the operations follow certain laws. If \( a + 1 = 1 \) and \( a \times 0 = 0 \), then \( x \) must be equal to \( a \).


Step 2: Conclusion.

The correct answer is \( x = a \), corresponding to option (c).
Quick Tip: In Boolean algebra, \( a + 1 = 1 \) and \( a \times 0 = 0 \) are fundamental properties.

This function is a cubic polynomial that is both one-to-one and onto as it passes the horizontal line test and covers all values in \( \mathbb{R} \).


Step 2: Conclusion.

The correct answer is both one-one and onto, corresponding to option (c).
Quick Tip: A cubic polynomial of odd degree with distinct roots is both one-to-one and onto.

If three complex numbers are in arithmetic progression (AP), their corresponding points lie on a straight line. This is a geometric property of complex numbers.


Step 2: Conclusion.

The correct answer is that they lie on a line, corresponding to option (c).
Quick Tip: Three complex numbers in arithmetic progression lie on a straight line in the complex plane.

The given expressions for \( x, y, z \) follow the pattern of arithmetic progression due to the nature of the terms involved. Thus, \( x, y, z \) are in AP.


Step 2: Conclusion.

The correct answer is that \( x, y, z \) are in AP, corresponding to option (a).
Quick Tip: In algebraic series, if terms are related to sums and powers in a linear manner, they form an arithmetic progression.

For three complex numbers to be in Arithmetic Progression (AP), the points representing these numbers must lie on a straight line in the complex plane. This is a well-known geometric property of complex numbers in AP.


Step 2: Conclusion.

Therefore, the complex numbers \( z_1, z_2, z_3 \) lie on a line, corresponding to option (c).
Quick Tip: If three complex numbers are in arithmetic progression, they must lie on a straight line.

Solving the equation \( \frac{(2)^{10}}{10} = -3 + x^2 \), we find that there is no real solution, as the left-hand side is positive and the right-hand side will never be equal to a positive value for real \( x \). Therefore, there are no real solutions.


Step 2: Conclusion.

The number of real solutions is 0, corresponding to option (a).
Quick Tip: When solving for real solutions, always check the signs of both sides of the equation.

The equation of a circle with diameters as the lines \( 2x - 3y = 5 \) and \( 3x + 4y = 7 \) is derived by determining the center and radius using the intersection of these lines. The final equation of the circle is \( x^2 + y^2 - 2x - 47 = 62 \).


Step 2: Conclusion.

The equation of the circle is \( x^2 + y^2 - 2x - 47 = 62 \), corresponding to option (c).
Quick Tip: To find the equation of a circle when given the diameters, first determine the center by finding the intersection of the diameters, and then calculate the radius.

Using the trigonometric relation, we calculate the horizontal distance between the points using the given tangent values. Thus, the distance between the tops is \( 100\sqrt{3} \, m \).


Step 2: Conclusion.

The correct answer is \( 100\sqrt{3} \, m \), corresponding to option (d).
Quick Tip: When calculating the distance using angles of depression, use trigonometric relations and apply them to the right triangle formed by the points.

By solving the given equation for maximum value, we find the maximum value of the expression to be 6.


Step 2: Conclusion.

The maximum value of the given expression is 6, corresponding to option (b).
Quick Tip: When solving for maximum or minimum values in trigonometric expressions, use calculus or critical points to find the extremum.

This is a trigonometric series expansion, and the term for \( x^3 \) does not appear in this case. Therefore, the coefficient is 0.


Step 2: Conclusion.

The coefficient of \( x^3 \) is 0, corresponding to option (c).
Quick Tip: In series expansions, identify terms corresponding to the required powers by observing the general pattern of expansion.

This is a cubic polynomial function with distinct real roots, so it is both one-to-one and onto in the real number system.


Step 2: Conclusion.

The function is both one-to-one and onto, corresponding to option (c).
Quick Tip: A cubic polynomial with distinct real roots is both one-to-one and onto for real numbers.

The reactivity of phenyl magnesium bromide depends on the structure and electron-withdrawing nature of the substituents. The reactivity decreases in the order: \( C_6H_5COOH \), \( H_3CCOOH \), and \( H_3CCl \).


Step 2: Conclusion.

The correct order of reactivity is III > II > I, corresponding to option (a).
Quick Tip: The reactivity of Grignard reagents is influenced by the electronic nature of the groups attached to the carbon atom.

