JEE Main 1 February 2023 Shift 2 question paper with solutions
JEE Main 1 February 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
Question 1
Physics · Physical World, Units and Measurements · Single correct
If the velocity of light c, universal gravitational constant G and planck's constant h are chosen as fundamental quantities. The dimensions of mass in the new system is:
Say dimensional formulae of mass is $H^x C^y G^z$. $$M^1 = (ML^2 T^{-1})^x (LT^{-1}) (M^{-1} L^3 T^{-2})^z$$ $$M^1 L^0 T^0 = M^{x-z} L^{2x+y+3z} T^{-x-y-2z}$$ On comparing both sides, $$x - z = 1$$ $$2x + y + 3z = 0$$ $$-x - y - 2z = 0$$ On solving above equations we get $$x = \frac{1}{2} y = \frac{1}{2} z = -\frac{1}{2}$$
Question 2
Physics · Motion in a Straight Line · Numerical
For a train engine moving with speed of $20 \, \mathrm{ms^{-1}}$. the driver must apply brakes at a distance of $500 \, \mathrm{m}$ before the station for the train to come to rest at the station. If the brakes were applied at half of this distance, the train engine would cross the station with speed $\sqrt{x} \, \mathrm{ms^{-1}}$. The value of $x$ is _____ (Assuming same retardation is produced by brakes)
As shown in the figure a block of mass 10 kg lying on a horizontal surface is pulled by a force F acting at an angle 30$^\circ$, with horizontal. For $\mu_s$ = 0.25, the block will just start to move for the value of F: [Given g = 10 ms$^{-2}$]
33.3 N
25.2 N
20 N
35.7 N
Answer: (b)
Solution
Given the problem, we start with the normal force equation: $$N = Mg - F \sin 30^\circ$$ Substituting the values, we have: $$= mg - \frac{F}{2} = 100 - \frac{F}{2} = \frac{200 - F}{2}$$ Next, we use the equation for the force component: $$F \cos 30^\circ = \mu N$$ Substituting the values, we get: $$\sqrt{3} \frac{F}{2} = 0.25 \times \left( \frac{200 - F}{2} \right)$$ Simplifying, we have: $$4 \sqrt{3} F = 200 - F$$ Solving for $F$, we find: $$F = \frac{200}{4 \sqrt{3} + 1} = 25.22$$
Question 4
Physics · Work, Energy and Power · Numerical
A block is fastened to a horizontal spring. The block is pulled to a distance $x = 10 \, \mathrm{cm}$ from its equilibrium position (at $x = 0$) on a frictionless surface from rest. The energy of the block at $x = 5 \, \mathrm{cm}$ is $0.25 \, \mathrm{J}$. The spring constant of the spring is $\mathrm{Nm}^{-1}$.
Answer: 50
Solution
Given $x_0 = 10 \, \mathrm{cm}$. Initial potential energy $U_i = \frac{1}{2} k x_0^2$. Initial kinetic energy $K_i = 0$. Final potential energy $U_f = \frac{1}{2} k \left( \frac{x_0}{2} \right)^2$. Final kinetic energy $K_f = 0.25 \, \mathrm{J}$. $$\frac{1}{2} k x_0^2 + 0 = \frac{1}{2} k \frac{x_0^2}{4} + 0.25$$ $$\frac{1}{2} k x_0^2 \frac{3}{4} = \frac{1}{4}$$ $$\frac{1}{2} k \frac{3}{100} = 1 \Rightarrow k = \frac{200}{3} \, \mathrm{N/m}$$ $= 67 \, \mathrm{N/m}$
Question 5
Physics · Work, Energy and Power · Numerical
A force $F = (5 + 3y^2)$ acts on a particle in the $y$-direction, where $F$ is newton and $y$ is in meter. The work done by the force during a displacement from $y = 2\, \mathrm{m}$ to $y = 5\, \mathrm{m}$ is ______ j.
Answer: 132
Solution
Given $F = 5 + 3y^2$. The work done $W$ is given by the integral: $$W = \int_{2}^{5} (5 + 3y^2) \, dy$$ Evaluating the integral, we have: $$= \left[ 5y + \frac{3y^3}{3} \right]_{2}^{5}$$ $$= 132 \, \mathrm{J}$$
Question 6
Physics · System of Particles and Rotational Motion · Single correct
Figures (a), (b), (c) and (d) show variation of force with time. The impulse is highest in figure.
Physics · System of Particles and Rotational Motion · Numerical
Moment of inertia of a disc of mass M and radius 'R' about any of its diameter is $\frac{MR^2}{4}$. The moment of inertia of this disc about an axis normal to the disc and passing through a point on its edge will be, $\frac{x}{2} MR^2$. The value of $x$ is _____.
