JEE Main 31 January 2023 Shift 1: 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 · Alternating Current · Single correct
If $R$, $X_L$, and $X_C$ represent resistance, inductive reactance and capacitive reactance. Then which of the following is dimensionless:
Answer: (b)
All three have same dimension therefore $$\frac{R}{\sqrt{X_L X_C}}$$ is dimensionless. Option 2
Physics · Motion in a Plane · Single correct
The initial speed of a projectile fired from ground is $u$. At the highest point during its motion, the speed of projectile is $\frac{\sqrt{3}}{2} u$. The time of flight of the projectile is:
Answer: (b)
Given $u \cos \theta = \frac{\sqrt{3}u}{2}$, it implies $\cos \theta = \frac{\sqrt{3}}{2}$. Therefore, $\theta = 30^\circ$. The time period $T$ is given by $$T = \frac{2u \sin 30^\circ}{g} = \frac{u}{g}.$$
Physics · Laws of Motion · Single correct
As shown in figure, a 70 kg garden roller is pushed with a force of $\vec{F} = 200 \, \mathrm{N}$ at an angle of $30^\circ$ with horizontal. The normal reaction on the roller is (Given $g = 10 \, \mathrm{m \, s^{-2}}$)
Answer: (c)
The normal force is given by the equation: $$N = mg + F \sin 30^\circ$$ Substituting the given values: $$= 700 + 200 \times \frac{1}{2} = 800 newton.$$
Physics · System of Particles and Rotational Motion · Single correct
100 balls each of mass m moving with speed v simultaneously strike a wall normally and reflected back with same speed, in time t s. The total force exerted by the balls on the wall is
Answer: (b)
Given $\vec{P}_i = N m v \hat{i}$ and $\vec{P}_f = -N m v \hat{i}$. $N$ is the number of balls striking the wall, $N = 100$. The change in momentum is $$\Delta \vec{P} = \vec{P}_f - \vec{P}_i = -2 N m v \hat{i} = -200 N m v \hat{i}.$$ The total force is given by $$\vec{F}_{Total} = \frac{\Delta \vec{P}}{\Delta t} = \frac{-200 m v t}{t}.$$ The magnitude of the force is $$|\vec{F}| = \frac{200 m v}{t}.$$
Physics · Gravitation · Single correct
At a certain depth "d" below surface of earth, value of acceleration due to gravity becomes four times that of its value at a height 3R above earth surface. Where R is Radius of earth (Take R = 6400 $\,$ $\mathrm{km}$). The depth d is equal to
Answer: (d)
Given $\frac{GM}{R^2}\left[1-\frac{d}{R}\right]=\frac{4GM}{(4R)^2}$. $1-\frac{d}{R}=\frac{1}{4}\Rightarrow\frac{d}{R}=\frac{3}{4}\Rightarrow d=\frac{3}{4}R$. $(d=4800\,\mathrm{km})$
Physics · Gravitation · Single correct
Spherical insulating ball and a spherical metallic ball of same size and mass are dropped from the same height. Choose the correct statement out of the following {Assume negligible air friction}
Answer: (b)
When metal is passing through magnetic field, eddy current will produce and it will oppose the motion, so it will take more time.
Physics · Mechanical Properties of Fluids · Single correct
If 1000 droplets of water of surface tension $0.07 \mathrm{N/m}$ having same radius $1 \mathrm{mm}$ each, combine to form a single drop. In the process the released surface energy is- $\($ Take $\pi$ = $\frac{22}{7}$ $\)$
Answer: (b)
Given $1000 \times \frac{4\pi}{3} (1)^3 = \frac{4\pi}{3} R^3$. $R = 10 \, \mathrm{mm}$. $T \times 1000 \times 4\pi \,(10^{-3})^2 - T \times 4\pi \,(10 \times 10^{-3})^2 = \Delta E$ $$\Delta E = 4 \times \pi \times 7 \times 10^{-2} [1000 - 100] \times 10^{-6}$$ $$\Delta E = 7.92 \times 10^{-4} \, \mathrm{J}$$ Option 2.
Physics · Thermodynamics · Single correct
The pressure of a gas changes linearly with volume from A to B as shown in figure. If no heat is supplied to or extracted from the gas then change in the internal energy of the gas will be
Answer: (d)
As $\Delta q = 0$ $\Delta u = -W$ $W = \int PdV$ $\Delta u = -W = 30 \times 10^3 \times 150 \times 10^{-6}$ $= 4500 \times 10^{-3}$ $= 4.5 \, \mathrm{J}$
Physics · Thermodynamics · Single correct
The correct relation between $\gamma = \frac{C_p}{c_v}$ and temperature $T$ is :
Answer: (b)
γ is independent of temperature. Option 2
Physics · Oscillations · Single correct
The maximum potential energy of a block executing simple harmonic motion is 25 J. $A$ is amplitude of oscillation. At $A/2$, the kinetic energy of the block is
Answer: (c)
Given $u_{max} = \frac{1}{2} m \omega^2 A^2 = 25 \, J$. KE at $\frac{A}{2} = \frac{1}{2} m v_1^2 = \frac{1}{2} m \omega^2 \left( A^2 - \frac{A^2}{4} \right)$. $$KE = \frac{1}{2} m \omega^2 \frac{3A^2}{4} = \frac{3}{4} \left( \frac{1}{2} m \omega^2 A^2 \right)$$ $$KE = \frac{3}{4} \times 25 = 18.75 \, J$$
Physics · Electric Charges and Fields · Single correct
Which of the following correctly represents the variation of electric potential (V) of a charged spherical conductor of radius (R) with radial distance (r) from the centre?
Answer: (c)
Physics · Current Electricity · Single correct
The drift velocity of electrons for a conductor connected in an electrical circuit is $V_d$. The conductor is now replaced by another conductor with same material and same length but double the area of cross section. The applied voltage remains same. The new drift velocity of electrons will be
Answer: (a)
Given $$V_d = \frac{eE}{m} \tau$$ that is independent of area.
Physics · Magnetism and Matter · Single correct
A bar magnet with a magnetic moment $5.0\text{ A }\text{m}^2$ is placed in parallel position relative to a magnetic field of $0.4\text{ T}$. The amount of required work done in turning the magnet from parallel to antiparallel position relative to the field direction is ________.
Answer: (a)
Given $u = -MB \cos \theta$. $W = \Delta u$. $W = -MB \cos 180^\circ (-mB \cos 0^\circ)$. $W = 2 \times MB = 2 \times 5 \times 0.4 = 4 \, \mathrm{J}$. Option 1
Physics · Moving Charges and Magnetism · Single correct
A rod with circular cross-section area $2 \, \mathrm{cm}^2$ and length $40 \, \mathrm{cm}$ is wound uniformly with 400 turns of an insulated wire. If a current of $0.4 \, \mathrm{A}$ flows in the wire windings, the total magnetic flux produced inside windings is $4\pi \times 10^{-6} \, \mathrm{Wb}$. The relative permeability of the rod is (Given: Permeability of vacuum $\mu_0 = 4\pi \times 10^{-7} \, \mathrm{NA}^{-2}$)
Answer: (c)
The magnetic flux $\phi$ is given by the formula $$\phi = \mu_r \mu_o \frac{N}{\ell} I \times A.$$ Given $\mu_r = 125$. The correct option is Option 3.
Physics · Wave Optics · Single correct
Two polaroide A and B are placed in such a way that the pass-axis of polaroids are perpendicular to each other. Now, another polaroid C is placed between A and B bisecting angle between them. If intensity of unpolarised light is $I_0$ then intensity of transmitted light after passing through polaroid B will be:
Answer: (c)
Given $I_A = \frac{I_o}{2}$. $I_C = \frac{I_o}{2} \cos^2 45 = \frac{I_o}{4}$. $I_B = I_C \cos^2 45 = \frac{I_o}{8}$. Option 3.
Physics · Wave Optics · Single correct
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R Assertion A: The beam of electrons shows wave nature and exhibit interference and diffraction. Reason R: Davisson Germer Experimentally verified the wave nature of electrons. In the light of the above statements. Choose the most appropriate answer from the options given below:
Answer: (d)
Conceptual
Physics · Dual Nature of Radiation and Matter · Single correct
If a source of electromagnetic radiation having power $15 \, \mathrm{kW}$ produces $10^{16}$ photons per second, the radiation belongs to a part of spectrum is. (Take Planck constant $h = 6 \times 10^{-34} \, \mathrm{Js}$)
Answer: (c)
Energy of one photon = $\frac{\text{Power}}{\text{Photon frequency}}$ \[ E = h \nu = \frac{15 \times 10^3}{10^{16}} \] \[ \nu = \frac{15 \times 10^{-13}}{6 \times 10^{-34}} = 2.5 \times 10^{21} \] So gamma Rays. Option 3.
Physics · Nuclei · Single correct
A free neutron decays into a proton but a free proton does not decay into neutron. This is because
Answer: (d)
As neutron has more rest mass than proton it will require energy to decay proton into neutron. Option 4.
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
The effect of increase in temperature on the number of electrons in conduction band ($n_e$) and resistance of a semiconductor will be as:
Answer: (c)
As temperature increases, more electrons excite to the conduction band and hence conductivity increases, therefore resistance decreases.
