JEE Main 30 January 2023 Shift 2: all 90 questions in paper order (Physics, Chemistry, Maths) with the answer key and worked solutions. Free to read; attempt it as a timed 3-hour mock test.
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Physics · Experimental Physics · Single correct
Match List I with List II. Choose the correct answer from the options given below:
Answer: (d)
Solution not Available
Physics · Motion in a Straight Line · Single correct
A vehicle travels $4\,\mathrm{km}$ with speed of $3\,\mathrm{km/h}$ and another $4\,\mathrm{km}$ with speed of $5\,\mathrm{km/h}$, then its average speed is :
Answer: (d)
Given $\($ $\frac{2}{V_{av}}$ = $\frac{1}{3}$ + $\frac{1}{5}$ = $\frac{8}{15}$ $\)$. Therefore, $\($ V_{av} = $\frac{15}{4}$ = 3.75 $\,$ km/h $\)$.
Physics · Motion in a Straight Line · Single correct
An object is allowed to fall from a height R above the earth, where R is the radius of earth. Its velocity when it strikes the earth's surface, ignoring air resistance, will be:
Answer: (b)
Loss in PE = Gain in KE $$\left( -\frac{GMm}{2R} \right) - \left( -\frac{GMm}{R} \right) = \frac{1}{2} mv^2$$ Therefore, $$v^2 = \frac{GM}{R} = gR$$ Thus, $$v = \sqrt{gR}$$
Physics · Laws of Motion · Numerical
A stone tied to 180 cm long string at its end is making 28 revolutions in horizontal circle in every minute. The magnitude of acceleration of stone is $\frac{1936}{x} \, \mathrm{ms^{-2}}$. The value of $x$ _______. (Take $\pi = \frac{22}{7}$)
Answer: 125
Given $$a = \omega^2 R = \left( \frac{28 \times 2\pi}{60} \right)^2 \times 1.8$$ Simplifying, we have $$= \left( \frac{56}{60} \times \frac{22}{7} \right)^2 \times 1.8$$ Further simplifying, $$= \frac{(44)^2}{225} \times 1.8$$ Calculating, $$= \frac{1936 \times 1.8}{225}$$ Finally, $$x = 125$$
Physics · Laws of Motion · Single correct
A block of $\sqrt{3} \, \mathrm{kg}$ is attached to a string whose other end is attached to the wall. An unknown force $F$ is applied so that the string makes an angle of $30^\circ$ with the wall. The tension $T$ is: (Given $g = 10 \, \mathrm{ms}^{-2}$)
Answer: (a)
Given $\theta = 30^\circ$. We have $\cos \theta = \frac{\sqrt{3}g}{T}$. Therefore, $$\frac{\sqrt{3}}{2} = \frac{\sqrt{3}g}{T}$$ which implies $T = 20 \, \mathrm{N}$.
Physics · Work, Energy and Power · Numerical
A body of mass 2 kg is initially at rest. It starts moving unidirectionally under the influence of a source of constant power P. Its displacement in 4s is $\frac{1}{3} \alpha^2 \sqrt{P}$ m. The value of $\alpha$ will be ……
Answer: 4
Given $\($ $\frac{1}{2}$ m V^2 = Pt $\)$. $\[$ V = $\sqrt{\frac{2Pt}{m}}$ $\]$ $\[$ $\frac{dx}{dt}$ = $\sqrt{\frac{2Pt}{m}}$ $\]$ $\[$ x = $\sqrt{\frac{2P}{m}}$ $\frac{2}{3}$ [t^{3/2}]_0^4 $\]$ $\[$ x = $\frac{16\sqrt{P}}{3}$ = $\frac{1}{3}$ $\times$ 16$\sqrt{P}$ $\]$ $\($ $\alpha$ = 4 $\)$
Physics · Laws of Motion · Single correct
A machine gun of mass $10\,\mathrm{kg}$ fires $20\,\mathrm{g}$ bullets at the rate of $180$ bullets per minute with a speed of $100\,\mathrm{m}\,\mathrm{s^{-1}}$ each. The recoil velocity of the gun is:
Answer: (d)
Given $$20 \times 10^{-3} \times \frac{180}{60} \times 100 = 10 \, \mathrm{V}$$ Therefore, $$v = 0.6 \, \mathrm{m/s}$$
Physics · System of Particles and Rotational Motion · Numerical
A uniform disc of mass 0.5 $\mathrm{kg}$ and radius $r$ is projected with velocity 18 $\mathrm{m/s}$ at $t = 0 \mathrm{s}$ on a rough horizontal surface. It starts off with a purely sliding motion at $t = 0 \mathrm{s}$. After 2 $\mathrm{s}$ it acquires a purely rolling motion (see figure). The total kinetic energy of the disc after 2 $\mathrm{s}$ will be _______ $\mathrm{J}$ (given, coefficient of friction is 0.3 and $g = 10 \mathrm{m/s^2}$).
Answer: 54
Given $a = -\mu_k g = -3$. Calculate $V = 18 - 3 \times 2$. Thus, $V = 12 \, \mathrm{m/s}$. The kinetic energy is given by $$\mathrm{KE} = \frac{1}{2} mv^2 + \frac{1}{2} mr^2 \frac{v^2}{r^2}.$$ Simplifying, $$\mathrm{KE} = \frac{3}{4} mv^2.$$ Finally, $$\mathrm{KE} = 3 \times 18 = 54 \, \mathrm{J}.$$
Physics · Mechanical Properties of Solids · Single correct
A force is applied to a steel wire ‘A’, rigidly clamped at one end. As a result elongation in the wire is 0.2 mm. If same force is applied to another steel wire ‘B’ of double the length and a diameter 2.4 times that of the wire ‘A’, the elongation in the wire ‘B’ will be (wires having uniform circular cross sections)
Answer: (d)
Given $$Y = \frac{F/A}{\Delta \ell / \ell}$$ Therefore, $$F = \frac{YA}{\ell} \Delta \ell$$ Equating, $$\left( \frac{A \Delta \ell}{\ell} \right)_1 = \left( \frac{A \Delta \ell}{\ell} \right)_2$$ Thus, $$\frac{\Delta \ell_2}{\Delta \ell_1} = \frac{A_1}{A_2} \times \frac{\ell_2}{\ell_1}$$ Substituting values, $$\frac{\Delta \ell_2}{0.2} = \frac{1}{2.4 \times 2.4} \times \frac{2}{1}$$ Finally, $$\Delta \ell_2 = 6.9 \times 10^{-2} \, \mathrm{mm}$$
Physics · Thermal Properties of Matter · Numerical
A faulty thermometer reads $5^{\circ} \mathrm{C}$ in melting ice and $95^{\circ} \mathrm{C}$ in steam. The correct temperature on absolute scale will be ...... K when the faulty thermometer reads $41^{\circ} \mathrm{C}$.
Answer: 313
$\dfrac{41^\circ - 5^\circ}{95^\circ - 5^\circ} = \dfrac{C - 0^\circ}{100^\circ - 0^\circ}$ $\Rightarrow C = \dfrac{36}{90} \times 100 = 40^\circ C = 313\,\text{K}$
Physics · Thermodynamics · Single correct
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: Efficiency of a reversible heat engine will be highest at $-273^\circ \mathrm{C}$ temperature of cold reservoir. Reason R: The efficiency of Carnot's engine depends not only on temperature of cold reservoir but it depends on the temperature of hot reservoir too and is given as $\eta = \left( 1 - \frac{T_2}{T_1} \right)$. In the light of the above statements, choose the correct answer from the options given below:
Answer: (d)
Both A and R are true and R is the correct explanation of A
Physics · Kinetic Theory · Single correct
A flask contains hydrogen and oxygen in the ratio of 2 : 1 by mass at temperature 27°C. The ratio of average kinetic energy per molecule of hydrogen and oxygen respectively is :
Answer: (b)
Given $$K_{av} = \frac{5}{2} kT$$. The ratio is 1:1.