Answer: 3
Solution
The moment of inertia is given by the formula: $$I = I_{cm} + Md^2$$ Substituting the values, we have: $$= \frac{MR^2}{2} + MR^2$$ Simplifying, we get: $$= \frac{3}{2} MR^2$$
Question 8
Physics · Gravitation · Single correct
The escape velocities of two planets A and B are in the ratio 1 : 2. If the ratio of their radii respectively is 1 : 3, then the ratio of acceleration due to gravity of planet A to the acceleration of gravity of planet B will be:
$\frac{4}{3}$
$\frac{3}{2}$
$\frac{2}{3}$
$\frac{3}{4}$
Answer: (d)
Solution
The escape velocity $V_e$ is given by $$V_e = \sqrt{\frac{2GM}{R}} = \sqrt{\frac{2G \rho \frac{4}{3} \pi R^3}{R}} = C \sqrt{\rho} R.$$ The ratio of escape velocities is $$\frac{V_{e_1}}{V_{e_2}} = \frac{R_1}{R_2} \sqrt{\frac{\rho_1}{\rho_2}} = \frac{1}{2}.$$ Therefore, $$\frac{R_1^2}{R_2^2} \times \frac{\rho_1}{\rho_2} = \frac{1}{4}.$$ Solving for $\frac{R_1}{R_2}$, we get $$\frac{R_1}{R_2} = \frac{1}{3}.$$ The gravitational acceleration $g$ is given by $$g = \frac{GM}{R^2} = \frac{G \frac{4}{3} \pi R^3 \times \rho}{R^2} = C \cdot \rho R.$$ The ratio of gravitational accelerations is $$\frac{g_1}{g_2} = \frac{\rho_1 R_1}{\rho_2 R_2} = \frac{1}{4} \frac{R_2^2}{R_1^2} \times \frac{R_1}{R_2} = \frac{1}{4} \times \frac{R_2}{R_1} = \frac{3}{4}.$$
Question 9
Physics · Gravitation · Single correct
For a body projected at an angle with the horizontal from the ground, choose the correct statement.
Gravitational potential energy is maximum at the highest point.
The horizontal component of velocity is zero at highest point.
The vertical component of momentum is maximum at the highest point.
The kinetic energy (K.E.) is zero at the highest point of projectile motion.
Answer: (a)
Solution
At highest point $V_y = 0$ $V_x = u_x = u \cos \theta$ $U_g = mgh$, it is maximum at $H_{max}$
Question 10
Physics · Mechanical Properties of Solids · Single correct
The Young's modulus of a steel wire of length 6 m and cross-sectional area 3 mm$^2$, is $2 \times 10^{11} \, \mathrm{N/m}^2$. The wire is suspended from its support on a given planet. A block of mass 4 kg is attached to the free end of the wire. The acceleration due to gravity on the planet is $\frac{1}{4}$ of its value on the earth. The elongation of wire is (Take $g$ on the earth $= 10 \, \mathrm{m/s}^2$):
Physics · Mechanical Properties of Fluids · Numerical
The surface of water in a water tank of cross section area $750 \, \mathrm{cm}^2$ on the top of a house is $h \, \mathrm{m}$ above the tap level. The speed of water coming out through the tap of cross section area $500 \, \mathrm{mm}^2$ is $30 \, \mathrm{cm/s}$. At that instant, $\frac{dh}{dt}$ is $x \times 10^{-3} \, \mathrm{m/s}$. The value of $x$ will be
For three low density gases A, B, C pressure versus temperature graphs are plotted while keeping them at constant volume, as shown in the figure. The temperature corresponding to the point 'K' is:
\[ -273^\circ\text{C} \]
\[ -100^\circ\text{C} \]
\[ -373^\circ\text{C} \]
\[ -40^\circ\text{C} \]
Answer: (a)
Solution
For isochoric process $$\frac{P}{T} = n \frac{R}{V} = constant$$ $$P = \frac{nR}{V} (t + 273)$$ If $P = 0$ then $t = -273^\circ \mathrm{C}$
Question 13
Physics · Thermodynamics · Single correct
A Carnot engine operating between two reservoirs has efficiency $\frac{1}{3}$. When the temperature of cold reservoir raised by $x$, its efficiency decreases to $\frac{1}{6}$. The value of $x$, if the temperature of hot reservoir is $99^\circ \, \mathrm{C}$, will be:
Choose the correct length (L) versus square of time period ($T^2$) graph for a simple pendulum executing simple harmonic motion.
Answer: (c)
Solution
The formula for the period of a pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g}}$. Squaring both sides, we have $$T^2 = \frac{4\pi^2}{g} \times \ell.$$ This shows that $T^2$ is proportional to $\ell$.
Question 15
Physics · Electric Charges and Fields · Numerical
A cubical volume is bounded by the surfaces $x = 0$, $x = a$, $y = 0$, $y = a$, $z = 0$, $z = a$. The electric field in the region is given by $\vec{E} = E_0 x \hat{i}$. Where $E_0 = 4 \times 10^4 \, \mathrm{NC}^{-1} \, \mathrm{m}^{-1}$. If $a = 2 \, \mathrm{cm}$, the charge contained in the cubical volume is $Q \times 10^{-14} \, \mathrm{C}$. The value of $Q$ is ____. Take $\epsilon_0 = 9 \times 10^{-12} \, \mathrm{C}^2/\mathrm{Nm}^2$.
Answer: 288
Solution
Given $\vec{E} = E_0 x \hat{i}$. The net flux $\phi_{net} = \phi_{ABCD} = E_0 a \cdot a^2$. The charge enclosed $\frac{q_{en}}{\varepsilon_0} = E_0 a^3$. Therefore, $q_{en} = E_0 \varepsilon_0 a^3 = 4 \times 10^4 \times 9 \times 10^{-12} \times 8 \times 10^{-6} = 288 \times 10^{-14} \, C$. Thus, $Q = 288$. Answer: 288.
Question 16
Physics · Electrostatic Potential and Capacitance · Single correct
Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R. Assertion A : Two metallic spheres are charged to the same potential. One of them is hollow and another is solid, and both have the same radii. Solid sphere will have lower charge than the hollow one. Reason R : Capacitance of metallic spheres depend on the radii of spheres. In the light of the above statements, choose the correct answer from the options given below.