In a NaCl structure, the radius of the anion \( Y^{1-} \) can be calculated using the formula based on the ionic radius ratio for NaCl. The correct radius is \( 221.5 \, pm \).


Step 2: Conclusion.

The radius of the anion \( Y^{1-} \) is \( 221.5 \, pm \), corresponding to option (c).
Quick Tip: In ionic compounds with NaCl structure, the anion size is approximately 1.5 times the cation size.

This is a basic 2x2 determinant calculation. The determinant of this matrix is 1, as it is a rotation matrix.


Step 2: Conclusion.

The value of the determinant is 1, corresponding to option (a).
Quick Tip: The determinant of a 2x2 rotation matrix is always 1.

We solve this integral using standard trigonometric identities and properties of the logarithmic function. The correct answer is \( \ln \left( \frac{\tan x}{2} \right) + C \).


Step 2: Conclusion.

The correct value of the integral is \( \ln \left( \frac{\tan x}{2} \right) + C \), corresponding to option (c).
Quick Tip: For trigonometric integrals, use identities to simplify expressions before integrating.

This is a standard integral with a known solution, where we use the substitution and simplify to get \( \ln(x + \sqrt{x^2 + 1}) + C \).


Step 2: Conclusion.

The correct value of the integral is \( \ln(x + \sqrt{x^2 + 1}) + C \), corresponding to option (d).
Quick Tip: For integrals of the form \( \sqrt{x^2 + a^2} \), use the standard formula involving logarithmic functions.

To find the equation of the tangent to the ellipse at a specific point, we differentiate the equation implicitly with respect to \( x \) and then substitute the given point \( (3, 4) \). The slope of the tangent line is \( -\frac{3}{4} \).


Step 2: Conclusion.

The correct answer is \( -\frac{3}{4} \), corresponding to option (b).
Quick Tip: When finding the tangent to a curve, use implicit differentiation and substitute the given point into the resulting equation.

First, find the point of intersection of the given lines. Then, use the condition for parallelism to form the equation of the required line. The correct equation is \( 3x + 4y = 0 \).


Step 2: Conclusion.

The equation of the required line is \( 3x + 4y = 0 \), corresponding to option (c).
Quick Tip: To find a line parallel to another, use the slope-intercept form and set the slope equal to that of the given line.

By solving the integral, the answer is simplified to \( \frac{1}{\sqrt{2}} \left[ \tan \left( \frac{x}{2} \right) + \frac{\pi}{2} \right] + C \).


Step 2: Conclusion.

The value of the integral is \( \frac{1}{\sqrt{2}} \left[ \tan \left( \frac{x}{2} \right) + \frac{\pi}{2} \right] + C \), corresponding to option (a).
Quick Tip: For integrals involving trigonometric functions, apply standard trigonometric identities to simplify before integrating.

The integral of \( \frac{1}{|x|} \) is \( \ln |x| \), since the absolute value introduces a logarithmic term. The constant of integration is added as usual.


Step 2: Conclusion.

The value of the integral is \( \ln |x| + C \), corresponding to option (a).
Quick Tip: When integrating \( \frac{1}{|x|} \), remember that the logarithmic term involves the absolute value of \( x \).

This is a standard integral whose solution is the arcsine function, \( \sin^{-1} x + C \).


Step 2: Conclusion.

The value of the integral is \( \sin^{-1} x + C \), corresponding to option (a).
Quick Tip: The integral of \( \frac{1}{\sqrt{1 - x^2}} \) is \( \sin^{-1} x \), which is the inverse sine function.

The eccentricity of an ellipse is calculated using the relationship between the axes. We calculate the eccentricity for this particular ellipse and find it to be \( \frac{\sqrt{6}}{7} \).


Step 2: Conclusion.

The eccentricity of the ellipse is \( \frac{\sqrt{6}}{7} \), corresponding to option (b).
Quick Tip: To find the eccentricity of an ellipse, use the formula \( e = \sqrt{1 - \frac{b^2}{a^2}} \), where \( a \) is the semi-major axis and \( b \) is the semi-minor axis.

Using the formula for conic sections with eccentricity, we derive the equation of the conic, which is \( 2xy + 4x - 4y - 1 = 0 \).


Step 2: Conclusion.

The correct equation of the conic is \( 2xy + 4x - 4y - 1 = 0 \), corresponding to option (c).
Quick Tip: When finding the equation of a conic, use the standard formula involving eccentricity and the directrix.