Physics · Communication Systems · Multiple correct
The amplitude of $15\sin(1000\pi t)$ is modulated by $10\sin(4\pi t)$ signal. The amplitude modulated signal contains frequency (ies) of (A) $500\text{ Hz}$ (B) $2\text{ Hz}$ (C) $250\text{ Hz}$ (D) $498\text{ Hz}$ (E) $502\text{ Hz}$ Choose the correct answer from the options given below:
Answer: (b)
Carrier wave frequency $$V_C = \frac{100\pi}{2\pi} = 500 \, Hz$$ Modulating wave frequency $$V_m = \frac{4\pi}{2\pi} = 2 \, Hz$$ Therefore, $V_C - V_m$, $V_C$, $V_C + V_m$ $$= 498 \, Hz, \, 500 \, Hz, \, 502 \, Hz$$
Physics · Motion in a Plane · Numerical
The speed of a swimmer is $4 \, \mathrm{km} \, \mathrm{h}^{-1}$ in still water. If the swimmer makes his strokes normal to the flow of river of width $1 \, \mathrm{km}$, he reaches a point $750 \, \mathrm{m}$ down the stream on the opposite bank. The speed of the river water is _________ $\mathrm{km} \, \mathrm{h}^{-1}$.
Answer: 3
Time to cross the river width $\omega = 1000 \, \mathrm{m}$ is $= \frac{1 \, \mathrm{km}}{4 \, \mathrm{km/h}}$. Drift $x = \mathrm{Vm/g} \times t$ where $\mathrm{Vm/g}$ is the velocity of the river with respect to the ground.
Physics · Laws of Motion · Numerical
A lift of mass $M = 500 \, \mathrm{kg}$ is descending with speed of $2 \, \mathrm{ms^{-1}}$. Its supporting cable begins to slip thus allowing it to fall with a constant acceleration of $2 \, \mathrm{ms^{-2}}$. The kinetic energy of the lift at the end of fall through to a distance of $6 \, \mathrm{m}$ will be ____ kJ.
Answer: 7
Given $v^2 = u^2 + 2as$. $$= 2^2 + 2 \times (2) \times (6)$$ $$= 4 + 24 = 28$$ The kinetic energy is given by $\mathrm{KE} = \frac{1}{2} mv^2$. $$= \frac{1}{2} (500) \times 28$$ $$= 7000 \, \mathrm{J}$$ $$= 7 \, \mathrm{kJ}$$
Physics · System of Particles and Rotational Motion · Numerical
A solid sphere of mass 1 kg rolls without slipping on a plane surface. Its kinetic energy is $7 \times 10^{-3}$ J. The speed of the centre of mass of the sphere is _____ cm s$^{-1}$.
Answer: 10
Given $\($ $\frac{1}{2}$ mv^2 + $\frac{1}{2}$ I $\omega$^2 = 7 $\times$ 10^{-3} $\)$. $\[$ $\frac{1}{2}$ mv^2 + $\frac{1}{2}$ $\left$( $\frac{2}{5}$ MR^2 $\right$) $\left$( $\frac{V}{R}$ $\right$)^2 = 7 $\times$ 10^{-3} $\]$ $\[$ $\frac{1}{2}$ MV^2 $\left$[ 1 + $\frac{2}{5}$ $\right$] = 7 $\times$ 10^{-3} $\]$ $\[$ $\frac{1}{2}$ (1)(V^2) $\left$( $\frac{7}{5}$ $\right$) = 7 $\times$ 10^{-3} $\]$ $\[$ V^2 = 10^{-2} $\]$ $\($ V = 10^{-1} = 0.1 $\,$ $\mathrm{m/s}$ = 10 $\,$ $\mathrm{cm/s}$ $\)$ Ans : 10
Physics · Mechanical Properties of Solids · Numerical
A thin rod having a length of 1 m and area of cross-section $3 \times 10^{-6} \, \mathrm{m}^2$ is suspended vertically from one end. The rod is cooled from $210^\circ \mathrm{C}$ to $160^\circ \mathrm{C}$. After cooling, a mass $M$ is attached at the lower end of the rod such that the length of rod again becomes 1 m. Young's modulus and coefficient of linear expansion of the rod are $2 \times 10^{11} \, \mathrm{N} \, \mathrm{m}^{-2}$ and $2 \times 10^{-5} \, \mathrm{K}^{-1}$, respectively. The value of $M$ is ________ kg. (Take $g = 10 \, \mathrm{m} \, \mathrm{s}^{-2}$)
Answer: 60
If $\Delta \ell$ is decrease in length of rod due to decrease in temperature $$\Delta \ell = \ell \alpha \Delta T$$ $$\alpha = 2 \times 10^{-5} \, \mathrm{K}^{-1}, \Delta T = (210 - 160)$$ $$= 50 \, \mathrm{K}$$ $$\Delta \ell = 1 \times 2 \times 10^{-5} \times 50 = 10^{-3} \, \mathrm{m}$$ Young Modulus $= Y = \frac{F/A}{\Delta \ell / \ell}$, $A = 3 \times 10^{-6} \, \mathrm{m}^2$ $$2 \times 10^{11} = \frac{Mg/3 \times 10^{-6}}{10^{-3}/1}$$ $$Mg = 2 \times 10^{11} \times 3 \times 10^{-9} = 6 \times 10^{-2}$$ $$M = 60 \, \mathrm{kg}$$ Ans is 60.
Physics · Oscillations · Numerical
In the figure given below, a block of mass $M = 490 \, \mathrm{g}$ placed on a frictionless table is connected with two springs having same spring constant $(K = 2 \, \mathrm{N} \, \mathrm{m}^{-1})$. If the block is horizontally displaced through 'X'm then the number of complete oscillations it will make in $14\pi$ seconds will be ________
Answer: 20
Keff = K + K as both springs are in use in parallel. = 2k = 2 $\times$ 2 = 4 \, $\mathrm{N/m}$ m = 490 \, $\mathrm{gm}$ = 0.49 \, $\mathrm{kg}$ T = 2$\pi$ $\sqrt{\frac{m}{\mathrm{Keff}}}$ = 2$\pi$ $\sqrt{\frac{0.49 \, \mathrm{kg}}{4}}$ = 2$\pi$ $\sqrt{\frac{49}{400}}$ = 2$\pi$ $\frac{7}{20}$ = $\frac{7\pi}{10}$ No. of oscillation in the 14$\pi$ is N = $\frac{time}{T}$ = $\frac{14\pi}{7\pi / 10}$ = 20 Ans in 20.
Physics · Electric Charges and Fields · Fill in the blank
Expression for an electric field is given by $$\vec{E} = 4000x^2 \hat{i}\text{ V}\text{m}^{-1}.$$ The electric flux through the cube of side $20\text{ cm}$ when placed in electric field (as shown in the figure) is ______ $\text{V}\text{ cm}$.
Answer: 640
Flux = $\vec{E} \cdot \vec{A}$ $$= 4000 \left(0.2\right)^2 \frac{\mathrm{V}}{\mathrm{m}} \cdot \left(0.2\right)^2 \mathrm{m}^2$$ $$= 4000 \times 16 \times 10^{-4} \, \mathrm{Vm}$$ $$= 640 \, \mathrm{Vcm}$$ Ans. 640
Physics · Current Electricity · Numerical
Two identical cells, when connected either in parallel or in series gives same current in an external resistance $5\,\Omega$. The internal resistance of each cell will be
Answer: 5
Parallel $$i = \frac{2\varepsilon}{5 + 2r} \ldots (1)$$ $$i = \frac{\varepsilon}{\frac{r}{2} + 5} \ldots (2)$$ Equating (1) and (2) $$\frac{2\varepsilon}{5 + 2r} = \frac{\varepsilon}{\frac{r}{2} + 5} \implies r + 10 = 5 + 2r$$ $$r = 5$$ Ans. 5
Physics · Electromagnetic Induction · Numerical
An inductor of $0.5 \, \mathrm{mH}$, a capacitor of $20 \, \mu\mathrm{F}$ and resistance of $20 \, \Omega$ are connected in series with a $220 \, \mathrm{V}$ ac source. If the current is in phase with the emf, the amplitude of current of the circuit is $\sqrt{x}$ A. The value of $x$ is -
Answer: 242
Given $X_L = X_C$. So, $Z = R = 20 \, \Omega$. $$i_{rms} = \frac{220}{20} = 11$$ $$i_{max} = 11 \sqrt{2} = \sqrt{242}$$ Answer: 242
Physics · Electromagnetic Waves · Numerical
In a medium the speed of light wave decreases to $0.2$ times to its speed in free space. The ratio of relative permittivity to the refractive index of the medium is $x : 1$. The value of $x$ is _____. (Given speed of light in free space $= 3 \times 10^8 \, \mathrm{m \, s^{-1}}$ and for the given medium $\mu_r = 1$)
Answer: 5
Given $V = \frac{C}{\mu}$, we have $\mu = \frac{C}{V} = \frac{C}{0.2C}$. Given $\mu = 5$. We know $\mu = \sqrt{\epsilon_r \mu_r}$. Thus, $\epsilon_r = \frac{\mu^2}{\mu_r}$. Therefore, $\epsilon_r = \mu = 5$.