A is false but R is true
Both A and R are true and R is the correct explanation of A
A is true but R is false
Both A and R are true but R is not the correct explanation of A
Answer: (a)
Solution
Potential of a conducting sphere is $$V = \frac{KQ}{R}$$ (Solid as well as hollow) $$V_1 = V_2 and R_1 = R_2$$ Therefore, $$Q_1 = Q_2$$
Question 17
Physics · Current Electricity · Single correct
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R. Assertion A: For measuring the potential difference across a resistance of 600 $\Omega$, the voltmeter with resistance 1000 $\Omega$ will be preferred over voltmeter with resistance 4000 $\Omega$. Reason R: Voltmeter with higher resistance will draw smaller current than voltmeter with lower resistance. In the light of the above statements, choose the most appropriate answer from the options given below.
A is not correct but R is correct
Both A and R are correct and R is the correct explanation of A
Both A and R are correct but R is not the correct explanation of A
A is correct but R is not correct
Answer: (a)
Solution
Error of voltmeter decreases with increase in its resistance.
Question 18
Physics · Current Electricity · Single correct
Equivalent resistance between the adjacent corners of a regular n-sided polygon of uniform wire of resistance R would be:
$\frac{(n-1)R}{n^2}$
$\frac{(n-1)R}{(2n-1)}$
$\frac{n^2 R}{n-1}$
$\frac{(n-1)R}{n}$
Answer: (a)
Solution
Suppose resistance of each arm is $r$, then $r = \frac{R}{n}$. $$R_{eq(AB)} = \frac{R_1 R_2}{R_1 + R_2}$$ $$\frac{r(n-1)r}{r + (n-1)r}$$ $$= \frac{r(n-1)r}{nr}$$ $$= \frac{n-1}{n} r$$ $$= \frac{(n-1)R}{n^2}$$
Question 19
Physics · Current Electricity · Numerical
In the given circuit the value of $\left| \frac{I_1 + I_3}{I_2} \right|$ is:
Answer: 2
Solution
Given the circuit, we have the following calculations: $$I_1 = I_2 = \frac{20 - 10}{10} = 1 \, \mathrm{A}$$ Thus, $I_3 = 1 \, \mathrm{A}$. The equation for the currents is: $$\left| \frac{I_1 + I_3}{I_2} \right| = 2$$
Question 20
Physics · Moving Charges and Magnetism · Single correct
A coil is placed in magnetic field such that plane of coil is perpendicular to the direction of magnetic field. The magnetic flux through a coil can be changed: A. By changing the magnitude of the magnetic field within the coil. B. By changing the area of coil within the magnetic field. C. By changing the angle between the direction of magnetic field and the plane of the coil. D. By reversing the magnetic field direction abruptly without changing its magnitude. Choose the most appropriate answer from the options given below:
A and B only
A, B and C only
A, B and D only
A and C only
Answer: (b)
Solution
Given $\phi = \mathbf{B} \cdot \mathbf{A}$. $$= BA \cos \theta$$ Most suitable ans is 2 [Otherwise ABCD]
Question 21
Physics · Moving Charges and Magnetism · Single correct
As shown in the figure, a long straight conductor with semicircular arc of radius $\frac{\pi}{10} \, \mathrm{m}$ is carrying current $I = 3 \, \mathrm{A}$. The magnitude of the magnetic field at the center $O$ of the arc is: (The permeability of the vacuum $= 4\pi \times 10^{-7} \, \mathrm{NA^{-2}}$)
$6 \, \mu\mathrm{T}$
$1 \, \mu\mathrm{T}$
$4 \, \mu\mathrm{T}$
$3 \, \mu\mathrm{T}$
Answer: (d)
Solution
The magnetic field at the center of a circular arc is given by $$B_C = \frac{\mu_0 I}{4 \pi R} \left( \pi \right) (B at centre of circular arc)$$ Simplifying, we have $$= \frac{\mu_0 I}{4 R} = \frac{4 \pi \times 10^{-7} \times 3}{4 \times \frac{\pi}{10}}$$ This results in $$= 3 \times 10^{-6} \, \mathrm{T} = 3 \, \mu \mathrm{T}$$
Question 22
Physics · Moving Charges and Magnetism · Numerical
A square shaped coil of area $70 \, \mathrm{cm}^2$ having 600 turns rotates in a magnetic field of $0.4 \, \mathrm{wbm}^{-2}$, about an axis which is parallel to one of the side of the coil and perpendicular to the direction of field. If the coil completes 500 revolution in a minute, the instantaneous emf when the plane of the coil is inclined at $60^\circ$ with the field, will be ____ V. (Take $\pi = \frac{22}{7}$)
Physics · Ray Optics and Optical Instruments · Single correct
Two objects A and B are placed at 15 cm and 25 cm from the pole in front of a concave mirror having radius of curvature 40 cm. The distance between images formed by the mirror is:
As shown in the figure, in Young's double slit experiment, a thin plate of thickness $t = 10 \, \mu \mathrm{m}$ and refractive index $\mu = 1.2$ is inserted in front of slit $S_1$. The experiment is conducted in air ($\mu = 1$) and uses a monochromatic light of wavelength $\lambda = 500 \, \mathrm{nm}$. Due to the insertion of the plate, central maxima is shifted by a distance of $x \beta_0$. $\beta_0$ is the fringe-width before the insertion of the plate. The value of the $x$ is _____.