Physics · Oscillations · Single correct
For a simple harmonic motion in a mass spring system shown, the surface is frictionless. When the mass of the block is 1 kg, the angular frequency is $\omega_1$. When the mass block is 2 kg the angular frequency is $\omega_2$. The ratio $\omega_2/\omega_1$ is:
Answer: (b)
The angular frequency $\omega$ is given by $$\omega = \sqrt{\frac{k}{m}}.$$ The ratio of angular frequencies is $$\frac{\omega_2}{\omega_1} = \sqrt{\frac{m_1}{m_2}} = \sqrt{\frac{1}{2}}.$$
Physics · Oscillations · Numerical
The velocity of a particle executing SHM varies with displacement $(x)$ as $4v^2 = 50 - x^2$. The time period of oscillations is $\frac{x}{7}$ s. The value of $x$ is ........ (Take $\pi = \frac{22}{7}$)
Answer: 88
Given $4v^2 = 50 - x^2$. Therefore, $$v = \frac{1}{2} \sqrt{50 - x^2}$$. We have $$\omega = \frac{1}{2}$$. The period $T$ is given by $$T = \frac{2\pi}{\omega} = 4\pi = \frac{88}{7}$$.
Physics · Electric Charges and Fields · Single correct
As shown in the figure, a point charge Q is placed at the centre of conducting spherical shell of inner radius a and outer radius b. The electric field due to charge Q in three different regions I, II and III is given by : (I : r b)
Answer: (b)
Electric field inside material of conductor is zero.
Physics · Electric Charges and Fields · Numerical
As shown in figure, a cuboid lies in a region with electric field $\mathbf{E} = 2x^2 \hat{i} - 4y \hat{j} + 6 \hat{k} \, \mathrm{N/C}$. The magnitude of charge within the cuboid is $n \varepsilon_0 \, \mathrm{C}$. The value of $n$ is ______ (if dimension of cuboid is $1 \times 2 \times 3 \, \mathrm{m^3}$)
Answer: 12
Given $\vec{E} = 2x^2 \hat{i} - 4y \hat{j} + 6 \hat{k}$. The net flux is calculated as follows: $$\phi_{net} = -8 \times 3 + 2 \times 6 = -12$$ Then, $$-12 = \frac{q}{\epsilon_0}$$ Therefore, $$|q| = 12 \epsilon_0$$
Physics · Current Electricity · Single correct
The equivalent resistance between A and B is
Answer: (a)
The equivalent resistance is calculated using the formula for resistors in parallel. $$\frac{1}{R_{eq}} = \frac{1}{2} + \frac{1}{12} + \frac{1}{4} + \frac{1}{6} + \frac{1}{2}$$ Simplifying the expression: $$= \frac{6 + 1 + 3 + 2 + 6}{12} = \frac{18}{12} = \frac{3}{2}$$ Therefore, the equivalent resistance is: $$R_{eq} = \frac{2}{3} \, \Omega$$
Physics · Current Electricity · Numerical
If the potential difference between B and D is zero, the value of $x$ is $\frac{1}{n} \, \Omega$. The value of $n$ is ........
Answer: 2
Given $\($ $\frac{2}{3}$ = $\frac{\overline{x+1}}{x}$ $\)$. Therefore, $\($ $\frac{2}{3}$ = $\frac{1}{x+1}$ $\)$. This implies $\($ x = 0.5 = $\frac{1}{2}$ $\)$. Finally, $\($ n = 2 $\)$.
Physics · Moving Charges and Magnetism · Single correct
As shown in the figure, a current of 2A flowing in an equilateral triangle of side $4\sqrt{3} \, \mathrm{cm}$. The magnetic field at the centroid O of the triangle is:
Answer: (d)
Given $d \tan 60^\circ = 2\sqrt{3}$. $d = 2 \, \mathrm{cm}$. $B = 3 \times \frac{\mu_0 i}{2 \pi d} \sin 60^\circ$. $$= 3 \times \frac{2 \times 10^{-7} \times 2}{2 \times 10^{-2}} \times \frac{\sqrt{3}}{2}$$ $$= 3 \sqrt{3} \times 10^{-5}$$
Physics · Moving Charges and Magnetism · Single correct
A current carrying rectangular loop PQRS is made of uniform wire. The length PR = QS = 5 $\mathrm{\ cm}$ and PQ = RS = 100 $\mathrm{\ cm}$. If ammeter current reading changes from I to 2I, the ratio of magnetic forces per unit length on the wire PQ due to wire RS in the two cases respectively $f^{I}_{PQ} : f^{2I}_{PQ}$ is:
Answer: (b)
Force $F$ is proportional to the product of currents $I_1$ and $I_2$. Therefore, $$F \propto I_1 I_2$$ The ratio of forces $F_1$ to $F_2$ is given as $$F_1 : F_2 = 1 : 4$$
Physics · Alternating Current · Numerical
In an ac generator, a rectangular coil of 100 turns each having area $14 \times 10^{-2} \, \mathrm{m}^2$ is rotated at $360 \, \mathrm{rev/min}$ about an axis perpendicular to a uniform magnetic field of magnitude $3.0 \, \mathrm{T}$. The maximum value of the emf produced will be ______ V. ( Take $\pi$ = $\frac{22}{7}$ )
Answer: 1584
The maximum electromotive force is given by the formula: $$\xi_{max} = NAB\omega$$ Substituting the given values: $$= 100 \times 14 \times 10^{-2} \times 3 \times \frac{360 \times 2\pi}{60}$$ Calculating the result: $$= 1584 \, \mathrm{V}$$
Physics · Alternating Current · Single correct
In the given circuit, rms value of current ($I_{rms}$) through the resistor $R$ is:
Answer: (a)
Given $$z = \sqrt{100^2 + (200 - 100)^2}$$ Simplifying, we have $$z = 100\sqrt{2} \, \Omega$$ The current is given by $$i_{rms} = \frac{V_{rms}}{z} = \frac{200\sqrt{2}}{100\sqrt{2}}$$ Thus, $$i_{rms} = 2 \, A$$
Physics · Ray Optics and Optical Instruments · Single correct
A thin prism $P_1$ with an angle $6^\circ$ and made of glass of refractive index $1.54$ is combined with another prism $P_2$ made from glass of refractive index $1.72$ to produce dispersion without average deviation. The angle of prism $P_2$ is :
Answer: (d)
Given $\delta_1 = \delta_2$ for no average deviation. Therefore, $6^\circ (1.54 - 1) = A(1.72 - 1)$. This implies $$A = \frac{6^\circ \times 0.54}{0.72}$$ which simplifies to $$\frac{18^\circ}{4} = 4.5^\circ.$$
Physics · Wave Optics · Numerical
In a Young's double slit experiment, the intensities at two points, for the path difference $\frac{\lambda}{4}$ and $\frac{\lambda}{3}$ ($\lambda$ being the wavelength of light used) are $I_1$ and $I_2$ respectively. If $I_0$ denotes the intensity produced by each one of the individual slits, then $$\frac{I_1 + I_2}{I_0} = \ldots$$
Answer: 3
Given $I = 4I_0 \cos^2 \left( \frac{\Delta \phi}{2} \right)$. For $I_1$, we have $I_1 = 4I_0 \cos^2 \left( \frac{\pi}{4} \right) = 2I_0$. For $I_2$, we have $I_2 = 4I_0 \cos^2 \left( \frac{2\pi}{3} \right) = I_0$. Therefore, $\($ $\frac{I_1 + I_2}{I_0}$ = 3 $\)$.