Physics · Atoms · Numerical
For hydrogen atom, $\lambda_1$ and $\lambda_2$ are the wavelengths corresponding to the transitions 1 and 2 respectively as shown in figure. The ratio of $\lambda_1$ and $\lambda_2$ is $\frac{x}{32}$. The value of $x$ is _____.
Answer: 27
Given $\($ $\frac{1}{\lambda}$ = Rz^2 $\left$[ $\frac{1}{n_1^2}$ - $\frac{1}{n_2^2}$ $\right$] $\)$. For $\($ $\frac{1}{\lambda_1}$ = Rz^2 $\left$[ $\frac{1}{1^2}$ - $\frac{1}{3^2}$ $\right$] = $\frac{8}{9}$ Rz^2 $\)$ $\($ $\ldots$ (1) $\)$. For $\($ $\frac{1}{\lambda_2}$ = Rz^2 $\left$[ $\frac{1}{1^2}$ - $\frac{1}{2^2}$ $\right$] = $\frac{3}{4}$ Rz^2 $\)$ $\($ $\ldots$ (2) $\)$. $\($ $\frac{1}{2}$ $\Rightarrow$ $\frac{\lambda_2}{\lambda_1}$ = $\frac{8}{9}$ $\times$ $\frac{4}{3}$ = $\frac{32}{27}$ $\)$. Thus, $\($ $\frac{\lambda_1}{\lambda_2}$ = $\frac{27}{32}$ $\)$. Answer: 27.
Chemistry · Structure of Atom · Single correct
Which transition in the hydrogen spectrum would have the same wavelength as the Balmer type transition from $n=4$ to $n=2$ of $\mathrm{He}^+$ spectrum
Answer: (a)
He$^+$ ion: $$\frac{1}{\lambda(\mathrm{H})} = \mathrm{R}(1)^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$$ $$\frac{1}{\lambda(\mathrm{He}^+)} = \mathrm{R}(2)^2 \left[ \frac{1}{2^2} - \frac{1}{4^2} \right]$$ Given $\lambda(\mathrm{H}) = \lambda(\mathrm{He}^+)$ $$\mathrm{R}(1)^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right] = \mathrm{R}(4) \left[ \frac{1}{2^2} - \frac{1}{4^2} \right]$$ $$\frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{1}{1^2} - \frac{1}{2^2}$$ On comparing $n_1 = 1$ and $n_2 = 2$ Ans. 1
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
The correct increasing order of the ionic radii is
Answer: (d)
In isoelectronic species, size is proportional to $\frac{1}{Z}$. Therefore, $\mathrm{Ca^{2+}} < \mathrm{K^+} < \mathrm{Cl^-} < \mathrm{S^{2-}}$ in terms of size.
Chemistry · Co-ordination Compounds · Single correct
Match List I with List II Choose the correct answer from the options given below:
Answer: (d)
The molecular geometry of the given compounds are as follows: (A) $\mathrm{XeF_4}$ is square planar. (B) $\mathrm{SF_4}$ is see-saw shaped. (C) $\mathrm{NH_4^+}$ is tetrahedral. (D) $\mathrm{BrF_3}$ is bent T-shaped.
Chemistry · Hydrogen · Single correct
$H_2O_2$ acts as a reducing agent in
Answer: (a)
The reaction is given by: $$\mathrm{NaOCl} + \mathrm{H_2O_2} \rightarrow 2\mathrm{NaCl} + \mathrm{H_2O} + \mathrm{O_2}$$ The oxidation state of chlorine in $\mathrm{NaOCl}$ is $+1$ and changes to $-1$ in $\mathrm{NaCl}$. The oxidation state of oxygen in $\mathrm{H_2O_2}$ is $-1$ and changes to $0$ in $\mathrm{O_2}$.
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Match items of column I and II
Answer: (c)
A. $\mathrm{H_2O/CH_2Cl_2} \rightarrow ii$, $\mathrm{CH_2Cl_2} > \mathrm{H_2O}$ (density) so they can be separated by differential solvent extraction. B. Due to H-bonding in $$\begin{array}{c} \mathrm{\begin{matrix} \end{matrix}} \end{array}$$ it can be separated from $$\begin{array}{c} \mathrm{\begin{matrix} \end{matrix}} \end{array}$$ by column chromatography. C. Kerosene / Naphthalene $\rightarrow iv$. Fractional distillation. Due to different B.P. of kerosene and Naphthalene it can be separated by fractional distillation. D. $\mathrm{C_6H_{12}O_6/NaCl} \rightarrow i$. Crystallization. NaCl (ionic compound) can be crystallized.
Chemistry · Hydrocarbons · Single correct
Choose the correct set of reagents for the following conversion trans ($\mathrm{Ph{-}CH{=}CH{-}CH_3}$) $\rightarrow$ cis ($\mathrm{Ph{-}CH{=}CH{-}CH_3}$)
Answer: (b)
The given reaction sequence involves the following steps: 1. The starting compound is phenylpropene, $Ph-CH=CH-CH_3$. 2. It reacts with $Br_2$ to form a dibromo compound, $Ph-CHBr-CHBr-CH_3$. 3. Treatment with alcoholic KOH leads to dehydrohalogenation, forming an alkyne, $Ph-C\equiv C-CH_3$. 4. Finally, hydrogenation in the presence of Lindlar's catalyst converts the alkyne to a cis-alkene, $Ph-CH=CH-CH_3$.
Chemistry · Electrochemistry · Single correct
Which one of the following statements is correct for electrolysis of brine solution?
Answer: (d)
Electrolysis of brine solution $NaCl(aq)\longrightarrow Na^+_{(aq)}+Cl^-_{(aq)}$ At anode : $2Cl^-_{(aq)}\longrightarrow Cl_{2(g)}+2e^-$ (Major) $2H_2O_{(\ell)}\longrightarrow O_{2(g)}+4H^+_{(aq)}+4e^-$ (Minor) At cathode : $2H_2O_{(\ell)}+2e^-\longrightarrow H_{2(g)}\uparrow+2OH^-_{(aq)}$ $2Na^++2OH^-\longrightarrow2NaOH$
Chemistry · Surface Chemistry · Single correct
Adding surfactants in non polar solvent, the micelles structure will look like
Answer: (c)
Non-Polar tail towards non-polar solvent
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
The methods NOT involved in concentration of ore are (A) Liquation (B) Leaching ( C ) Electrolysis (D) Hydraulic washing (E) Froth floatation Choose the correct answer from the options given below :
Answer: (c)
Methods involved in concentration of ore are (i) Hydraulic Washing (ii) Froth Flotation (iii) Magnetic Separation
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
Identify X, Y and Z in the following reaction. (Equation not balanced) $$\mathrm{ClO} + \mathrm{NO}_2 \rightarrow \mathrm{X} \xrightarrow{\mathrm{H_2O}} \mathrm{Y} + \mathrm{Z}$$
Answer: (c)
The reaction is given as: $$\mathrm{ClO} + \mathrm{NO_2} \rightarrow \mathrm{ClONO_2} \xrightarrow{+\mathrm{H_2O}} \mathrm{HOCl} + \mathrm{HNO_3}$$
Chemistry · Redox Reactions · Single correct
When $\mathrm{Cu}^{2+}$ ion is treated with KI, a white precipitate, X appears in solution. The solution is titrated with sodium thiosulphate, the compound Y is formed. X and Y respectively are
Answer: (b)
$\mathrm{Cu^{2+}+2KI \rightarrow CuI_2\downarrow + 2K^+}$ (Unstable) $\mathrm{I^-}$ is a strong reducing agent; it reduces $\mathrm{Cu^{2+}}$ to $\mathrm{Cu^+}$. $2\mathrm{CuI_2} \rightarrow \mathrm{Cu_2I_2}\downarrow + \mathrm{I_2}$ (White precipitate) $\;X$ $\mathrm{KI + I_2 \rightarrow KI_3}$ (Brown solution) $\mathrm{I_3^- \rightleftharpoons I_2 + I^-}$ $\mathrm{KI_3 + 2Na_2S_2O_3 \rightarrow KI + NaI + Na_2S_4O_6}$ $\;Y$
Chemistry · The d-and f-Block Elements · Single correct
The correct order of basicity of oxides of vanadium is
Answer: (a)
With increase in % of oxygen, acidic nature of oxide of an element increases and basic nature decreases.
Chemistry · The d-and f-Block Elements · Single correct
$\mathrm{Nd^{2+}} =$ _____
Answer: (b)
The electronic configuration of Nd (60) is $[\mathrm{Xe}] \, 4f^4 \, 5d^0 \, 6s^2$. For $\mathrm{Nd}^{2+}$, the configuration is $[\mathrm{Xe}] \, 4f^4 \, 5d^0 \, 5s^0$.
Chemistry · Co-ordination Compounds · Single correct
Cobalt chloride when dissolved in water forms pink colored complex X which has octahedral geometry. This solution on treating with cone HCl forms deep blue complex, Y which has a Z geometry. X, Y and Z, respectively, are
Answer: (a)
The reaction is as follows: $$\mathrm{CoCl_2 + 6H_2O \rightarrow [Co(H_2O)_6]Cl_2}$$ Pink (X) octahedral. Upon addition of concentrated HCl, $$\mathrm{[CoCl_4]^{2-}}$$ is formed, which is a blue solution (Y) and tetrahedral (Z).