Physics · Dual Nature of Radiation and Matter · Single correct
The threshold frequency of metal is $f_0$. When the light of frequency $2f_0$ is incident on the metal plate, the maximum velocity of photoelectron is $v_1$. When the frequency of incident radiation is increased to $5f_0$, the maximum velocity of photoelectrons emitted is $v_2$. The ratio of $v_1$ to $v_2$ is:
$\frac{v_1}{v_2} = \frac{1}{2}$
$\frac{v_1}{v_2} = \frac{1}{8}$
$\frac{v_1}{v_2} = \frac{1}{16}$
$\frac{v_1}{v_2} = \frac{1}{4}$
Answer: (a)
Solution
Given $K_{max} = hf - hf_0$. For $f = 2f_0$, $$\frac{1}{2} m V_1^2 = 2hf_0 - hf_0 = hf_0.$$ For $f = 5f_0$, $$\frac{1}{2} m V_2^2 = 5hf_0 - hf_0 = 4hf_0.$$ Therefore, $$\frac{V_1}{V_2} = \frac{1}{2}.$$
Question 27
Physics · Atoms · Single correct
An electron of a hydrogen like atom, having $Z = 4$, jumps from $4^{th}$ energy state to $2^{nd}$ energy state, The energy released in this process, will be: (Given $Rch = 13.6 \, eV$) Where $R$ = Rydberg constant $c$ = Speed of light in vacuum $h$ = Planck's constant
13.6 eV
10.5 eV
3.4 eV
40.8 eV
Answer: (d)
Solution
The change in energy is given by the formula: $$\Delta E = 13.6 Z^2 \left[ \frac{1}{2^2} - \frac{1}{4^2} \right] eV$$ Substituting the values, we have: $$= 13.6 \times (4)^2 \left( \frac{1}{4} - \frac{1}{16} \right) eV.$$ Simplifying further: $$= 13.6 [4 - 1] eV$$ $$= 13.6 \times 3 = 40.8 eV$$
Question 28
Physics · Atoms · Numerical
Nucleus a having $Z = 17$ and equal number of protons and neutrons has $1.2 \, \mathrm{MeV}$ binding energy per nucleon. Another nucleus B of $Z = 12$ has total $26$ nucleons and $1.8 \, \mathrm{MeV}$ binding energy per nucleons. The difference of binding energy of B and A will be _____ MeV.
Answer: 6
Solution
Total binding energy = $1.2 \times 34 = 40.8 \, \mathrm{MeV}$ For B mass number = 26 total binding energy = $1.8 \times 26 \, \mathrm{MeV}$ $$= 46.8 \, \mathrm{MeV}$$ Difference of BE = $6 \, \mathrm{MeV}$
Question 29
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
Choose the correct statement about Zener diode:
It works as a voltage regulator in reverse bias and behaves like simple pn junction diode in forward bias.
It works as a voltage regulator in both forward and reverse bias.
It works a voltage regulator only in forward bias.
It works as a voltage regulator in forward bias and behaves like simple pn junction diode in reverse bias.
Answer: (a)
Solution
Works as voltage regulator in reverse bias and as simple P-n junction in forward bias.
Question 30
Physics · Communication Systems · Single correct
In an amplitude modulation, a modulating signal having amplitude of $X \, \mathrm{V}$ is superimposed with a carrier signal of amplitude $Y \, \mathrm{V}$ in first case. Then, in second case, the same modulating signal is superimposed with different carrier signal of amplitude $2Y \, \mathrm{V}$. The ratio of modulation index in the two case respectively will be:
Chemistry · Some Basic Concepts of Chemistry · Numerical
The molality of a 10$\%$ (v/v) solution of di-bromine solution in CCl_4 (carbon tetrachloride) is 'x'. $x = \times 10^{-2}$ M. (Nearest integer) [Given : molar mass of $Br_2 = 160$ g mol$^{-1}$ atomic mass of $C = 12$ g mol$^{-1}$ atomic mass of $Cl = 35.5$ g mol$^{-1}$ density of dibromine $= 3.2$ g cm$^{-3}$ density of $CCl_4 = 1.6$ g cm$^{-3}$]
Answer: 139
Solution
10 ml solute in 90 ml solvent. Mass of solute = $10 \times 3.2 = 32 \, \mathrm{g}$. Mass of solvent = $90 \times 1.6 \, \mathrm{g}$. $$m = \frac{32 \times 1000}{160 \times 90 \times 1.6} = 1.388$$ $$m = 138.8 \times 10^{-2} = 139$$
Question 32
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
Which one of the following sets of ions represents a collection of isoelectronic species? (Given : Atomic Number : F : 9 , Cl : 17, Na = 11, Mg = 12, Al = 13, K = 19, Ca = 20, Sc = 21)
0.3 g of ethane undergoes combustion at $27^\circ \mathrm{C}$ in a bomb calorimeter. The temperature of calorimeter system (including the water) is found to rise by $0.5^\circ \mathrm{C}$. The heat evolved during combustion of ethane at constant pressure is ________ kJ mol$^{-1}$. (Nearest integer) [Given : The heat capacity of the calorimeter system is $20 \, \mathrm{kJ} \, \mathrm{K}^{-1}$, $R = 8.3 \, \mathrm{JK}^{-1} \, \mathrm{mol}^{-1}$. Assume ideal gas behaviour. Atomic mass of C and H are 12 and 1 g mol$^{-1}$ respectively]
Answer: 1006
Solution
(Bomb calorimeter $\rightarrow$ constant volume) Heat released By combustion of 1 mole $C_2H_6 \, (\Delta U) = \dfrac{20 \times 0.5}{0.3} \times 30 = 1000\,\mathrm{kJ}$ $C_2H_6(g) + \dfrac{7}{2}O_2(g) \rightarrow 2CO_2(g) + 3H_2O(l)$ $\Delta n_g = 2 - \left(1 + \dfrac{7}{2}\right) = -\dfrac{5}{2}$ $\Delta H = \Delta U + \Delta nRT$ $= -1000 - \dfrac{5}{2} \times 8.3 \times 300 \times 10^{-3}\,\mathrm{kJ}$ $= -1000 - 6.225$ $= -1006\,\mathrm{kJ}$ So heat released $= 1006\,\mathrm{kJ\,mol^{-1}}$
Question 35
Chemistry · Equilibrium · Single correct
The effect of addition of helium gas to the following reaction in equilibrium state, is : $$\mathrm{PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)}$$
the equilibrium will shift in the forward direction and more of $\mathrm{Cl_2}$ and $\mathrm{PCl_3}$ gases will be produced.