Physics · Electric Charges and Fields · Single correct
A point source of 100 W emits light with 5$\%$ efficiency. At a distance of 5 m from the source, the intensity produced by the electric field component is :
Answer: (b)
Given $$I_{EF} = \frac{1}{2} \times \frac{5}{4\pi \times 5^2}$$ This simplifies to $$= \frac{1}{40\pi} \, \mathrm{W/m^2}$$
Physics · Dual Nature of Radiation and Matter · Single correct
An electron accelerated through a potential difference $V_1$ has a de-Broglie wavelength of $\lambda$. When the potential is changed to $V_2$, its de-Broglie wavelength increases by 50$\%$. The value of $\left( \frac{V_1}{V_2} \right)$ is equal to:
Answer: (b)
Given $$ KE=\frac{P^2}{2m}, \qquad P=\frac{h}{\lambda} $$ $$ eV_1=\frac{\left(\frac{h}{\lambda}\right)^2}{2m} $$ $$ eV_2=\frac{\left(\frac{h}{1.5\lambda}\right)^2}{2m} $$ $$ \frac{V_1}{V_2}=(1.5)^2=\frac{9}{4} $$
Physics · Nuclei · Single correct
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: The nuclear density of nuclides $^{10}_{5}\mathrm{B}$, $^{6}_{3}\mathrm{Li}$, $^{56}_{26}\mathrm{Fe}$, $^{20}_{10}\mathrm{Ne}$ and $^{209}_{83}\mathrm{Bi}$ can be arranged as $\rho^{N}_{\mathrm{Bi}} > \rho^{N}_{\mathrm{Fe}} > \rho^{N}_{\mathrm{Ne}} > \rho^{N}_{\mathrm{B}} > \rho^{N}_{\mathrm{Li}}$. Reason R: The radius $R$ of nucleus is related to its mass number $A$ as $R = R_0 A^{1/3}$, where $R_0$ is a constant. In the light of the above statement, choose the correct answer from the options given below:
Answer: (b)
Nuclear density is independent of $A$.
Physics · Nuclei · Fill in the blank
A radioactive nucleus decays by two different processes. The half-life of the first process is 5 minutes and that of the second process is 30 $\mathrm{s}$. The effective half-life of the nucleus is calculated to be $\frac{\alpha}{11}$ $\mathrm{s}$. The value of $\alpha$ is ______.
Answer: 300
Given \( \frac{dN_1}{dt} = -\lambda_1 N \) and \( \frac{dN_2}{dt} = -\lambda_2 N \). \( \frac{dN}{dt} = -(\lambda_1 + \lambda_2) N \). Therefore, \( \lambda_{eq} = \lambda_1 + \lambda_2 \). \( \frac{1}{t_{1/2}} = \frac{1}{300} + \frac{1}{30} = \frac{11}{300} \). Thus, \( t_{1/2} = \frac{300}{11} \).
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
The output Y for the inputs A and B of circuit is given by Truth table of the shown circuit is :
Answer: (d)
Given circuit represent XOR
Physics · Experimental Physics · Single correct
Match List I with List II Choose the correct answer from the options given below :
Answer: (d)
Solution not Available
Chemistry · Structure of Atom · Single correct
The wave function ($\Psi$) of 2s is given by $$\Psi_{2s} = \frac{1}{2\sqrt{2\pi}} \left( \frac{1}{a_0} \right)^{1/2} \left( 2 - \frac{r}{a_0} \right) e^{-r/2a_0}$$ At $r = r_0$, radial node is formed. Thus, $r_0$ in terms of $a_0$
Answer: (d)
At node $\Psi_{2s} = 0$ $$2 - \frac{r_0}{a_0} = 0$$ Therefore, $$r_0 = 2a_0$$
Chemistry · Structure of Atom · Single correct
Maximum number of electrons that can be accommodated in shell with $n = 4$ are:
Answer: (b)
The number of electrons in the orbitals of sub-shell of $n = 4$ are 4s: 2 4p: 6 4d: 10 4f: 14 (Total) 32
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
Bond dissociation energy of $\text{E}-\text{H}$ bond of the "$\text{H}_2\text{E}$" hydrides of group 16 elements (given below), follows order.
Answer: (a)
Bond dissociation energy of E–H bond in hydrides of group 16 follows the order $\mathrm{H_2O} > \mathrm{H_2S} > \mathrm{H_2Se} > \mathrm{H_2Te}$
Chemistry · Thermodynamics · Numerical
1 mole of ideal gas is allowed to expand reversibly and adiabatically from a temperature of $27^{\circ} \mathrm{C}$. The work done is $3 \, \mathrm{kJ \, mol^{-1}}$. The final temperature of the gas is _____ $\mathrm{K}$ (Nearest integer). Given $C_v = 20 \, \mathrm{J \, mol^{-1} \, K^{-1}}$.
Answer: 150
$q = 0$ $\Delta U = w$ $1 \times 20 \times (T_2 - 300) = -3000$ $T_2 - 300 = -150$ $T_2 = 150\,\mathrm{K}$
Chemistry · Equilibrium · Numerical
Consider the following equation: $2SO_2(g)+O_2(g)\rightleftharpoons2SO_3(g)$, $\Delta H=-190\,kJ$. The number of factors which will increase the yield of $SO_3$ at equilibrium from the following is ______. (A) Increasing temperature (B) Increasing pressure (C) Adding more $SO_2$ (D) Adding more $O_2$ (E) Addition of catalyst
Answer: (c)
The yield of $\mathrm{SO_3}$ at equilibrium will be due to B. Increasing pressure C. Adding more $\mathrm{SO_2}$ D. Adding more $\mathrm{O_2}$
Chemistry · Hydrogen · Numerical
The strength of 50 volume solution of hydrogen peroxide is _____ g/L (Nearest integer). Given: Molar mass of $\mathrm{H_2O_2}$ is $34 \, \mathrm{g \, mol^{-1}}$ Molar volume of gas at STP = $22.7 \, \mathrm{L}$.
Answer: 150
Molarity = $\frac{50}{11.35}$ Therefore, Strength in gm/L = $\frac{50}{11.35}$ $\times$ 34
Chemistry · The s-Block Elements · Single correct
Chlorides of which metal are soluble in organic solvents:
Answer: (d)
BeCl$_2$ having covalent nature is soluble in organic solvent.
Chemistry · The s-Block Elements · Single correct
Which of the following reaction is correct?
Answer: (c)
The reaction given is: $$4 \mathrm{LiNO_3} \xrightarrow{\Delta} 2 \mathrm{Li_2O} + 4 \mathrm{NO_2} + \mathrm{O_2}$$
Chemistry · The p-Block Elements (Group-13 and 14) · Single correct
Boric acid in solid, whereas $\mathrm{BF}_3$ is gas at room temperature because of
Answer: (c)
Boric acid has strong hydrogen bonding while $\mathrm{BF_3}$
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
The most stable carbocation for the following is:
Answer: (a)
The $+M$ effect of $\mathrm{NH_2}$ is stabilizing the carbocation.
Chemistry · Equilibrium · Single correct
The correct order of $pK_a$ values for the following compounds is:
Answer: (b)
Due to $-M$ effect of $-\mathrm{NO_2}$ group, it increases acidity. $+M$ effect of $\mathrm{N(CH_3)_2}$ decreases acidity. Hyperconjugation of isopropyl decreases acidity. Therefore, order of acidic strength.
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Match List I with List II:
Answer: (b)
List I contains mixtures and List II contains the corresponding separation techniques. The correct matches are as follows: 1. $\mathrm{CHCl_3} + \mathrm{C_6H_5NH_2}$ is separated by Distillation. 2. $\mathrm{C_6H_{14}} + \mathrm{C_5H_{12}}$ is separated by Fractional distillation. 3. $\mathrm{C_6H_5NH_2} + \mathrm{H_2O}$ is separated by Steam distillation. 4. Organic compound in $\mathrm{H_2O}$ is separated by Differential extraction.
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R. Reason R : Zn-Hg/HCl is used to reduce carbonyl group to $-CH_2-$ group. In the light of the above statements, choose the correct answer from the options given below:
Answer: (a)
The acid sensitive alcohol group reacts with HCl, hence Clemmenson reduction is not suitable for above conversion.
Chemistry · Environmental Chemistry · Single correct
The water quality of a pond was analysed and its BOD was found to be 4. The pond has
Answer: (c)
Clean water has BOD value of $< 5$ while polluted water has BOD of $15$ or more.