Chemistry · Haloalkanes and Haloarenes · Single correct
The correct order of melting point of dichlorobenzenes is
Answer: (d)
The boiling points and melting points for the compounds are given as follows: Boiling point in Kelvin: 453, 446, 448 Melting point in Kelvin: 256, 249, 323 The compound with the highest melting point is more symmetrical and has better crystal fitting, leading to maximum packing efficiency. The compound with the highest boiling point has the maximum dipole moment $\mu_{max}$.
Chemistry · Alcohols, Phenols and Ethers · Single correct
An organic compound ‘A’ with empirical formula $C_6H_6O$ gives sooty flame on burning. Its reaction with bromine solution in low polarity solvent results in high yield of B. B is
Answer: (a)
Aromatic compounds burn with a sooty flame.
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Consider the following reaction $\mathrm{Propanal+Methanal} \xrightarrow{\substack{(i)\ \mathrm{dil.NaOH}\\ (ii)\ \Delta\\ (iii)\ \mathrm{NaCN}\\ (iv)\ \mathrm{H_3O^+}}} \mathrm{Product\ B\ (C_5H_8O_3)}$ The correct statement for product B is. It is
Answer: (c)
The reaction starts with $\mathrm{CH_3CH_2CHO}$ and $\mathrm{HCHO}$ in the presence of $\mathrm{OH^-}$ and heat. This forms $\mathrm{CH_3C(ONa)CH_2CN}$. Upon treatment with $\mathrm{H_3O^+}$, it converts to $\mathrm{CH_3C^*H(OH)CH_2COOH}$. The carboxylic acid will give $\mathrm{CO_2}$ gas with $\mathrm{NaHCO_3}$ solution.
Chemistry · Chemical Bonding and Molecular Structure · Single correct
Answer: (d)
The given reaction sequence involves the reduction of a nitro group to an amino group followed by acetylation. The first step is the reduction of the nitro group $\mathrm{NO_2}$ to an amino group $\mathrm{NH_2}$ using $\mathrm{H_2/Pd}$ in $\mathrm{C_2H_5OH}$. The second step involves the acetylation of the amino group to form an acetamide $\mathrm{NHCOCH_3}$ using acetic anhydride $(\mathrm{CH_3CO})_2\mathrm{O}$ in the presence of pyridine.
Chemistry · Chemistry in Everyday Life · Single correct
Which of the following artificial sweeteners has the highest sweetness value in comparison to cane sugar?
Answer: (c)
Sweetness value order with respect to cane sugar: Alitame > Sucralose > Saccharin > Aspartame
Chemistry · Biomolecules · Single correct
A protein 'X' with molecular weight of 70,000 u, on hydrolysis gives amino acids. One of these amino acid is
Answer: (b)
Only in option (2) $\alpha$-Amino acid is given all the
Chemistry · Some Basic Concepts of Chemistry · Numerical
On complete combustion, $0.492 \, \mathrm{g}$ of an organic compound gave $0.792 \, \mathrm{g}$ of $\mathrm{CO}_2$. The $\%$ of carbon in the organic compound is _____ (Nearest integer)
Answer: 44
Weight of C in 0.792 gm $\mathrm{CO_2}$ is calculated as follows: $$\frac{12}{44} \times 0.792 = 0.216$$ The percentage of C in the compound is: $$\frac{0.216}{0.492} \times 100$$ $$= 43.90\%$$ Ans: 44
Chemistry · Some Basic Concepts of Chemistry · Numerical
Zinc reacts with hydrochloric acid to give hydrogen and zinc chloride. The volume of hydrogen gas produced at STP from the reaction of $11.5 \, \mathrm{g}$ of zinc with excess HCl is _______ $\mathrm{L}$ (Nearest integer) (Given: Molar mass of Zn is $65.4 \, \mathrm{g \, mol^{-1}}$ and Molar volume of $\mathrm{H_2}$ at STP $= 22.7 \, \mathrm{L}$)
Answer: 4
The reaction is given by $\($ $\mathrm{Zn}$ + 2$\mathrm{HCl}$ $\rightarrow$ $\mathrm{ZnCl_2}$ + $\mathrm{H_2}$ $\uparrow$ $\)$. Moles of Zn used are $\($ $\frac{11.5}{65.4}$ $\)$, which equals the moles of $\($ $\mathrm{H_2}$ $\)$ evolved. The volume of $\($ $\mathrm{H_2}$ $\)$ is $\($ $\frac{11.5}{65.4}$ $\times$ 22.7$\,$ $\mathrm{L}$ = 3.99$\,$ $\mathrm{L}$ $\)$.
Chemistry · Thermodynamics · Numerical
The enthalpy change for the conversion of $\frac{1}{2}\text{Cl}_2 \text{ (g)}$ to $\text{Cl}^- \text{ (aq)}$ is $(\text{ }- \text{ })\text{ kJmol}^{-1}$ (Nearest integer) $\text{Given : }\Delta_{\text{dis}} \text{H}_{\text{Cl}_2(\text{ g})}^0 = 240\text{kJmol}^{-1}$, $\Delta_{\text{eg}}\text{H}_{\text{Cl}(\text{ g})}^0 = -350\text{kJmol}^{-1}$, $\Delta_{\text{hyd}}\text{H}_{\text{Cl}^-_{(\text{g})}}^0 = -380\text{kJmol}^{-1}$
Answer: 610
The reaction is given by: $$\frac{1}{2} \mathrm{Cl_2}_{(g)} \rightarrow \mathrm{Cl}_{(g)} \rightarrow \mathrm{Cl^-}_{(g)} \rightarrow \mathrm{Cl^-}_{(aq.)}$$ The change in enthalpy is calculated as: $$\Delta H^\circ = \frac{1}{2} \times 240 + (-350) + (-380)$$ $$= -610 ans.$$
Chemistry · Equilibrium · Numerical
For the reaction \[ \mathrm{SO_2(g) + \frac{1}{2}O_2(g) \rightleftharpoons SO_3(g)} \] $K_p = 2 \times 10^{12}$ at $27^\circ\mathrm{C}$ and $1\,\mathrm{atm}$ pressure. The $K_c$ for the same reaction is _______ $\times 10^{13}$. (Nearest integer) Given: $R = 0.082 \, \mathrm { L \ atm\,K^{-1}} mol^{-1}$.
Answer: 1
The reaction is given by $\mathrm{SO_2}_{(g)} + \frac{1}{2} \mathrm{O_2}_{(g)} \rightleftharpoons \mathrm{SO_3}_{(g)}$. The equilibrium constant $K_P$ is $2 \times 10^{12}$ at $300 \, \mathrm{K}$. The relationship between $K_P$ and $K_C$ is given by $K_P = K_C \times (RT)^{\Delta n_g}$. Substituting the values, we have $$2 \times 10^{12} = K_C \times (0.082 \times 300)^{-1/2}.$$ Solving for $K_C$, we find $$K_C = 9.92 \times 10^{12}.$$ Therefore, $$K_C = 0.992 \times 10^{13}.$$
Chemistry · States of Matter · Numerical
The total pressure of a mixture of non-reacting gases X ($0.6 \, \mathrm{g}$) and Y ($0.45 \, \mathrm{g}$) in a vessel is $740 \, \mathrm{mm}$ of Hg. The partial pressure of the gas X is ________ $\mathrm{mm}$ of Hg. (Nearest Integer) (Given: molar mass X $= 20$ and Y $= 45 \, \mathrm{g} \, \mathrm{mol}^{-1}$)
Answer: 555
Given $P_X = \chi_X P_T$. $$= \frac{\frac{0.6}{20}}{\frac{0.6}{20} + \frac{0.45}{45}} \times 740$$ $P_X = 555 \, \mathrm{mm \, Hg}$
Chemistry · Solutions · Numerical
At 27°C, a solution containing 2.5 g of solute in 250.0 mL of solution exerts an osmotic pressure of 400 Pa. The molar mass of the solute is ____ g mol⁻¹ (Nearest integer) (Given : R = 0.083 L bar K⁻¹ mol⁻¹)
Answer: 62250
Given $\pi = CRT$. $$\frac{400 \, \mathrm{Pa}}{10^5} = \frac{\frac{2.5 \, \mathrm{g}}{M_o}}{\frac{250}{1000} \, \mathrm{L}} \times 0.83 \, \frac{\mathrm{L} - \mathrm{bar}}{\mathrm{K} \cdot \mathrm{mol}} \times 300 \, \mathrm{K}$$ $M_o = 62250$
Chemistry · Electrochemistry · Numerical
The logarithm of equilibrium constant for the reaction $\mathrm{Pd}^{2+} + 4\mathrm{Cl}^- \rightleftharpoons \mathrm{PdCl}_4^{2-}$ is _____ (Nearest integer) Given: $\frac{2.303RT}{F} = 0.06\, \mathrm{V}$ $\mathrm{Pd}^{2+}_{(aq)} + 2e^- \rightleftharpoons \mathrm{Pd}(s)$ $E^\circ = 0.83\, \mathrm{V}$ $\mathrm{PdCl}_4^{2-} (aq) + 2e^- \rightleftharpoons \mathrm{Pd}(s) + 4\mathrm{Cl}^-_{(aq)}$ $E^0 = 0.65\text{ V}$
Answer: 6
Given $E^\circ = 0.65 \, \mathrm{V}$. $$\Delta G^\circ = -RT \ell n K$$ $$-nFE^\circ_{cell} = -RT \times 2.303 (\log_{10} K)$$ $$E^\circ_{cell} \times n = \log K$$ $$0.06 \ldots (1)$$ $$\mathrm{Pd}^{+2} (\mathrm{aq.}) + 2e^- \rightleftharpoons \mathrm{Pd} (\mathrm{s}), E^\circ_{cat, red^n} = 0.83$$ $$\mathrm{Pd} (\mathrm{s}) + 4\mathrm{Cl}^- (\mathrm{aq.}) \rightleftharpoons \mathrm{PdCl}_4^{2-} (\mathrm{aq.}) + 2e^-, E^\circ_{Anode, Oxid^n} = 0.65$$ Net Reaction: $$\mathrm{Pd}^{2+} (\mathrm{aq.}) + 4\mathrm{Cl}^- (\mathrm{aq.}) \rightleftharpoons \mathrm{PdCl}_4^{2-} (\mathrm{aq.})$$ $$E^\circ_{cell} = E^\circ_{cat, red^n} - E^\circ_{Anode, Oxid^n}$$ $$E^\circ_{cell} = 0.83 - 0.65 = 0.18 \ldots (2)$$ Also $n = 2$. Using equation (1), (2) $\&$ (3)
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
$A \rightarrow B$ The rate constants of the above reaction at $200 \, \mathrm{K}$ and $300 \, \mathrm{K}$ are $0.03 \, \mathrm{min}^{-1}$ and $0.05 \, \mathrm{min}^{-1}$ respectively. The activation energy for the reaction is _____ $\mathrm{J}$ (Nearest integer) (Given: $\ln 10 = 2.3$ $R = 8.3 \, \mathrm{J} \, \mathrm{K}^{-1} \, \mathrm{mol}^{-1}$ $\log 5 = 0.70$ $\log 3 = 0.48$ $\log 2 = 0.30$)
Answer: 2520
Given the equation: $$\log \frac{K_{300}}{K_{200}} = \frac{E_a}{2.3 \times 8.314} \left( \frac{1}{T_1} - \frac{1}{T_2} \right)$$ Substitute the values: $$\log \frac{0.05}{0.03} = \frac{E_a}{2.305 \times 8.314} \times \left[ \frac{1}{200} - \frac{1}{300} \right]$$ Solving for $E_a$: $$E_a = 2519.88 \, \mathrm{J} \Rightarrow E_a = 2520 \, \mathrm{J}$$
Chemistry · Redox Reactions · Fill in the blank
The oxidation sate of phosphorus in hypophosphoric acid is _____.