the equilibrium will go backward due to suppression of dissociation of $\mathrm{PCl_5}$.
helium will deactivate $\mathrm{PCl_5}$ and reaction will stop.
addition of helium will not affect the equilibrium.
Answer: (d)
Solution
Given the reaction: $$\mathrm{PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)}$$ Case 1: At constant pressure, volume will increase so the reaction will shift in the forward direction. Then the answer will be A. Case 2: At constant volume, no change in active mass so the reaction will not shift in any direction. Then the answer will be B.
Question 36
Chemistry · Analytical Chemistry · Single correct
Given below are two statements: one is labelled as \textbf{Assertion (A)} and the other is labelled as \textbf{Reason (R)}. \textbf{Assertion (A):} An aqueous solution of $KOH$ when used for volumetric analysis, its concentration should be checked before the use. \textbf{Reason (R):} On aging, $KOH$ solution absorbs atmospheric $CO_2$. In the light of the above statements, choose the correct answer from the options given below.
is not correct but ($R$) is correct
Both (A) and ($R$) are correct but ($R$) is not the correct explanation of (A)
Both (A) and ($R$) are correct and ($R$) is the correct explanation of (A)
is correct but ($R$) is not correct
Answer: (c)
Solution
KOH absorbs $\mathrm{CO_2}$. So its concentration should be checked.
Question 37
Chemistry · Hydrogen · Single correct
O-O bond length in $\mathrm{H_2O_2}$ is $X$ than the O-O bond length in $\mathrm{F_2O_2}$. The O-H bond length in $\mathrm{H_2O_2}$ is $Y$ than that of the O-F bond in $\mathrm{F_2O_2}$. Choose the correct option for $X$ and $Y$ from the given below.
X - shorter, Y - shorter
X - shorter, Y - longer
X - longer, Y - longer
X - longer, Y - shorter
Answer: (d)
Solution
According to bent rule more electronegative atom occupy less s-characters so bond length increases. O–H bond will be short than O–F bond due to small size of H than F.
Question 38
Chemistry · Hydrogen · Single correct
The starting material for convenient preparation of deuterated hydrogen peroxide ($D_2O_2$) in laboratory is:
$K_2S_2O_8$
2-ethylanthraquinol
$BaO_2$
$BaO$
Answer: (a)
Solution
The chemical reaction is given by: $$(\mathrm{K_2S_2O_8(s)} + 2\mathrm{D_2O(l)} \rightarrow 2\mathrm{KDSO_4(aq.)} + \mathrm{D_2O_2})$$
Question 39
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R ). Assertion (A): Gypsum is used for making fireproof wall boards. Reason (R ): Gypsum is unstable at high temperatures. In the light of the above statements, choose the correct answer from the options given below:
Both (A) and (R ) are correct but (R ) is not the correct explanation of (A).
is correct but (R ) is not correct.
is not correct but (R ) is correct.
Both (A) and (R ) are correct and (R ) is the correct explanation of (A).
Answer: (a)
Solution
Gypsum is used for making fireproof wall boards.
Question 40
Chemistry · The p-Block Elements (Group-13 and 14) · Single correct
The correct order of bond enthalpy $(kJ mol^{-1})$ is
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Given below are two statements : Statement I : Sulphanilic acid gives esterification test for carboxyl group. Statement II : Sulphanilic acid gives red colour in Lassigne’s test for extra element detection. In the light of the above statements, choose the most appropriate answer from the options given below :
Statement I is correct but Statement II is incorrect.
Both Statement I and Statement II are incorrect.
Both Statement I and Statement II are correct.
Statement I is incorrect but Statement II is correct.
Answer: (d)
Solution
Sulphanilic acid does not show esterification test. Presence of both sulphur and nitrogen give red colour in Lassaigne's test.
Testosterone, which is a steroidal hormone, has the following structure. The total number of asymmetric carbon atom/s in testosterone is
Answer: 6
Solution
The structure shown is testosterone. It contains several chiral centers, indicated by the asterisks.
Question 43
Chemistry · Hydrocarbons · Single correct
Answer: (a)
Solution
Question 44
Chemistry · Environmental Chemistry · Single correct
The industrial activity held least responsible for global warming is :
manufacturing of cement
steel manufacturing
Electricity generation in thermal power plants
Industrial production of urea
Answer: (d)
Solution
In urea production $\mathrm{NH_3}$ and $\mathrm{CO_2}$ consumed so least responsible for global warming.