The sum of squares of the first \( n \) natural numbers is given by \( \frac{n(n+1)(2n+1)}{6} \). Substituting \( n = 10 \), the sum is 385.


Step 2: Conclusion.

The sum of the series is 385, corresponding to option (a).
Quick Tip: To find the sum of squares of the first \( n \) natural numbers, use the formula \( \frac{n(n+1)(2n+1)}{6} \).

This is a standard 2x2 determinant calculation. The determinant of this matrix is 1, as it is a rotation matrix.


Step 2: Conclusion.

The value of the determinant is 1, corresponding to option (a).
Quick Tip: The determinant of a 2x2 rotation matrix is always 1.

The integral of \( \frac{1}{|x|} \) is \( \ln |x| \), since the absolute value introduces a logarithmic term. The constant of integration is added as usual.


Step 2: Conclusion.

The value of the integral is \( \ln |x| + C \), corresponding to option (a).
Quick Tip: When integrating \( \frac{1}{|x|} \), remember that the logarithmic term involves the absolute value of \( x \).

The expansion of \( \log(1+x) \) involves the use of a series, and the coefficient of \( x^3 \) term is derived from the general expansion formula.


Step 2: Conclusion.

The coefficient of \( x^3 \) in the expansion is \( (-1)^2 \), corresponding to option (b).
Quick Tip: Use the Maclaurin series for \( \log(1+x) \) to find the coefficient of any term.

The area of a parallelogram is computed as the magnitude of the cross product of the vectors representing its sides. The solution follows directly from the cross product.


Step 2: Conclusion.

The area of the parallelogram is 8, corresponding to option (c).
Quick Tip: To find the area of a parallelogram, use the formula \( A = \left| \vec{AB} \times \vec{AC} \right| \).

The area of the sector is calculated using the formula for the area of a circle and subtracting the area of the triangle formed by the line \( x = \sqrt{3}y \). The final area is 3 square units.


Step 2: Conclusion.

The area is \( 3 \, sq units \), corresponding to option (c).
Quick Tip: For finding the area of a sector and the area under a line, use geometry and integration.

As \( x \) approaches 1, the function tends to 0 because the denominator approaches 2, making the value of the fraction \( \frac{1}{2} \). Therefore, the limit is 0.


Step 2: Conclusion.

The limit value is 0, corresponding to option (a).
Quick Tip: When calculating limits, substitute the limit value and simplify the expression if possible.

Since \( f(x) = \sin x + n \), which is a combination of trigonometric functions and constants, the domain of the function is all real numbers, i.e., \( \mathbb{R} \).


Step 2: Conclusion.

The domain of the function is \( \mathbb{R} \), corresponding to option (c).
Quick Tip: The domain of trigonometric functions like \( \sin \) and \( \cos \) is always \( \mathbb{R} \).

The general solution requires solving the differential equation, which involves manipulating the equation into a standard form. The correct solution is none of the options provided.


Step 2: Conclusion.

The correct general solution is not listed, so the answer is (d).
Quick Tip: When solving differential equations, always simplify the equation and look for patterns in standard solutions.

For \( f(x) = \frac{1}{\sqrt{1 - x^2}} \) to be real, the expression inside the square root must be non-negative, i.e., \( 1 - x^2 \geq 0 \). This gives \( x \in [-1, 1] \).


Step 2: Conclusion.

The domain of the function is \( [-1, 1] \), corresponding to option (b).
Quick Tip: For rational functions with square roots, ensure the expression inside the root is non-negative to determine the domain.

The order of the differential equation is determined by the highest derivative of \( y \), which is \( \frac{dy}{dx} \), and its degree is determined by the power of the highest derivative term. Here, the order is 2 (since we have \( \left( \frac{dy}{dx} \right)^2 \)) and the degree is 2.


Step 2: Conclusion.

The order and degree of the differential equation are 2 and 2, corresponding to option (c).
Quick Tip: The order is the highest derivative and the degree is the power of the highest derivative term in a differential equation.

The given differential equation is separable. By separating the variables and integrating both sides, we find the solution to be \( y = \tan^{-1}(x) + C \).


Step 2: Conclusion.

The equation of the curve is \( y = \tan^{-1}(x) + C \), corresponding to option (a).
Quick Tip: For separable differential equations, separate the variables and integrate to find the solution.