Answer: 4
The compound is $\mathrm{H_4P_2O_6}$. The structure is shown with two phosphorus atoms each double bonded to an oxygen atom and single bonded to a hydroxyl group. The oxidation state (O.S.) of phosphorus (P) is $+4$.
Chemistry · Amines · Numerical
How many of the transformation given below would result in aromatic amines?
Answer: (3)
Maths · Complex Numbers and Quadratic Equations · Single correct
The number of real roots of the equation $$\sqrt{x^2 - 4x + 3} + \sqrt{x^2 - 9} = \sqrt{4x^2 - 14x + 6}$$, is:
Answer: (b)
Given $$\sqrt{(x-1)(x-3)} + \sqrt{(x-3)(x+3)}$$ This simplifies to $$\sqrt{4 \left( x - \frac{12}{4} \right) \left( x - \frac{2}{4} \right)}$$ This implies $$\sqrt{x-3} = 0 \implies x = 3$$ which is in the domain. Alternatively, $$\sqrt{x-1} + \sqrt{x+3} = \sqrt{4x-2}$$ This gives $$2\sqrt{(x-1)(x+3)} = 2x - 4$$ Expanding, we have $$x^2 + 2x - 3 = x^2 - 4x + 4$$ Solving for $x$, $$6x = 7$$ Thus, $$x = \frac{7}{6}$$ (rejected)
Maths · Complex Numbers and Quadratic Equations · Single correct
For all $z \in \mathbb{C}$ on the curve $C_1 : |z| = 4$, let the locus of the point $z + \frac{1}{z}$ be the curve $C_2$. Then
Answer: (a)
Let $w = z + \frac{1}{z} = 4e^{i\theta} + \frac{1}{4}e^{-i\theta}$. Therefore, $$w = \frac{17}{4} \cos \theta + i \frac{15}{4} \sin \theta.$$ So locus of $w$ is ellipse $$\frac{x^2}{\left(\frac{17}{4}\right)^2} + \frac{y^2}{\left(\frac{15}{4}\right)^2} = 1.$$ Locus of $z$ is circle $x^2 + y^2 = 16$.
Maths · Inverse Trigonometric Functions · Single correct
If the sum and product of four positive consecutive terms of a G.P., are 126 and 1296, respectively, then the sum of common ratios of all such GPs is
Answer: (a)
Given the sequence $a, ar, ar^2, ar^3$ with $(a, r > 0)$. $a^4 r^6 = 1296$ $a^2 r^3 = 36$ $$a = \frac{6}{r^{3/2}}$$ The sum $a + ar + ar^2 + ar^3 = 126$. $$\frac{1}{r^{3/2}} + \frac{r}{r^{3/2}} + \frac{r^2}{r^{3/2}} + \frac{r^3}{r^{3/2}} = \frac{126}{6} = 21$$ $$\left( r^{-3/2} + r^{3/2} \right) + \left( r^{1/2} + r^{-1/2} \right) = 21$$ $$r^{1/2} + r^{-1/2} = A$$ $$r^{-3/2} + r^{3/2} + 3A = A^3$$ $$A^3 - 3A + A = 21$$ $$A^3 - 2A = 21$$ $$A = 3$$ $$\sqrt{r} + \frac{1}{\sqrt{r}} = 3$$ $$r + 1 = 3\sqrt{r}$$ $$r^2 + 2r + 1 = 9r$$ $$r^2 - 7r + 1 = 0$$
Maths · Conic Sections · Single correct
Let a circle $C_1$ be obtained on rolling the circle $x^2 + y^2 - 4x - 6y + 11 = 0$ upwards 4 units on the tangent $T$ to it at the point $(3, 2)$. Let $C_2$ be the image of $C_1$ in $T$. Let $A$ and $B$ be the centers of circles $C_1$ and $C_2$ respectively, and $M$ and $N$ be respectively the feet of perpendiculars drawn from $A$ and $B$ on the $x$-axis. Then the area of the trapezium $AMNB$ is:
Answer: (b)
Given $C = (2, 3)$, $r = \sqrt{2}$. Centre of $G = A = 2 + 4 \frac{1}{\sqrt{2}}$, $$3 + \frac{4}{\sqrt{2}} = \left(2 + 2\sqrt{2}, 3 + 2\sqrt{2}\right)$$ $A\left(2 + 2\sqrt{2}, 3 + 2\sqrt{2}\right)$ $B\left(4 + 2\sqrt{2}, 1 + 2\sqrt{2}\right)$ $$\frac{x - \left(2 + 2\sqrt{2}\right)}{1} = \frac{y - \left(3 + 2\sqrt{2}\right)}{-1} = 2$$ Therefore, the area of the trapezium: $$\frac{1}{2}\left(4 + 4\sqrt{2}\right)2 = 4\left(1 + \sqrt{2}\right)$$
Maths · Conic Sections · Single correct
If the maximum distance of a normal to the ellipse $\frac{x^2}{4}+\frac{y^2}{b^2}=1$, $b<2$, from the origin is $1$, then the eccentricity of the ellipse is:
Answer: (b)
Equation of normal is $2x \sec \theta - b \csc \theta = 4 - b^2$. Distance from $(0, 0) = \frac{4 - b^2}{\sqrt{4 \sec^2 \theta + b^2 \csc^2 \theta}}$. Distance is maximum if $4 \sec^2 \theta + b^2 \csc^2 \theta$ is minimum. Therefore, $\tan^2 \theta = \frac{b}{2}$. $$\Rightarrow \frac{4 - b^2}{\sqrt{4 \cdot \frac{b+2}{2} + b^2 \cdot \frac{b+2}{b}}} = 1$$ $$\Rightarrow 4 - b^2 = b + 2 \Rightarrow b = 1 \Rightarrow e = \frac{\sqrt{3}}{2}$$
Maths · Mathematical Reasoning · Single correct
(S1)(p $\Rightarrow$ q) $\lor$ (p $\land$ ($\sim$ q)) is a tautology (S2)(($\sim$ p) $\Rightarrow$ ($\sim$ q)) $\land$ (($\sim$ p) $\lor$ q) is a Contradiction. Then
Answer: (b)
The truth tables for the given logical expressions are as follows: For the first table: \begin{tabular}{|l|l|l|l|l|l|} \hline p & q & p $\Rightarrow$ q & $\neg$ q & p $\land$ $\neg$ q & (p $\Rightarrow$ q) $\lor$ (p $\land$ $\neg$ q) \\ \hline T & T & T & F & F & T \\ \hline T & F & F & T & T & T \\ \hline F & T & T & F & F & T \\ \hline F & F & T & T & F & T \\ \hline \end{tabular} $_$ For the second table: \begin{tabular}{|l|l|l|l|l|l|} \hline $\neg$ p & $\neg$ q & $\neg$ p $\Rightarrow$ $\neg$ q & $\neg$ p $\lor$ q & (($\neg$ p) $\Rightarrow$ ($\neg$ q)) $\land$ ($\neg$ p) $\lor$ q \\ \hline F & F & T & T & T \\ \hline F & T & T & F & F \\ \hline T & F & F & T & F \\ \hline T & T & T & T & T \\ \hline \end{tabular}
Maths · Relations and Functions (Advanced) · Single correct
Let R be a relation on $\mathbb{N} \times \mathbb{N}$ defined by $(a, b) \, R \, (c, d)$ if and only if $ad(b - c) = bc(a - d)$. Then R is
Answer: (a)
$(a,b)\ R\ (c,d)\Rightarrow ad(b-c)=bc(a-d)$ Symmetric: $(c,d)\ R\ (a,b)\Rightarrow cb(d-a)=da(c-b)$ $\Rightarrow$ Symmetric. Reflexive: $(a,b)\ R\ (a,b)\Rightarrow ab(b-a)\neq ba(a-b)$ $\Rightarrow$ Not reflexive. Transitive: $(2,3)\ R\ (3,2)$ and $(3,2)\ R\ (5,30)$ but $((2,3),(5,30))\notin R$ $\Rightarrow$ Not transitive.