Question 45
Chemistry · The Solid State · Numerical
A metal M crystallizes into two lattices :- face centred cubic (fcc) and body centred cubic (bcc) with unit cell edge length of 2.0 and 2.5 $\AA$ respectively. The ratio of densities of lattices fcc to bcc for the metal M is _______. (Nearest integer)
Answer: 4
Solution
The formula for density is given by $$d = \frac{Z \times M}{N_A a^3}$$ For FCC and BCC structures, the ratio of densities is $$\frac{d_{FCC}}{d_{BCC}} = \frac{\frac{4 \times M_w}{N_A \times (2)^3}}{\frac{2 \times M_w}{N_A \times (2.5)^3}} = 3.90$$
Question 46
Chemistry · Solutions · Numerical
$20\%$ of acetic acid is dissociated when its $5 \, \mathrm{g}$ is added to $500 \, \mathrm{mL}$ of water. The depression in freezing point of such water is _______ $\times 10^{-3} \, ^\circ \mathrm{C}$. Atomic mass of C, H and O are $12, 1$ and $16 \, \mathrm{a.m.u.}$ respectively. [Given: Molal depression constant and density of water are $1.86 \, \mathrm{K \, kg \, mol^{-1}}$ and $1 \, \mathrm{g \, cm^{-3}}$ respectively.]
Answer: 372
Solution
Given the equation for the van't Hoff factor: $$i = 1 + (n - 1) \alpha$$ Substituting the given values: $$(i = 1 + 0.2 \times (2 - 1) = 1.2)$$ The change in freezing point is given by: $$\Delta T_f = i \times K_f \times m$$ Substituting the values: $$\Delta T_f = 1.2 \times 1.86 \times \frac{5 \times 1000}{60 \times 500}$$ Calculating the change in freezing point: $$\Delta t_f = 3.72$$ Expressing the change in freezing point in scientific notation: $$\Delta T_f = 372 \times 10^{-2}$$
Question 47
Chemistry · Electrochemistry · Numerical
1 $\times 10^{-5}$ $\mathrm{M}$ $\mathrm{AgNO_3}$ is added to 1 $\mathrm{L}$ of saturated solution of $\mathrm{AgBr}$. The conductivity of this solution at 298 $\mathrm{K}$ is $\times 10^{-8}$ $\mathrm{S}$ $\mathrm{m}^{-1}$. [Given: $K_{sp}$($\mathrm{AgBr}$) = 4.9 $\times 10^{-13}$ at 298 $\mathrm{K}$] $\lambda^0_\mathrm{{Ag}^+}$ = 6 $\times 10^{-3}$ $\mathrm{S}$ $\mathrm{m}^2$ $\mathrm{mol}^{-1}$ $\lambda^0_\mathrm{{Br}^-}$ = 8 $\times 10^{-3}$ $\mathrm{S}$ $\mathrm{m}^2$ $\mathrm{mol}^{-1}$ $\lambda^0_\mathrm{{NO_3}^-}$ = 7 $\times 10^{-3}$ $\mathrm{S}$ $\mathrm{m}^2$ $\mathrm{mol}^{-1}$
Chemistry · Chemical Kinetics and Nuclear Chemistry · Single correct
The graph which represents the following reaction is: $$(\mathrm{C_6H_5})_3\mathrm{C-Cl} \xrightarrow[Pyridine]{OH^-} (\mathrm{C_6H_5})_3\mathrm{C-OH}$$
Answer: (c)
Solution
It is SN1 reaction so rate of reaction depends on the concentration of alkyl halide only.
Question 49
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
A $\rightarrow$ B The above reaction is of zero order. Half life of this reaction is 50 min. The time taken for the concentration of A to reduce to one-fourth of its initial value is _______ min. (Nearest integer)
Answer: 75
Solution
Assume reaction starts with $1$ mole A. $t_{1/2} = \frac{a}{2k}$, $K = \frac{1}{2 \times 50}$. For $75\%$ completion: \[ a - \frac{a}{4} = kt \] \[ t = \frac{3a}{4k} = \frac{3}{4} \times \frac{100}{a} = 75 \]
Question 50
Chemistry · Surface Chemistry · Single correct
In the figure, a straight line is given for Freundlich Adsorption ($y = 3x + 2.505$). The value of $\frac{1}{n}$ and $\log K$ are respectively.
0.3 and $\log 2.505$
0.3 and 0.7033
3 and 2.505
3 and 0.7033
Answer: (c)
Solution
Given $\frac{x}{m} = Kp^{1/n}$. Taking the logarithm, we have $\log \frac{x}{m} = \log k + \frac{1}{n} \log P$. Therefore, $Y = 3x + 2.505$, $\frac{1}{n} = 3$, $\log K = 2.505$.
Question 51
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
Among following compounds, the number of those present in copper matte is ________.
CuCO_3
Cu_2S
Cu_2O
FeO
Answer: (c)
Solution
FeS and Cu_2S, present in copper matte
Question 52
Chemistry · The d-and f-Block Elements · Single correct
Given below are two statements : one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : $\mathrm{Cu}^{2+}$ in water is more stable than $\mathrm{Cu}^{+}$. Reason (R) : Enthalpy of hydration for $\mathrm{Cu}^{2+}$ is much less than that of $\mathrm{Cu}^{+}$. In the light of the above statements, choose the correct answer from the options given below :
Both (A) and (R) are correct and (R) is the correct explanation of (A).
is correct but (R) is not correct.