Chemistry · The Solid State · Numerical
Iron oxide $\mathrm{FeO}$ crystallises in a cubic lattice with a unit cell edge length of $5.0\,\mathrm{\AA}$. If the density of $\mathrm{FeO}$ in the crystal is $4.0\,\mathrm{g\,cm^{-3}}$, then the number of $\mathrm{FeO}$ units present per unit cell is _____ (Nearest integer). Given: Molar masses of Fe and O are $56$ and $16\,\mathrm{g\,mol^{-1}}$, respectively. $N_A = 6.0 \times 10^{23}\,\mathrm{mol^{-1}}$
Answer: 4
Given $$d = \frac{Z \times M}{N_0 \times a^3}$$ $$4 = \frac{Z \times 72}{6 \times 10^{23} \times 125 \times 10^{-24}}$$ $$Z = 4.166 \approx 4$$
Chemistry · Solutions · Numerical
Lead storage battery contains 38$\%$ by weight solution of $\mathrm{H_2SO_4}$. The van't Hoff factor is 2.67 at this concentration. The temperature in Kelvin at which the solution in the battery will freeze is _____ (Nearest integer). Given $K_f = 1.8 \, \mathrm{K \, kg \, mol^{-1}}$
Answer: 243
The formula for freezing point depression is given by $\Delta T_f = i \cdot K_f \cdot m$. Therefore, $$\Delta T_f = 2.67 \times 1.8 \times \frac{38}{98} \times \frac{1000}{62}$$ Thus, $$\Delta T_f = 30.05$$
Chemistry · Electrochemistry · Numerical
The electrode potential of the following half cell at 298 K. $X | X^{2+} (0.001 M) || Y^{2+} (0.01 M) | Y$ is _____ $\times 10^{-2}$ V (Nearest integer). Given: $E^0_{x^{2+}|x} = -2.36 V$ $E^0_{y^{2+}|y} = +0.36$V $\frac{2.303RT}{F}$ = 0.06 V
Answer: 275
The reaction is given by: $$\mathrm{X + Y^{2+} \rightarrow Y + X^{2+}}$$ The standard cell potential is calculated as: $$E^0_{Cell} = 0.36 - (-2.36) = 2.72 \, \mathrm{V}$$ The cell potential is calculated using the Nernst equation: $$E_{Cell} = 2.72 - \frac{0.06}{2} \log \frac{0.001}{0.01}$$ Simplifying gives: $$= 2.72 + 0.03 = 2.75 \, \mathrm{V}$$ Finally, the cell potential is: $$= 275 \times 10^{-2} \, \mathrm{V}$$
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
An organic compound undergoes first order decomposition. If the time taken for the $60\%$ decomposition is $540 \, \mathrm{s}$, then the time required for $90\%$ decomposition will be is ______ $\mathrm{s}$. (Nearest integer). Given: $\ln 10 = 2.3$; $\log 2 = 0.3$
Answer: 1350
Given $\frac{t_1}{t_2} = \frac{1}{K} \ln \frac{a_0}{0.4a_0}$ and $t_2 = \frac{1}{K} \ln \frac{a_0}{0.1a_0}$. \[ \frac{540}{t_2} = \frac{\ln \frac{10}{4}}{\ln 10} \] \[ \frac{540}{t_2} = \frac{\log 10 - \log 4}{\log 10} \] \[ \frac{540}{t_2} = \frac{1 - 0.6}{1} \] \[ \Rightarrow \frac{540}{t_2} = 0.4 \] \[ \Rightarrow t_2 = \frac{540}{0.4} = 1350 \text{ sec} \]
Chemistry · Surface Chemistry · Numerical
The graph of $\log \frac{x}{m}$ vs $\log p$ for an adsorption process is a straight line inclined at an angle of $45^\circ$ with intercept equal to $0.6020$. The mass of gas adsorbed per unit mass of adsorbent at the pressure of $0.4 \, \mathrm{atm}$ is _______ $\times 10^{-1}$ (Nearest integer) Given: $\log 2 = 0.3010$
Answer: 16
Given $\log \frac{x}{m} = \log k + \frac{1}{n} \log P$. Since $\frac{1}{n} = \tan 45^\circ = 1$, we have $\log k = 0.6020 = \log 4$. Therefore, $\frac{K}{x} = 4$, which implies $\frac{x}{m} = K \cdot P^{1/n}$. Thus, $\frac{x}{m} = 4(0.4) = 1.6$. Finally, $\frac{x}{m} = 1.6 = 16 \times 10^{-1}$.
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
Given below are two statements: Statement I: During Electrolytic refining, the pure metal is made to act as anode and its impure metallic form is used as cathode. Statement II: During the Hall-Heroult electrolysis process, purified $\mathrm{Al_2O_3}$ is mixed with $\mathrm{Na_3AlF_6}$ to lower the melting point of the mixture. In the light of the above statements, choose the most appropriate answer from the options given below:
Answer: (a)
In Electrolytic refining, the pure metal is used as cathode and impure metal is used as anode. $\mathrm{Na_3AlF_6}$ is added during electrolysis of $\mathrm{Al_2O_3}$ to lower the melting point and increase conductivity.
Chemistry · Analytical Chemistry · Single correct
Formulae for Nessler's reagent is:
Answer: (c)
Nessler's reagent is $\mathrm{K_2HgI_4}$.
Chemistry · Redox Reactions · Single correct
KMnO_4 oxidises $I^-$ in acidic and neutral/faintly alkaline solution, respectively to
Answer: (a)
In acidic medium $$2\mathrm{MnO_4^-} + 10\mathrm{I^-} + 16\mathrm{H^+} \rightarrow 2\mathrm{Mn^{2+}} + 5\mathrm{I_2} + 8\mathrm{H_2O}$$ In neutral/faintly alkaline solution $$2\mathrm{MnO_4^-} + \mathrm{I^-} + \mathrm{H_2O} \rightarrow 2\mathrm{MnO_2} + 2\mathrm{OH^-} + \mathrm{IO_3^-}$$
Chemistry · Co-ordination Compounds · Single correct
Match List I with List II:
Answer: (d)
For $[\mathrm{Fe(NH_3)_6}]^{+2}$, $\Delta_0 < P$, hence the pairing of electrons does not occur in $t_{2g}$. Therefore complex is outer orbital and its hybridisation is $sp^3d^2$.
Chemistry · Co-ordination Compounds · Single correct
$1\,\mathrm{L}$ of $0.02\,\mathrm{M}$ solution of $[\mathrm{Co(NH_3)_5SO_4}]\mathrm{Br}$ is mixed with $1\,\mathrm{L}$ of $0.02\,\mathrm{M}$ solution of $[\mathrm{Co(NH_3)_5Br}]\mathrm{SO_4}$. The resulting solution is divided into two equal parts $(X)$ and treated with excess $\mathrm{AgNO_3}$ solution and $\mathrm{BaCl_2}$ solution, respectively, as shown below: $1\,\mathrm{L}$ solution $(X) + \mathrm{AgNO_3}$ solution (excess) $\rightarrow Y$ $1\,\mathrm{L}$ solution $(X) + \mathrm{BaCl_2}$ solution (excess) $\rightarrow Z$ The number of moles of $Y$ and $Z$, respectively, are:
Answer: (b)
The reaction of $\left[ \mathrm{Co(NH_3)_5SO_4} \right]\mathrm{Br}$ with $\mathrm{AgNO_3}$ produces $\mathrm{AgBr}$ as a precipitate. The initial amount is $0.01 \, \mathrm{mol}$ and $\mathrm{AgNO_3}$ is in excess. The reaction of $\left[ \mathrm{Co(NH_3)_5Br} \right]\mathrm{SO_4}$ with $\mathrm{BaCl_2}$ produces $\mathrm{BaSO_4}$ as a precipitate. The initial amount is $0.01 \, \mathrm{mol}$ and $\mathrm{BaCl_2}$ is in excess.
Chemistry · Co-ordination Compounds · Single correct
The Cl - Co - Cl bond angle values in a fac-[$\mathrm{Co(NH_3)_3Cl_3}$] complex is/are:
Answer: (b)
The Cl - Co - Cl bond angle in above octahedral complex is $90^\circ$.
Chemistry · Haloalkanes and Haloarenes · Single correct
Decreasing order towards $S_N1$ reaction for the following compounds is:
Answer: (c)
The rate of $S_{N}1$ reaction depends upon stability of carbocation which follows the order $$OMe > H > Cl > NO_2$$. Therefore, the reactivity order is as shown.