Maths · Matrices · Single correct
Let $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{bmatrix}$. Then the sum of the diagonal elements of the matrix $(A + I)^{11}$ is equal to:
Answer: (c)
Given $$A^2 = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{bmatrix}$$ $$= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 4 & -1 \\ 0 & 12 & -3 \end{bmatrix} = A$$ Therefore, $$A^3 = A^4 = \ldots = A$$ $$(A + I)^{11} = \binom{11}{0} A^{11} + \binom{11}{1} A^{10} + \ldots + \binom{11}{10} A + \binom{11}{11} I$$ $$= (\binom{11}{0} + \binom{11}{1} + \ldots + \binom{11}{10}) A + I$$ $$= (2^{11} - 1) A + I = 2047 A + I$$ Thus, the sum of diagonal elements is $$= 2047 (1 + 4 - 3) + 3$$ $$= 4094 + 3 = 4097$$
Maths · Determinants · Single correct
For the system of linear equations $$x + y + z = 6$$ $$\alpha x + \beta y + 7z = 3$$ $$x + 2y + 3z = 14,$$ which of the following is NOT true?
Answer: (d)
By equation 1 and 3 $$y + 2z = 8$$ $$y = 8 - 2z$$ And $$x = -2 + z$$ Now putting in equation 2 $$\alpha(z - 2) + \beta(-2z + 8) + 7z = 3$$ $$\Rightarrow (\alpha - 2\beta + 7)z = 2\alpha - 8\beta + 3$$ So equations have unique solution if $$\alpha - 2\beta + 7 \neq 0$$ And equations have no solution if $$\alpha - 2\beta + 7 = 0 and 2\alpha - 8\beta + 3 \neq 0$$ And equations have infinite solution if $$\alpha - 2\beta + 7 = 0 and 2\alpha - 8\beta + 3 = 0$$
Maths · Inverse Trigonometric Functions · Single correct
If $\sin^{-1} \frac{\alpha}{17} + \cos^{-1} \frac{4}{5} - \tan^{-1} \frac{77}{36} = 0$, $0 < \alpha < 13$, then $\sin^{-1}(\sin \alpha) + \cos^{-1}(\cos \alpha)$ is equal to
Answer: (a)
Given $\cos^{-1} \frac{4}{5} = \tan^{-1} \frac{3}{4}$. Therefore, $\sin^{-1} \frac{\alpha}{17} = \tan^{-1} \frac{77}{36} - \tan^{-1} \frac{3}{4} = \tan^{-1} \left( \frac{\frac{77}{36} - \frac{3}{4}}{1 + \frac{77}{36} \cdot \frac{3}{4}} \right)$. $\sin^{-1} \frac{\alpha}{17} = \tan^{-1} \frac{8}{15} = \sin^{-1} \frac{8}{17}$. Thus, $\frac{\alpha}{17} = \frac{8}{17} \Rightarrow \alpha = 8$. Therefore, $\sin^{-1} (\sin 8) + \cos^{-1} (\cos 8) = 3\pi - 8 + 8 - 2\pi = \pi$.
Maths · Conic Sections · Single correct
Let $y=f(x)$ represent a parabola with focus $\left(-\dfrac12,0\right)$and directrix $y=-\dfrac12$. Then \[ S= \left\{ x\in\mathbb R: \tan^{-1}\!\left(\sqrt{f(x)}\right) +\sin^{-1}\!\left(\sqrt{\,f(x)+1\,}\right) =\dfrac{\pi}{2} \right\}: \]
Answer: (a)
Given $\($ $\left$( x + $\frac{1}{2}$ $\right$)^2 = $\left$( y + $\frac{1}{4}$ $\right$) $\)$ and $\($ y = (x^2 + x) $\)$. $\($ $\tan$^{-1} $\sqrt{x(x+1)}$ + $\sin$^{-1} $\sqrt{x^2 + x + 1}$ = $\frac{\pi}{2}$ $\)$. $\($ 0 $\leq$ x^2 + x + 1 $\leq$ 1 $\)$. $\($ x^2 + x $\leq$ 0 $\)$ $\($ $\ldots$ (1) $\)$ Also $\($ x^2 + x $\geq$ 0 $\)$ $\($ $\ldots$ (2) $\)$ Therefore, $\($ x^2 + x = 0 $\Rightarrow$ x = 0, -1 $\)$. $\($ S $\)$ contains 2 elements.
Maths · Relations and Functions · Single correct
If the domain of the function $f(x) = \frac{[x]}{1 + x^2}$, where $[x]$ is greatest integer $\leq x$, is $[2, 6)$, then its range is
Answer: (d)
The function $f(x)$ is defined piecewise over different intervals. For $x \in [2,3)$, $f(x) = \frac{2}{1+x^2}$. For $x \in [3,4)$, $f(x) = \frac{3}{1+x^2}$. For $x \in [4,5)$, $f(x) = \frac{4}{1+x^2}$. For $x \in [5,6)$, $f(x) = \frac{5}{1+x^2}$. The graph shows the behavior of $f(x)$ over these intervals. The point $(\frac{5}{37}, \frac{2}{5})$ is highlighted on the graph.