(1) is not correct but (R) is correct.
Both (A) and (R) are correct but (R) is not the correct explanation of (A).
Answer: (a)
Solution
The reaction is given by $$2\mathrm{Cu}^+ \rightarrow \mathrm{Cu}^{2+} + \mathrm{Cu}$$. The stability of $\mathrm{Cu}^{2+}(\mathrm{aq})$ rather than $\mathrm{Cu}^+(\mathrm{aq})$, is due to the much more negative $\Delta_{\mathrm{hyd}}H$ of $\mathrm{Cu}^{2+}(\mathrm{aq})$ than $\mathrm{Cu}^+(\mathrm{aq})$, which more than compensates for the second ionisation enthalpy of Cu.
Question 53
Chemistry · The d-and f-Block Elements · Single correct
Which element is not present in Nessler’s reagent ?
Mercury
Potassium
Iodine
Oxygen
Answer: (d)
Solution
Nessler's Reagent is $\mathrm{K_2[HgI_4]}$.
Question 54
Chemistry · Co-ordination Compounds · Single correct
The complex cation which has two isomers is
$[\mathrm{Co(H_2O)_6}]^{3+}$
$[\mathrm{Co(NH_3)_5Cl}]^{2+}$
[\mathrm{Co(NH_3)_5NO_2}]^{2+}$
$[\mathrm{Co(NH_3)_5Cl}]^{+}$
Answer: (c)
Solution
For the complex $[\mathrm{Co(NH_3)_5NO_2}]^{2+}$, two linkage isomers are possible.
Question 55
Chemistry · Co-ordination Compounds · Numerical
The spin only magnetic moment of $[\mathrm{Mn}(\mathrm{H}_2\mathrm{O})_6]^{2+}$ complexes is _____ B.M. (Nearest integer)
Answer: 6
Solution
Given: Atomic no. of Mn is 25. $$[\mathrm{Mn(H_2O)_6}]^{2+}$$ $$\mathrm{Mn^{2+} = 3d^5}$$ $$\mu = \sqrt{5(5 + 2)} = 5.91 \mathrm{BM}$$
Question 56
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
The structures of major products A, B and $C$ in the following reaction are in sequence.
Answer: (d)
Solution
Question 57
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
In a reaction, reagent 'X' and 'Y' respectively are:
($\mathrm{CH_3CO_2OH/H^+}$ and $\mathrm{CH_3OH/H^+}$, $\Delta$)
($\mathrm{CH_3CO_2OH}$ and $\mathrm{(CH_3CO_2)_2O/H^+}$)
($\mathrm{CH_3OH/H^+, \Delta}$ and $\mathrm{CH_3OH/H^+}$)
$\mathrm{CH_3OH/H^+, \Delta}$ and $\mathrm{(CH_3CO_2)_2O/H^+}$
Answer: (a)
Solution
Question 58
Chemistry · Chemistry in Everyday Life · Numerical
Among the following, the number of tranquilizer/s is/are _________.
Chloroliazepoxide
Veronal
Valium
Salvarsan
Answer: (c)
Solution
(chlordiazepoxide, Veronal, Valium is tranquilizer whereas salvarsan is antibiotic.
Question 59
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
All structures given below are of vitamin C. Most stable of them is :
Answer: (a)
Solution
H-bonding stabilised vitamin C.
Question 60
Chemistry · Amines · Single correct
Given below are two statements : one is labelled as \textbf{Assertion (A)} and the other is labelled as \textbf{Reason (R)}. \textbf{Assertion (A) :} $\alpha$-halocarboxylic acid on reaction with dil. $\text{NH}_3$ gives good yield of $\alpha$-amino carboxylic acid whereas the yield of amines is very low when prepared from alkyl halides. \textbf{Reason (R) :} Amino acids exist in zwitter ion form in aqueous medium. In the light of the above statements, choose the \textbf{correct} answer from the options given below :
Both (A) and (R ) are correct and (R ) is the correct explanation of (A).
Both (A) and (R ) are correct but (R ) is not the correct explanation of (A).
(A) is correct but (R ) is not correct.
(A) is not correct but (R ) is correct.
Answer: (a)
Solution
Fact based
Maths
Question 61
Maths · Complex Numbers and Quadratic Equations · Numerical
The number of integral values of $k$, for which one root of the equation $2x^2 - 8x + k = 0$ lies in the interval $(1, 2)$ and its other root lies in the interval $(2, 3)$, is:
Answer: 3
Solution
Question 62
Maths · Complex Numbers and Quadratic Equations · Single correct
Let $a,b$ be two real numbers such that $ab<0$. If the complex number $\frac{1+ai}{b+i}$ is of unit modulus and $a+ib$ lies on the circle $|z-1|=|2z|$, then a possible value of $\frac{1+[a]}{4b}$, where $[t]$ is the greatest integer function, is:
-$\frac{1}{2}$
-1
1
$\frac{1}{2}$
Answer: (a)
Solution
Given $ab < 0$ and $\($ $\frac{1 + ai}{b + i}$ = 1 $\)$. We have $|1 + ai| = |b + i|$. This implies $a^2 + 1 = b^2 + 1 \Rightarrow a = \pm b \Rightarrow b = -a$ as $ab < 0$. The pair $(a, b)$ lies on $|z - 1| = |2z|$. $|a + ib - 1| = 2|a + ib|$. $(a - 1)^2 + b^2 = 4(a^2 + b^2)$. $(a - 1)^2 = a^2 = 4(2a^2)$. $1 - 2a = 6a^2 \Rightarrow 6a^2 + 2a - 1 = 0$. $a = \frac{-2 \pm \sqrt{28}}{12} = \frac{-1 \pm \sqrt{7}}{6}$. $a = \frac{\sqrt{7} - 1}{6}$ and $b = \frac{1 - \sqrt{7}}{6}$. $[a] = 0$. Therefore, $\frac{1 + [a]}{4b} = \frac{6}{4(1 - \sqrt{7})} = -\left(\frac{1 + \sqrt{7}}{4}\right)$. Or $[a] = 0$. Similarly, it is not matching with $a = \frac{-1 - \sqrt{7}}{6}$. No answer is matching.