Chemistry · Amines · Single correct
In the above conversion of compound (X) to product (Y), the sequence of reagents to be used will be:
Answer: (d)
The reaction sequence starts with the nitration of toluene to form nitrotoluene. This is followed by reduction using $Fe/H^+$ to convert the nitro group to an amino group, forming aminotoluene. Next, bromination with $Br_2$ in aqueous solution introduces bromine atoms to the aromatic ring. The amino group is then converted to a diazonium salt using $HNO_2$. Finally, the diazonium group is replaced by hydrogen using $H_3PO_2$, resulting in the formation of bromotoluene.
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Numerical
Number of compounds from the following which will not dissolve in cold $\mathrm{NaHCO_3}$ and $\mathrm{NaOH}$ solutions but will dissolve in hot $\mathrm{NaOH}$ solution is
Answer: 3
Compound 2, 3, 7
Chemistry · Chemistry in Everyday Life · Single correct
Given below are two statements: One is labelled as Assertion A and the other labelled as Reason R. Assertion A: Antihistamines do not affect the secretion of acid in stomach. Reason R: Antiallergic and antacid drugs work on different receptors. In the light of the above statements, choose the correct answer from the options given below:
Answer: (b)
Antiallergic and antacid drugs work on different receptors. NCERT(XII) vol. 2 page no. 451-452
Chemistry · Biomolecules · Fill in the blank
A short peptide on complete hydrolysis produces 3 moles of glycine (G), two moles of leucine (L) and two moles of valine (V) per mole of peptide. The number of peptide linkages in it are _______.
Answer: 6
Number of peptide linkage = (amino acid - 1) $$= 7 - 1 = 6$$
Maths · Complex Numbers and Quadratic Equations · Fill in the blank
If the value of real number $a > 0$ for which $x^2 - 5ax + 1 = 0$ and $x^2 - ax - 5 = 0$ have a common real roots is $\frac{3}{\sqrt{2\beta}}$ then $\beta$ is equal to .
Answer: 13
Two equations have a common root. Therefore, $(4a)(26a)=(-6)^2=36$. This implies $a^2=\frac{9}{26}$. Since $a>0$, $a=\frac{3}{\sqrt{26}}$. But $a=\frac{3}{\sqrt{2\beta}}$. Hence, $\ \beta=13$.
Maths · Permutations and Combinations · Single correct
The number of ways of selecting two numbers $a$ and $b$, $a \in \{2, 4, 6, \ldots, 100\}$ and $b \in \{1, 3, 5, \ldots, 99\}$ such that 2 is the remainder when $a + b$ is divided by 23 is
Answer: (c)
Given $a \in \{2, 4, 6, 8, 10, \ldots, 100\}$ and $b \in \{1, 3, 5, 7, 9, \ldots, 99\}$. Now, $a + b \in \{25, 71, 117, 163\}$. (i) $a + b = 25$, number of ordered pairs $(a, b)$ is 12. (ii) $a + b = 71$, number of ordered pairs $(a, b)$ is 35. (iii) $a + b = 117$, number of ordered pairs $(a, b)$ is 42. (iv) $a + b = 163$, number of ordered pairs $(a, b)$ is 19. Therefore, total = 108 pairs.
Maths · Permutations and Combinations · Numerical
The number of seven digits odd numbers, that can be formed using all the seven digits 1, 2, 2, 2, 3, 3, 5 is _____.
Answer: 240
Digits are 1, 2, 2, 2, 3, 3, 5 If unit digit 5, then total numbers = $$\frac{6!}{3!2!}$$ If unit digit 3, then total numbers = $$\frac{6!}{3!}$$ If unit digit 1, then total numbers = $$\frac{6!}{3!2!}$$ Therefore, total numbers = 60 + 60 + 120 = 240
Maths · Sequences and Series · Single correct
Let a, b, c > 1, $a^3$, $b^3$ and $c^3$ be in A.P., and $\log_b a$, $\log_c b$ and $\log_a c$ be in G.P. If the sum of first 20 terms of an A.P., whose first term is $\frac{a + 4b + c}{3}$ and the common difference is $\frac{a - 8b + c}{10}$ is $-444$, then $abc$ is equal to
Answer: (b)
As $a^3$, $b^3$, $c^3$ be in A.P. $\rightarrow a^3 + c^3 = 2b^3$ $\ldots$ (1) $\log_b a$, $\log_a c$, $\log_c b$ are in G.P. $$\frac{\log b}{\log a} \cdot \frac{\log c}{\log b} = \left(\frac{\log a}{\log c}\right)^2$$ $$\therefore (\log a)^3 = (\log c)^3 \Rightarrow a = c$$ $\ldots$ (2) From (1) and (2) $$a = b = c$$ $T_1 = \frac{a + 4b + c}{3} = 2a; \ d = \frac{a - 8b + c}{10} = \frac{-6a}{10} = \frac{-3}{5}a$ $$\therefore S_{20} = \frac{20}{2} \left[ 4a + 19 \left( \frac{-3}{5}a \right) \right]$$ $$= 10 \left[ \frac{20a - 57a}{5} \right]$$ $$= -74a$$ $$\therefore -74a = -444 \Rightarrow a = 6$$ $$\therefore abc = 6^3 = 216$$
Maths · Sequences and Series · Fill in the blank
The 8^{th} common term of the series S_1 = 3 + 7 + 11 + 15 + 19 + $\ldots$, S_2 = 1 + 6 + 11 + 16 + 21 + $\ldots$ is ______.
Answer: 151
Given $T_8 = 11 + (8 - 1) \times 20$. $$= 11 + 140 = 151$$
Maths · Binomial Theorem · Single correct
Let $x = \left( 8\sqrt{3} + 13 \right)^{13}$ and $y = \left( 7\sqrt{2} + 9 \right)^{9}$. If $[t]$ denotes the greatest integer $\leq t$, then
Answer: (a)
$x=(8\sqrt{3}+13)^{13}$ $={}^{13}C_{0}(8\sqrt{3})^{13} +{}^{13}C_{1}(8\sqrt{3})^{12}(13) +\cdots$ $x'=(8\sqrt{3}-13)^{13}$ $={}^{13}C_{0}(8\sqrt{3})^{13} -{}^{13}C_{1}(8\sqrt{3})^{12}(13) +\cdots$ $x-x'$ $=2\left[{}^{13}C_{1}(8\sqrt{3})^{12}(13) +{}^{13}C_{3}(8\sqrt{3})^{10}(13)^{3}+\cdots\right]$ Therefore $x-x'$ is even integer, hence [x] is even Now, $y=(7\sqrt{2}+9)^{9}$ $={}^{9}C_{0}(7\sqrt{2})^{9} +{}^{9}C_{1}(7\sqrt{2})^{8}(9) +\cdots +{}^{9}C_{2}(7\sqrt{2})^{7}(9)^{2} +\cdots$ $y'=(7\sqrt{2}-9)^{9}$ $={}^{9}C_{0}(7\sqrt{2})^{9} -{}^{9}C_{1}(7\sqrt{2})^{8}(9) +\cdots +{}^{9}C_{2}(7\sqrt{2})^{7}(9)^{2} -\cdots$ $y-y'$ $=2\left[ {}^{9}C_{1}(7\sqrt{2})^{8}(9) +{}^{9}C_{3}(7\sqrt{2})^{6}(9)^{3} +\cdots \right]$ $y-y'=$ is an even integer,hence $[y]$ is even.
Maths · Binomial Theorem · Numerical
$50^{th}$ root of a number x is 12 and $50^{th}$ root of another number y is 18. Then the remainder obtained on dividing (x + y) by 25 is .
Answer: 23
Given $x + y = 12^{50} + 18^{50} = (150 - 6)^{25} + (325 - 1)^{25}$. This simplifies to $25K - (6^{25} + 1) = 25K - ((5 + 1)^{25} + 1)$. Therefore, $= 25K_1 - 2$. The remainder is 23.
Maths · Three Dimensional Geometry · Numerical
Let $P(a_1, b_1)$ and $Q(a_2, b_2)$ be two distinct points on a circle with center $C(\sqrt{2}, \sqrt{3})$. Let $O$ be the origin and $OC$ be perpendicular to both $CP$ and $CQ$. If the area of the triangle $OCP$ is $\frac{\sqrt{35}}{2}$, then $a_1^2 + a_2^2 + b_1^2 + b_2^2$ is equal to _____.