Maths · Differential Equations · Single correct
Let $$y = f(x) = \sin^3\left(\frac{\pi}{3}\left(\cos\left(\frac{\pi}{3\sqrt{2}}\left(-4x^3 + 5x^2 + 1\right)^{\frac{3}{2}}\right)\right)\right)$$ . Then, at $x = 1$,
Answer: (b)
Given $1 < x < 1$. $y = \sin^5(\pi/3 \cos g(x))$ $$g(x) = \frac{\pi}{3\sqrt{2}} \left( -4x^3 + 5x^2 + 1 \right)^{3/2}$$ $$g(1) = 2\pi/3$$ $$y' = 3 \sin^2 \left( \frac{\pi}{3} \cos g(x) \right) \times \cos \left( \frac{\pi}{3} \cos g(x) \right) \times \frac{\pi}{3} (-\sin g(x)) g'(x)$$ $$y'(1) = 3 \sin^2 \left( -\frac{\pi}{6} \right) \cdot \cos \left( \frac{\pi}{6} \right) \cdot \frac{\pi}{3} \left( -\sin \frac{2\pi}{3} \right) g'(1)$$ $$g'(x) = \frac{\pi}{3\sqrt{2}} \left( -4x^3 + 5x^2 + 1 \right)^{1/2} (-12x^2 + 10x)$$ $$g'(1) = \frac{\pi}{2\sqrt{2}} (\sqrt{2}) (-2) = -\pi$$ $$y'(1) = \frac{9}{4} \cdot \frac{\sqrt{3}}{2} \cdot \frac{-\sqrt{3}}{2} (-\pi) = \frac{3\pi^2}{16}$$ $$y(1) = \sin^3 (\pi/3 \cos 2\pi/3) = -\frac{1}{8}$$ $$2y'(1) + 3\pi^2 y(1) = 0$$
Maths · Applications of Derivatives · Single correct
A wire of length 20 $\mathrm{m}$ is to be cut into two pieces. A piece of length $\ell_1$ is bent to make a square of area $A_1$ and the other piece of length $\ell_2$ is made into a circle of area $A_2$. If $2A_1 + 3A_2$ is minimum then $(\pi \ell_1) : \ell_2$ is equal to:
Answer: (a)
Given $\ell_1 + \ell_2 = 20$ which implies $\frac{d\ell_2}{d\ell_1} = -1$. $A_1 = \left( \frac{\ell_1}{4} \right)^2$ and $A_2 = \pi \left( \frac{\ell_2}{2\pi} \right)^2$. Let $S = 2A_1 + 3A_2 = \frac{\ell_1^2}{8} + \frac{3\ell_2^2}{4\pi}$. Differentiating $S$ with respect to $\ell$, we have: $$\frac{ds}{d\ell} = 0 \implies \frac{2\ell_1}{8} + \frac{6\ell_2}{4\pi} \cdot \frac{d\ell_2}{d\ell_1} = 0$$ This implies: $$\frac{\ell_1}{4} = \frac{6\ell_2}{4\pi} \implies \frac{\pi \ell_1}{\ell_2} = 6$$
Maths · Integrals · Single correct
Let $\alpha \in (0, 1)$ and $\beta = \log_e(1 - \alpha)$. Let $$P_n(x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \ldots + \frac{x^n}{n}, \ x \in (0, 1).$$ Then the integral $\int_0^\alpha \frac{t^{50}}{1-t} \, dt$ is equal to
Answer: (b)
The integral is given by $$\int_0^\alpha \frac{t^{50} - 1 + 1}{1-t} = -\int_0^\alpha (1 + t + \ldots + t^{49}) + \int_0^\alpha \frac{1}{1-t} \, dt$$ which simplifies to $$= -\left( \frac{\alpha^{50}}{50} + \frac{\alpha^{49}}{49} + \ldots + \frac{\alpha^1}{1} \right) + \left( \frac{\ln(1-f)}{-1} \right)_0^\alpha$$ resulting in $$= -P_{50}(\alpha) - \ln(1-\alpha)$$ and finally $$= -P_{50}(\alpha) - \beta$$
Maths · Integrals · Single correct
The value of $$\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{(2 + 3 \sin x)}{\sin x (1 + \cos x)} \, dx$$ is equal to
Answer: (c)
Given $$\int_{\pi/3}^{\pi/2} \left( \frac{2 + 3 \sin x}{\sin x (1 + \cos x)} \right) dx = 2 \int_{\pi/3}^{\pi/2} \frac{dx}{\sin x + \sin x \cos x} + 3$$ We have $$3 \int_{\pi/3}^{\pi/2} \frac{dx}{1 + \cos x} = \int_{\pi/3}^{\pi/2} \frac{1 - \cos x}{\sin^2 x} dx$$ This equals $$= \int_{\pi/3}^{\pi/2} (\csc^2 x - \cot x \csc x) dx$$ Which simplifies to $$= (\cos \mathrm{ecx} - \cot x) \bigg|_{\pi/3}^{\pi/2} = (1) - \left( \frac{2}{\sqrt{3}} - \frac{1}{\sqrt{3}} \right) = 1 - \frac{1}{\sqrt{3}}$$ Now, $$\int_{\pi/3}^{\pi/2} \frac{dx}{\sin x (1 + \cos x)} = \int \frac{dx}{(2 \tan x/2)(1 + 1 - \tan^2 x/2)}$$ This becomes $$= \int \frac{(1 + \tan^2 x/2) \sec^2 x/2 \, dx}{2 \tan x/2}$$ Let $\tan x/2 = t$ and $\sec x/2 \, dx = dt$. Then, $$\frac{1}{2} \int \frac{1 + t^2}{t} \, dt = \frac{1}{2} \left[ \ln t + t^2 \right]_1^{1/\sqrt{3}}$$ This evaluates to $$= \frac{1}{2} \left[ \left( 0 + \frac{1}{2} \right) - \left( \ln \frac{1}{\sqrt{3}} + \frac{1}{6} \right) \right] = \left( \frac{1}{3} + \ln \sqrt{3} \right) \frac{1}{2}$$ Simplifying further, $$= \left( \frac{1}{6} + \frac{1}{2} \ln \sqrt{3} \right)$$ Thus, $$2 \left( \frac{1}{6} + \frac{1}{2} \ln \sqrt{3} \right) + 3 \left( 1 - \frac{1}{\sqrt{3}} \right)$$ Finally, $$= \frac{1}{3} + \ln \sqrt{3} + 3 - \sqrt{3} = \frac{10}{3} + \ln \sqrt{3} - \sqrt{3}$$
Maths · Integrals · Single correct
Let a differentiable function $f$ satisfy $$f(x) + \int_{3}^{x} \frac{f(t)}{t} \, dt = \sqrt{x+1}, \; x \geq 3.$$ Then $12f(8)$ is equal to:
Answer: (c)
Differentiate with respect to $x$. $$f'(x) + \frac{f(x)}{x} = \frac{1}{2\sqrt{x} + 1}$$ Integrating factor is $I.F. = e^{\int \frac{1}{x} \, dx} = e^{\ln x} = x$. $$xf(x) = \int \frac{x}{2\sqrt{x} + 1} \, dx$$ Let $x + 1 = t^2$. $$= \int \frac{t^2 - 1}{2t} \, 2t \, dt$$ $$xf(x) = \frac{t^3}{3} - t + c$$ $$xf(x) = \frac{(x+1)^{3/2}}{3} - \sqrt{x+1} + c$$ Also, putting $x = 3$ in the given equation $f(3) + 0 = \sqrt{4}$. $$f(3) = 2$$ $$\Rightarrow C = 8 - \frac{8}{3} = \frac{16}{3}$$ $$f(x) = \frac{(x+1)^{3/2}}{3} - \sqrt{x+1} + \frac{16}{3}$$ $$f(8) = \frac{9 - 3 + \frac{16}{3}}{8} = \frac{34}{24}$$ $$\Rightarrow 12 \, f(8) = 17$$
Maths · Vector Algebra · Single correct
Let $\vec{a} = 2\hat{i} + \hat{j} + \hat{k}$, and $\vec{b}$ and $\vec{c}$ be two nonzero vectors such that $\left|\vec{a} + \vec{b} + \vec{c}\right| = \left|\vec{a} + \vec{b} - \vec{c}\right|$ and $\vec{b}\cdot\vec{c} = 0$. Consider the following two statements: (A) $\left|\vec{a} + \lambda\vec{c}\right| \ge \left|\vec{a}\right|$ for all $\lambda \in \mathbb{R}$. (B) $\vec{a}$ and $\vec{c}$ are always parallel
Answer: (c)
Given $|\bar{a} + \bar{b} + \bar{c}|^2 = |\bar{a} + \bar{b} - \bar{c}|^2$. $$2 \bar{a} \cdot \bar{b} + 2 \bar{b} \cdot \bar{c} + 2 \bar{c} \cdot \bar{a} = 2 \bar{a} \cdot \bar{b} - 2 \bar{b} \cdot \bar{c} - 2 \bar{c} \cdot \bar{a}$$ $$4 \bar{a} \cdot \bar{c} = 0$$ B is incorrect. $$|\bar{a} + \lambda \bar{c}|^2 \geq |\bar{a}|^2$$ $$\lambda^2 c^2 \geq 0$$ True $\forall \lambda \in \mathbb{R}$ (A) is correct.
Maths · Three Dimensional Geometry · Single correct
Let the shortest distance between the lines L : $\frac{x-5}{-2} = \frac{y-\lambda}{0} = \frac{z+\lambda}{1}$, $\lambda \geq 0$ and $L_1 : x+1 = y-1 = 4-z$ be $2\sqrt{6}$. If $(\alpha, \beta, \gamma)$ lies on $L$, then which of the following is NOT possible?
Answer: (a)
The cross product $\vec{b}_1 \times \vec{b}_2$ is given by the determinant: $$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 0 & 1 \\ 1 & 1 & -1 \end{vmatrix} = -\hat{i} - \hat{j} - 2\hat{k}.$$ The vector difference $\vec{a}_2 - \vec{a}_1$ is: $$6\hat{i} + (\lambda - 1)\hat{j} + (-\lambda - 4)\hat{k}.$$ The equation $2\sqrt{6} = \frac{-6 - \lambda + 1 + 2\lambda + 8}{\sqrt{1 + 1 + 4}}$ simplifies to: $$|\lambda + 3| = 12 \Rightarrow \lambda = 9, -15.$$ The values of $\alpha$ and $\gamma$ are given by: $$\alpha = -2k + 5, \gamma = k - \lambda where k \in \mathbb{R}.$$ Therefore, $$\Rightarrow \alpha + 2\gamma = 5 - 2\lambda = -13, 35.$$
Maths · Probability · Single correct
A bag contains 6 balls. Two balls are drawn from it at random and both are found to be black. The probability that the bag contains at least 5 black balls is
Answer: (a)
The expression is given by $$\frac{{^5C_2 + ^6C_2}}{{^2C_2 + ^3C_2 + ^4C_2 + ^5C_2 + ^8C_2}} = \frac{10 + 15}{1 + 3 + 6 + 10 + 15}.$$ Simplifying, we have $$\frac{25}{35} = \frac{5}{7}.$$
Maths · Permutations and Combinations · Numerical
Let 5 digit numbers be constructed using the digits 0, 2, 3, 4, 7, 9 with repetition allowed, and are arranged in ascending order with serial numbers. Then the serial number of the number 42923 is
Answer: 2997
Given the sequence of operations: $$2 + \frac{6}{6} + \frac{6}{6} + \frac{6}{6} = 1296$$ $$3 + \frac{6}{6} + \frac{6}{6} + \frac{6}{6} = 1296$$ $$40 + \frac{6}{6} + \frac{6}{6} = 216$$ $$420 + \frac{6}{6} = 36$$ $$\frac{422}{6} + \frac{6}{6} = 36$$ $$423 + \frac{6}{6} = 36$$ $$424 + \frac{6}{6} = 36$$ $$427 + \frac{6}{6} = 36$$ $$429 + \frac{0}{6} = 6$$ $$429 \times 2 \times 0 = 1$$ $$429 \times 2 \times 2 = 1$$ $$429 \times 2 \times 3 = 1$$ The final result is: $$= 2997$$
Maths · Sequences and Series · Subjective
Let $a_1, a_2, \ldots, a_n$ be in A.P. If $a_5 = 2a_7$ and $a_{11} = 18$, then $$12 \left( \frac{1}{\sqrt{a_{10}} + \sqrt{a_{11}}} + \frac{1}{\sqrt{a_{11}} + \sqrt{a_{12}}} + \ldots + \frac{1}{\sqrt{a_{17}} + \sqrt{a_{18}}} \right)$$ is equal to .