Answer: 24
Given $\($ $\frac{1}{2}$ $\times$ $\mathrm{PC}$ $\times$ $\sqrt{5}$ = $\frac{\sqrt{35}}{2}$ $\)$; $\($ $\mathrm{PC}$ = $\sqrt{7}$ $\)$. $\($ a_1^2 + b_1^2 + a_2^2 + b_2^2 = $\mathrm{OP}$^2 + $\mathrm{OQ}$^2 $\)$ $\($ = 2 $\times$ (5 + 7) = 24 $\)$
Maths · Sequences and Series · Single correct
The parabolas : $ax^2 + 2bx + cy = 0$ and $dx^2 + 2ex + fy = 0$ intersect on the line $y = 1$. If $a$, $b$, $c$, $d$, $e$, $f$ are positive real numbers and $a$, $b$, $c$ are in G.P., then
Answer: (c)
Given $ax^4 + 2bx + c = 0$. This implies $ax^2 + 2\sqrt{ac}x + c = 0$ (since $b^2 = ac$). Therefore, $\left( x\sqrt{a} + \sqrt{c} \right)^2 = 0$. This gives $x^2 - \frac{\sqrt{c}}{\sqrt{a}} = 0$ $\ldots$ (1). Now, consider $dx^2 + 2ex + f = 0$. This implies $d\left( \frac{c}{a} \right) + 2e\left[ -\frac{\sqrt{c}}{\sqrt{a}} \right] + f = 0$. Therefore, $\frac{dc}{a} + f = 2e\sqrt{\frac{c}{a}}$. This implies $\frac{d}{a} + \frac{f}{c} = 2e\sqrt{\frac{1}{ac}}$. Therefore, $\frac{d}{a} + \frac{f}{c} = \frac{2e}{b}$ [as $b = \sqrt{ae}$]. Thus, $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in A.P.
Maths · Conic Sections · Single correct
Let A be a point on the x-axis. Common tangents are drawn from A to the curves $x^2 + y^2 = 8$ and $y^2 = 16x$. If one of these tangents touches the two curves at Q and R, then $(QR)^2$ is equal to
Answer: (d)
Given $y = mx + \frac{4}{m}$. $$\frac{\frac{4}{m}}{\sqrt{1 + m^2}} = 2\sqrt{2} \therefore m = \pm 1$$ $y = \pm x \pm 4$. Point of contact on parabola. Let $m = 1$, $\($ $\frac{a}{m^2}$, $\frac{2a}{m}$ $\)$. $R (4, 8)$ Point of contact on circle $Q (-2, 2)$ $$\therefore (QR)^2 = 36 + 36 = 72$$
Maths · Limits and Derivatives · Single correct
Let $f$, $g$ and $h$ be the real valued functions defined on $\mathbb{R}$ as $f(x) = \begin{cases} \frac{x}{|x|}, & x \neq 0 \\ 1, & x = 0 \end{cases}$, $g(x) = \begin{cases} \frac{\sin(x+1)}{(x+1)}, & x \neq -1 \\ 1, & x = -1 \end{cases}$ and $h(x) = 2[x] - f(x)$, where $[x]$ is the greatest integer $\leq x$. Then the value of $\lim_{x \to 1} g(h(x-1))$ is
Answer: (a)
LHL = $\lim$_{k $\to$ 0} g(h(-k)), $\;$ k > 0 = $\lim$_{k $\to$ 0} g(-2 + 1) $\therefore$ f(x) = -1 $\;$ $\forall$ $\;$ x 0 = $\lim$_{k $\to$ 0} g(-1) $\therefore$ f(x) = 1, $\;$ $\forall$ $\;$ x > 0 = 1
Maths · Mathematical Reasoning · Single correct
Consider the following statements: P : I have fever Q : I will not take medicine R : I will take rest The statement “If I have fever, then I will take medicine and I will take rest” is equivalent to:
Answer: (a)
Given $P \to (\sim Q \land R)$. This is equivalent to $\sim P \lor (\sim Q \land R)$. Applying distribution, we get $(\sim P \lor \sim Q) \land (\sim P \lor R)$.
Maths · Statistics · Single correct
Let S be the set of all values of $a_1$ for which the mean deviation about the mean of 100 consecutive positive integers $a_1, a_2, a_3, \ldots, a_{100}$ is 25. Then S is
Answer: (c)
Let $a_1$ be any natural number. $a_1, a_1 + 1, a_1 + 2, \ldots, a_1 + 99$ are values of $a_i$'s. $$\bar{x} = \frac{a_1 + (a_1 + 1) + (a_1 + 2) + \ldots + a_1 + 99}{100}$$ $$= \frac{100a_1 + (1 + 2 + \ldots + 99)}{100} = a_1 + \frac{99 \times 100}{2 \times 100}$$ $$= a_1 + \frac{99}{2}$$ Mean deviation about mean = $$\frac{\sum_{i=1}^{100} |x_i - \bar{x}|}{100}$$ $$2 \left( \frac{99}{2} + \frac{97}{2} + \frac{95}{2} + \ldots + \frac{1}{2} \right)$$ $$= \frac{100}{100}$$ $$= \frac{1 + 3 + \ldots + 99}{100}$$ $$= \frac{50}{2} [1 + 99]$$ $$= \frac{100}{100}$$ $$= 25$$
Maths · Matrices · Single correct
If $P$ is a $3 \times 3$ real matrix such that $P^T = aP + (a - 1)I$, where $a > 1$, then
Answer: (d)
Given $\mathbf{P}^1 = a \mathbf{P} + (a-1) \mathbf{I}$. Therefore, $\mathbf{P} = a \mathbf{P}^\mathrm{T} + (a-1) \mathbf{I}$. This implies $\mathbf{P}^\mathrm{T} - \mathbf{P} = a (\mathbf{P} - \mathbf{P}^\mathrm{T})$. Hence, $\mathbf{P} = \mathbf{P}^\mathrm{T}$, as $a \neq -1$. Now, $\mathbf{P} = a \mathbf{P} + (a-1) \mathbf{I}$. Therefore, $\mathbf{P} = -\mathbf{I} \implies |\mathbf{P}| = 1$. Thus, $|\mathrm{Adj} \mathbf{P}| = 1$.
Maths · Determinants · Single correct
For $\alpha, \beta \in \mathbb{R}$, suppose the system of linear equations $$x - y + z = 5$$ $$2x + 2y + \alpha z = 8$$ $$3x - y + 4z = \beta$$ has infinitely many solutions. Then $\alpha$ and $\beta$ are the roots of
Answer: (c)
The determinant of the matrix is given by: $$\begin{vmatrix} 1 & -1 & 1 \\ 2 & 2 & \alpha \\ 3 & -1 & 4 \end{vmatrix} = 0;$$ Expanding the determinant, we have: $$8 + \alpha - 2(-4 + 1) + 3(-\alpha - 2) = 0$$ Simplifying, we get: $$8 + \alpha + 6 - 3\alpha - 6 = 0$$ Solving for $\alpha$, we find: $$\alpha = 4$$
Maths · Inverse Trigonometric Functions · Single correct
Let $a_1 = 1, a_2, a_3, a_4, \ldots$ be consecutive natural numbers. Then $\tan^{-1}\left(\frac{1}{1+a_1a_2}\right) + \tan^{-1}\left(\frac{1}{1+a_2a_3}\right) + \ldots + \tan^{-1}\left(\frac{1}{1+a_{2021}a_{2022}}\right)$ is equal to
Answer: (c)
Given $a_2 - a_1 = a_3 - a_2 = \ldots = a_{2022} - a_{2021} = 1$. Therefore, $$\tan^{-1} \left( \frac{a_2 - a_1}{1 + a_1 a_2} \right) + \tan^{-1} \left( \frac{a_3 - a_2}{1 + a_2 a_3} \right) + \ldots + \tan^{-1} \left( \frac{a_{2022} - a_{2021}}{1 + a_{2021} a_{2022}} \right)$$ $$= \left[ (\tan^{-1} a_2) - \tan^{-1} a_1 \right] + \left[ \tan^{-1} a_3 - \tan^{-1} a_2 \right] + \ldots$$ $$+ \left[ \tan^{-1} a_{2022} - \tan^{-1} a_{2021} \right]$$ $$= \tan^{-1} a_{2022} - \tan^{-1} a_1$$ $$= \tan^{-1} (2022) - \tan^{-1} 1 = \tan^{-1} 2022 - \frac{\pi}{4} (option 3)$$ $$= \left( \frac{\pi}{2} - \cot^{-1} (2022) \right) - \frac{\pi}{4}$$ $$= \frac{\pi}{4} - \cot^{-1} (2022) (option 1)$$
Maths · Relations and Functions · Single correct
The range of the function $f\left( x \right) = \sqrt{3-x} + \sqrt{2+x}$ is
Answer: (a)
Given $$y^2 = 3 - x + 2 + x + 2 \sqrt{(3-x)(2+x)}$$ Simplifying, we have $$= 5 + 2 \sqrt{6 + x - x^2}$$ Rewriting, $$y^2 = 5 + 2 \sqrt{\frac{25}{4} - \left(x - \frac{1}{2}\right)^2}$$ The maximum value of $y$ is $$y_{\max} = \sqrt{5 + 5} = \sqrt{10}$$ The minimum value of $y$ is $$y_{\min} = \sqrt{5}$$
Maths · Relations and Functions · Numerical
Let A = $\{$1, 2, 3, 5, 8, 9$\}$. Then the number of possible functions $f : A \rightarrow A$ such that $f(m \cdot n) = f(m) \cdot f(n)$ for every $m, n \in A$ with $m \cdot n \in A$ is equal to .
Answer: 432
Given $f(1) = 1$; $f(9) = f(3) \times f(3)$. i.e., $f(3) = 1$ or $3$. Total function $= 1 \times 6 \times 2 \times 6 \times 6 \times 1 = 432$
Maths · Applications of Derivatives · Single correct
If the functions $f(x) = \frac{x^3}{3} + 2bx + \frac{ax^2}{2}$ and $g(x) = \frac{x^3}{3} + ax + bx^2$, $a \neq 2b$ have a common extreme point, then $a + 2b + 7$ is equal to
Answer: (d)
Given $f'(x) = x^2 + 2b + ax$ and $g'(x) = x^2 + a + 2bx$. The equation $(2b - a) - x(2b - a) = 0$ implies $x = 1$ is the common root. Put $x = 1$ in $f'(x) = 0$ or $g'(x) = 0$. $$1 + 2b + a = 0$$ $$7 + 2b + a = 6$$
Maths · Integrals · Numerical
If $\int \sqrt{\sec 2x - 1} \, dx = \alpha \log_e \left| \cos 2x + \beta + \sqrt{\cos 2x \left(1 + \cos \frac{1}{\beta} x \right)} \right|$ + constant, then $\beta - \alpha$ is equal to _____.
Answer: 1
The integral \[ \int \sqrt{\sec 2x - 1}\, dx \] is transformed as follows: \[ \int \sqrt{\frac{1-\cos 2x}{\cos 2x}}\, dx \] This simplifies to \[ \sqrt{2}\int \frac{\sin x}{\sqrt{2\cos^2 x-1}}\, dx. \] Let \[ \cos x=t, \] then \[ -\sin x\,dx=dt. \] Thus, the integral becomes \[ -\sqrt{2}\int \frac{dt}{\sqrt{2t^2-1}}. \] This results in \[ -\ln \left|\sqrt{2}\cos x+\sqrt{\cos 2x}\right|+c. \] Further simplification gives \[ -\frac{1}{2}\ln \left|2\cos^2 x+\cos 2x +2\sqrt{2}\cos x\,\sqrt{\cos 2x}\right|+c. \] Finally, \[ -\frac{1}{2}\ln \left|\cos 2x+\frac{1}{2} +\sqrt{\cos 2x}\,\sqrt{1+\cos 2x}\right|+c. \]
Maths · Applications of Integrals · Single correct
So, it is true for every natural no. $a_1$ \[ \lim_{n \to \infty} \frac{3}{n} \left\{ 4 + \left( 2 + \frac{1}{n} \right)^2 + \left( 2 + \frac{2}{n} \right)^2 + \ldots + \left( 3 - \frac{1}{n} \right)^2 \right\} \] is equal to
Answer: (d)
The limit is given by $$\lim_{n \to \infty} \frac{3}{n} \sum_{r=0}^{n-1} \left(2 + \frac{r}{n}\right)^2$$ which is equal to $$3 \int_{0}^{1} (2 + x)^2 \, dx = 27 - 8 = 19.$$
Maths · Complex Numbers and Quadratic Equations · Single correct
Let q be the maximum integral value of p in [0, 10] for which the roots of the equation $x^2 - px + \frac{5}{4}p = 0$ are rational. Then the area of the region $\{(x, y) : 0 \leq y \leq (x-q)^2, 0 \leq x \leq q\}$ is
Answer: (a)
Given the equation $x^2 - px + \frac{5p}{4} = 0$. The discriminant is $D = p^2 - 5p = p(p - 5)$. Therefore, $q = 9$. The inequality is $0 \leq y \leq (x - 9)^2$. The area is given by $$Area = \int_{0}^{9} (x - 9)^2 \, dx = 243.$$
Maths · Applications of Integrals · Numerical
Let A be the area of the region $$\left\{ (x, y) : y \geq x^2, y \geq (1-x)^2, y \leq 2x(1-x) \right\}$$ Then 540 A is equal to
Answer: 25
The area is given by $$A = 2 \int_{1/3}^{1/2} \left( 2x - 2x^2 - (1-x)^2 \right) \, dx$$ which simplifies to $$= 2 \left[ 2x^2 - x^3 - x \right]_{1/3}^{1/2}$$ Thus, $$\therefore A = \frac{5}{108} \implies 540A = \frac{5}{108} \times 540 = 25$$
Maths · Differential Equations · Single correct
The solution of the differential equation $$\frac{dy}{dx} = -\left(\frac{x^2 + 3y^2}{3x^2 + y^2}\right), \ y(1) = 0$$ is
Answer: (c)
Put $y = vx$. $$v + x \frac{dv}{dx} = -\left( \frac{1 + 3v^2}{3 + v^2} \right)$$ $$x \frac{dv}{dx} = -\frac{(v+1)^3}{3 + v^2}$$ $$\frac{(3 + v^2)}{(v+1)^3} dv + \frac{dx}{x} = 0$$ $$\int \frac{4 dv}{(v+1)^3} + \int \frac{dv}{v+1} - \int \frac{2 dv}{(v+1)^2} + \int \frac{dx}{x} = 0$$ $$\frac{-2}{(v+1)^2} + \ln(v+1) + \frac{2}{v+1} + \ln x = c$$ $$\frac{-2x^2}{(x+y)^2} + \ln \left( \frac{x+y}{x} \right) + \frac{2x}{x+y} + \ln x = c$$ $$\frac{2xy}{(x+y)^2} + \ln(x+y) = c$$ Therefore, $c = 0$, as $x = 1$, $y = 0$. $$\frac{2xy}{(x+y)^2} + \ln(x+y) = 0$$
Maths · Vector Algebra · Single correct
Let $\lambda \in \mathbb{R}$, $\vec{a} = \lambda\hat{i} + 2\hat{j} - 3\widehat{k}$, $\vec{b} = \hat{i} - \lambda\hat{j} + 2\widehat{k}$. If $\left( \left( \vec{a} + \vec{b} \right) \times \left( \vec{a} \times \vec{b} \right) \right) \times \left( \vec{a} - \vec{b} \right) = 8\hat{i} - 40\hat{j} - 24\widehat{k}$, then $\left| \lambda \left( \vec{a} + \vec{b} \right) \times \left( \vec{a} - \vec{b} \right) \right|^2$ is equal to
Answer: (a)
Given $\vec{a} = \lambda \hat{i} + 2 \hat{j} - 3 \hat{k}$ and $\vec{b} = \hat{i} - \lambda \hat{j} + 2 \hat{k}$. $$\Rightarrow (\vec{b} - \vec{a}) \times \left( (\vec{a} + \vec{b}) \times (\vec{a} \times \vec{b}) \right) = 8 \hat{i} - 40 \hat{j} - 24 \hat{k}$$ $$\Rightarrow \left( (\vec{a} - \vec{b}) \cdot (\vec{a} + \vec{b}) \right) (\vec{a} \times \vec{b}) = 8 \hat{i} - 40 \hat{j} - 24 \hat{k}$$ $$\Rightarrow 8 \left( \vec{a} \times \vec{b} \right) = 8 \hat{i} - 40 \hat{j} - 24 \hat{k}$$ Now, $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \lambda & 2 & -3 \\ 1 & -\lambda & 2 \end{vmatrix}$ $$= (4 - 3 \lambda) \hat{i} - (2 \lambda + 3) \hat{j} + (-\lambda^2 - 2) \hat{k}$$ $$\Rightarrow \lambda = 1$$ $$\therefore \vec{a} = \hat{i} + 2 \hat{j} - 3 \hat{k}$$ $$\vec{b} = \hat{i} - \hat{j} + 2 \hat{k}$$ $$\Rightarrow \vec{a} + \vec{b} = 2 \hat{i} + \hat{j} - \hat{k}, \vec{a} - \vec{b} = 3 \hat{j} - 5 \hat{k}$$ $$\Rightarrow (\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 0 & 3 & -5 \end{vmatrix} = 2 \hat{i} + 10 \hat{j} + 6 \hat{k}$$ $$\therefore required answer = 4 + 100 + 36 = 140$$