Answer: 8
Given $2a_7 = a_5$ (given). $2(a_1 + 6d) = a_1 + 4d$. $a_1 + 8d = 0 \ldots (1)$ $a_1 + 10d = 18 \ldots (2)$ By (1) and (2) we get $a_1 = -72$, $d = 9$. $a_{18} = a_1 + 17d = -72 + 153 = 81$. $a_{10} = a_1 + 9d = 9$. $$12 \left( \frac{\sqrt{a_{11}} - \sqrt{a_{10}}}{d} + \frac{\sqrt{a_{12}} - \sqrt{a_{11}}}{d} + \ldots + \frac{\sqrt{a_{18}} - \sqrt{a_{17}}}{d} \right)$$ $$12 \left( \frac{\sqrt{a_{18}} - \sqrt{a_{10}}}{d} \right) = \frac{12(9 - 3)}{9} = \frac{12 \times 6}{6} = 8$$
Maths · Permutations and Combinations · Numerical
Number of 4-digit numbers that are less than or equal to 2800 and either divisible by 3 or by 11, is equal to ________.
Answer: 710
1000 to 2799 divisible by 3. 1002 + (n - 1) $\times$ 3 = 2799 $n = 600$ Divisible by 11 $$1 - 2799 \rightarrow \left\lfloor \frac{2799}{11} \right\rfloor = [254] = 254$$ $$1 - 999 \rightarrow \left\lfloor \frac{999}{11} \right\rfloor = 90$$ 1000 to 2799 = 254 - 90 = 164 Divisible by 33 $$1 - 2799 \rightarrow \left\lfloor \frac{2799}{33} \right\rfloor = 84$$ $$1 - 999 \rightarrow \left\lfloor \frac{999}{33} \right\rfloor = 30$$ 1000 to 2799 $\rightarrow$ 54 Therefore, $n(3) + n(11) - n(33)$ 600 + 164 - 54 = 710
Maths · Binomial Theorem · Fill in the blank
Let $\alpha > 0$ be the smallest number such that the expansion of $\left(x^{\frac{2}{3}} + \frac{2}{x^3}\right)^{30}$ has a term $\beta x^{-\alpha}$, $\beta \in \mathbb{N}$. Then $\alpha$ is equal to ______.
Answer: 2
Given $$T_{r+1} = {^{30}C_r} \left(x^{2/3}\right)^{30-r} \left(\frac{2}{x^3}\right)^r$$ This simplifies to $$= {^{30}C_r} \cdot 2^r \cdot x^{\frac{60 - 11r}{3}}$$ For the exponent of $x$ to be less than 0: $$\frac{60 - 11r}{3} 60 \implies r = 6$$ Thus, $$T_7 = {^{30}C_6} \cdot 2^6 \cdot x^{-2}$$ We have also observed $\beta = {^{30}C_6} (2)^6$ is a natural number. Therefore, $\alpha = 2$.
Maths · Binomial Theorem · Fill in the blank
The remainder on dividing $5^{99}$ by 11 is _____.
Answer: 9
$5^{99}$$=5^{4}\cdot5^{95}$ $=625\left(5^{5}\right)^{19}$ $=625\left(3125\right)^{19}$ $=625\left(3124+1\right)^{19}$ $=625\times11k\times19\times625$ $=11k_1+616+9$ $=11(k_2)+9$
Maths · Statistics · Subjective
If the variance of the frequency distribution is 3, then $\alpha$ is equal to
Answer: 5
Given the table, we calculate $\sigma_x^2 = \sigma_d^2$ using the formula: $$\sigma_x^2 = \sigma_d^2 = \frac{\sum f_i d_i^2}{\sum f_i} - \left(\frac{\sum f_i d_i}{\sum f_i}\right)^2$$ Substituting the values, we have: $$\frac{150}{45 + \alpha} - 0 = 3$$ This simplifies to: $$\Rightarrow 150 = 135 + 3\alpha$$ $$\Rightarrow 3\alpha = 15 \Rightarrow \alpha = 5$$
Maths · Applications of Integrals · Numerical
Let for $x \in \mathbb{R}$ $$f(x) = \frac{x + |x|}{2}$$ and $$g(x) = \begin{cases} x, & x < 0 \\ x^2, & x \geq 0 \end{cases}.$$ Then area bounded by the curve $y = (f \circ g)(x)$ and the lines $y = 0$, $2y - x = 15$ is equal to .
Answer: 72
Given $$f(x) = \frac{x + |x|}{2} = \begin{cases} x & x \geq 0 \\ 0 & x < 0 \end{cases}$$ $$g(x) = \begin{cases} x^2 & x \geq 0 \\ x & x < 0 \end{cases}$$ $$f(g(x)) = f\left[g(x)\right] = \begin{cases} g(x) & g(x) \geq 0 \\ 0 & g(x) < 0 \end{cases}$$ $$f(g(x)) = \begin{cases} x^2 & x \geq 0 \\ 0 & x < 0 \end{cases}$$ $$2y - x = 15$$ $$A = \int_{0}^{3} \left(\frac{x + 15}{2} - x^2\right) \, dx + \frac{1}{2} \times \frac{15}{2} \times 15$$ $$\left. \frac{x^2}{4} + \frac{15x}{2} - \frac{x^3}{3} \right|_{0}^{3} + \frac{225}{4}$$ $$= \frac{9}{4} + \frac{45}{2} - 9 + \frac{225}{4} = \frac{99 - 36 + 225}{4}$$ $$= \frac{288}{4} = 72$$
Maths · Vector Algebra · Numerical
Let $\vec{a}$ and $\vec{b}$ be two vector such that $|\vec{a}| = \sqrt{14}$, $|\vec{b}| = \sqrt{6}$ and $|\vec{a} \times \vec{b}| = \sqrt{48}$. Then $(\vec{a} \cdot \vec{b})^2$ is equal to _________.
Answer: 36
Given $|\vec{a}| = \sqrt{14}$, $|\vec{b}| = \sqrt{6}$, $|\vec{a} \times \vec{b}| = \sqrt{48}$. $|\vec{a} \times \vec{b}|^2 + |\vec{a} \cdot \vec{b}|^2 = |\vec{a}|^2 \times |\vec{b}|^2$ $$\Rightarrow (\vec{a} \cdot \vec{b})^2 = 84 - 48 = 36$$
Maths · Three Dimensional Geometry · Fill in the blank
Let the line L: $\frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-3}{1}$ intersect the plane $2x + y + 3z = 16$ at the point P. Let the point Q be the foot of perpendicular from the point R$(1, -1, -3)$ on the line L. If $\alpha$ is the area of triangle PQR, then $\alpha^2$ is equal to ________.
Answer: 180
Any point on L $((2\lambda + 1), (-\lambda - 1), (\lambda + 3))$. $2(2\lambda + 1) + (-\lambda - 1) + 3(\lambda + 3) = 16$ $6\lambda + 10 = 16 \Rightarrow \lambda = 1$ Therefore, $P = (3, -2, 4)$. DR of $QR = \langle 2\lambda, -\lambda, \lambda + 6 \rangle$ DR of $L = \langle 2, -1, 1 \rangle$ $4\lambda + \lambda + \lambda + 6 = 0$ $6\lambda + 6 = 0 \Rightarrow \lambda = -1$ $Q = (-1, 0, 2)$ $R(1, -1, -3)$ $\overrightarrow{QR} = 2\hat{i} - \hat{j} - 5\hat{k}$ $\overrightarrow{QP} = 4\hat{i} - 2\hat{j} + 2\hat{k}$ $\overrightarrow{QR} \times \overrightarrow{QP} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & -5 \\ 4 & -2 & 2 \end{vmatrix} = -12\hat{i} - 24\hat{j}$ $\frac{1}{2} \sqrt{144 + 576} = \frac{\sqrt{720}}{4} = 180$
Maths · Three Dimensional Geometry · Fill in the blank
Let $\theta$ be the angle between the planes $P_1 = \vec{r} \cdot (\hat{i} + \hat{j} + 2\hat{k}) = 9$ and $P_2 = \vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = 15$. Let $L$ be the line that meets $P_2$ at the point $(4, -2, 5)$ and makes an angle $\theta$ with the normal of $P_2$. If $\alpha$ is the angle between $L$ and $P_2$, then $\tan^2\theta \cot^2\alpha$ is equal to ______.
Answer: 9
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