Maths · Vector Algebra · Single correct
Let $\vec{a}$ and $\vec{b}$ be two vectors. Let |$\vec{a}$| = 1, |$\vec{b}$| = 4 and $\vec{a}$ $\cdot$ $\vec{b}$ = 2. If $\vec{c}$ = (2$\vec{a}$ $\times$ $\vec{b}$) - 3$\vec{b}$, then the value of $\vec{b}$ $\cdot$ $\vec{c}$ is
Answer: (b)
Given $\vec{c} = (2 \vec{a} \times \vec{b}) - 3 \vec{b}$. Calculate $\vec{b} \cdot \vec{c}$. $$\vec{b} \cdot \vec{c} = \vec{b} \cdot (2 \vec{a} \times \vec{b}) - 3 \vec{b} \cdot \vec{b}$$ $$= -3 |\vec{b}|^2$$ $$= -48$$
Maths · Three Dimensional Geometry · Single correct
A vector $\vec{v}$ in the first octant is inclined to the x-axis at $60^\circ$, to the y-axis at $45^\circ$ and to the z-axis at an acute angle. If a plane passing through the points $\left(\sqrt{2}, -1, 1\right)$ and $(a, b, c)$, is normal to $\vec{v}$, then
Answer: (c)
Given $\hat{v} = \cos 60^\circ \hat{i} + \cos 45^\circ \hat{j} + \cos \gamma \hat{k}$. $$\Rightarrow \frac{1}{4} + \frac{1}{2} + \cos^2 \gamma = 1 (\gamma \rightarrow Acute)$$ $$\Rightarrow \cos \gamma = \frac{1}{2}$$ $$\Rightarrow \gamma = 60^\circ$$ Equation of plane is $$\frac{1}{2} \left( x - \sqrt{2} \right) + \frac{1}{\sqrt{2}} \left( y + 1 \right) + \frac{1}{2} (z - 1) = 0$$ $$\Rightarrow x + \sqrt{2} y + z = 1$$ $(a, b, c)$ lies on it. $$\Rightarrow a + \sqrt{2} b + c = 1$$
Maths · Three Dimensional Geometry · Single correct
If a plane passes through the points $(-1, k, 0)$, $(2, k, -1)$, $(1, 1, 2)$ and is parallel to the line $$\frac{x-1}{1} = \frac{2y+1}{2} = \frac{z+1}{-1}$$, then the value of $$\frac{k^2+1}{(k-1)(k-2)}$$ is
Answer: (d)
Given the equations $\($ $\frac{x-1}{1}$ = $\frac{2y+1}{2}$ = $\frac{z+1}{-1}$ $\)$ and $\($ $\frac{x-1}{1}$ = $\frac{y+1}{1}$ = $\frac{z+1}{-1}$ $\)$. Points: $\($ A(-1, k, 0) $\)$, $\($ B(2, k, -1) $\)$, $\($ C(1, 1, 2) $\)$. $\($ $\overrightarrow{CA}$ = -2$\hat{i}$ + (k-1)$\hat{j}$ - 2$\hat{k}$ $\)$ $\($ $\overrightarrow{CB}$ = $\hat{i}$ + (k-1)$\hat{j}$ - 3$\hat{k}$ $\)$ The cross product $\($ $\overrightarrow{CA}$ $\times$ $\overrightarrow{CB}$ = $\begin{vmatrix}$ $\hat{i}$ & $\hat{j}$ & $\hat{k}$ $\\$ -2 & k-1 & -2 $\\$ 1 & k-1 & -3 $\end{vmatrix}$ $\)$ $\($ = $\hat{i}$(3k + 3 + 2k - 2) - $\hat{j}$(6 + 2) + $\hat{k}$(2k + 2 - k + 1) $\)$ $\($ = (1-k)$\hat{i}$ - 8$\hat{j}$ + (3-3k)$\hat{k}$ $\)$ The line $\($ $\frac{x-1}{1}$ = $\frac{y+1}{2}$ = $\frac{z+1}{-1}$ $\)$ is perpendicular to the normal vector. Therefore, $\($ 1 $\cdot$ (1-k) + 1 $\cdot$ (-8) + (-1)(3-3k) = 0 $\)$ $\($ $\Rightarrow$ 1-k-8-3+3k = 0 $\)$ $\($ $\Rightarrow$ 2k = 10 $\Rightarrow$ k = 5 $\)$ Thus, $\($ $\frac{k^2+1}{(k-1)(k-2)}$ = $\frac{26}{4 \cdot 3}$ = $\frac{13}{6}$ $\)$
Maths · Three Dimensional Geometry · Fill in the blank
Let a line L pass through the point P(2, 3, 1) and be parallel to the line $x + 3y - 2z - 2 = 0 = x - y + 2z$. If the distance of L from the point (5, 3, 8) is $\alpha$, then $3\alpha^2$ is equal to ____.
Answer: 158
The equation of the line is $\($ $\frac{x-2}{1}$ = $\frac{y-3}{-1}$ = $\frac{z-1}{-1}$ $\)$. Let $\($ Q $\)$ be $\($ (5, 3, 8) $\)$ and the foot of the perpendicular from $\($ Q $\)$ on this line be $\($ R $\)$. Now, $\($ R $\equiv$ (k+2, -k+3, -k+1) $\)$. The direction ratios of $\($ QR $\)$ are $\($ (k-3, -k, -k-7) $\)$. Therefore, $$ (1)(k-3) + (-1)(-k) + (-1)(-k-7) = 0 $$ Solving gives $\($ k = -$\frac{4}{3}$ $\)$. Thus, $$ \alpha^2 = \left( \frac{13}{3} \right)^2 + \left( \frac{4}{3} \right)^2 + \left( \frac{17}{3} \right)^2 = \frac{474}{9} $$ Therefore, $$ 3\alpha^2 = 158 $$
Maths · Probability · Numerical
A bag contains six balls of different colours. Two balls are drawn in succession with replacement. The probability that both the balls are of the same colour is $p$. Next four balls are drawn in succession with replacement and the probability that exactly three balls are of the same colours is $q$. If $p : q = m : n$, where $m$ and $n$ are coprime, then $m + n$ is equal to
Answer: 14
Given $\($ p = $\frac{{^6C_1}}{{6 \times 6}}$ = $\frac{1}{6}$ $\)$. $\($ q = $\frac{{^6C_1 \times ^5C_1 \times 4}}{{6 \times 6 \times 6 \times 6}}$ = $\frac{5}{54}$ $\)$. Therefore, $\($ p : q = 9 : 5 $\Rightarrow$ m + n = 14 $\)$.