Question 63
Maths · Permutations and Combinations · Numerical
Number of integral solutions to the equation $x + y + z = 21$, where $x \geq 1$, $y \geq 3$, $z \geq 4$, is equal to ______.
Answer: 105
Solution
The combination $^{15}C_2$ is calculated as follows: $$^{15}C_2 = \frac{15 \times 14}{2} = 105.$$
Question 64
Maths · Permutations and Combinations · Numerical
The total number of six digit numbers, formed using the digits 4, 5, 9 only and divisible by 6, is ______.
The sum $\sum_{n=1}^{\infty} \frac{2n^2 + 3n + 4}{(2n)!}$ is equal to:
$\frac{11e}{2} + \frac{7}{2e}$
$\frac{13e}{4} + \frac{5}{4e} - 4$
$\frac{11e}{2} + \frac{7}{2e} - 4$
$\frac{13e}{4} + \frac{5}{4e}$
Answer: (b)
Solution
Question 66
Maths · Sequences and Series · Fill in the blank
The sum of the common terms of the following three arithmetic progressions. 3, 7, 11, 15, $\ldots$, 399, 2, 5, 8, 11, $\ldots$, 359 and 2, 7, 12, 17, $\ldots$, 197, is equal to _____.
Answer: 321
Solution
The sequences are: $3, 7, 11, 15, \ldots, 399$ with $d_1 = 4$ $2, 5, 8, 11, \ldots, 359$ with $d_2 = 3$ $2, 7, 12, 17, \ldots, 197$ with $d_3 = 5$ The least common multiple of $d_1$, $d_2$, and $d_3$ is $\mathrm{LCM}(d_1, d_2, d_3) = 60$. The common terms are $47, 107, 167$. The sum is $321$.
Question 67
Maths · Binomial Theorem · Numerical
If the term without $x$ in the expansion of $$\left(\frac{2}{x^3} + \frac{\alpha}{x^3}\right)^{22}$$ is 7315, then $|\alpha|$ is equal to .
Answer: 1
Solution
The expression for $T_{r+1}$ is given by $T_{r+1}=$ ${}^{22}C_{r} \left(\frac{2}{3}\right)^{22-r} (\alpha)^{r} x^{-3r}$ Simplifying, we have $= {}^{22}C_{r}\, x^{\frac{44}{3}-\frac{2r}{3}-3r} (\alpha)^{r}$ Equating the powers of $x,$ we get $\frac{44}{3}=\frac{11r}{3}$ Solving for $r,$ we find $r=4$ Substituting $r=4$ into the expression, we have ${}^{22}C_{4}\cdot\alpha^{4}=7315$ Calculating the binomial coefficient, $\frac{22\times21\times20\times19}{24}\cdot\alpha^{4}=7315$ Solving for $\alpha,$ we find $\alpha=1$
Question 68
Maths · Binomial Theorem · Fill in the blank
Let the sixth term in the binomial expansion of $$\left( \sqrt{2^{\log_2 \left(10 - 3^x\right)}} + \sqrt[5]{2^{(x-2)\log_2 3}} \right)^m$$, in the increasing powers of $$2^{(x-2)\log_2 3}$$, be 21. If the binomial coefficients of the second, third and fourth terms in the expansion are respectively the first, third and fifth terms of an A.P., then the sum of the squares of all possible values of x is _____.
Answer: 4
Solution
Given $T_6={}^{m}C_{5}\left(10-3^x\right)^{\frac{m-5}{2}}\cdot\left(3^{x-2}\right)=21$ ${}^{m}C_{1},\ {}^{m}C_{2},\ {}^{m}C_{3}$ are in A.P. Therefore, $2\cdot{}^{m}C_{2}={}^{m}C_{1}+{}^{m}C_{3}$ Solving for $m,$ we get $m=2$ (rejected), $7$ Put in equation (1): $21\cdot(10-3^x)\cdot\frac{3^x}{9}=21$ $3^{2x}-10\cdot3^x+9=0$ $(3^x-1)(3^x-9)=0$ Therefore, $3^x=3^0,\ 3^2$ Therefore, $x=0,\ 2$
Question 69
Maths · Conic Sections · Numerical
If the x-intercept of a focal chord of the parabola $y^2 = 8x + 4y + 4$ is 3, then the length of this chord is equal to .
Answer: 16
Solution
Given $y^2 = 8x + 4y + 4$. Rewriting, we have $(y - 2)^2 = 8(x + 1)$. This is in the form $y^2 = 4ax$. Here, $a = 2$, $X = x + 1$, $Y = y - 2$. The focus is $(1, 2)$. The equation of the line is $y - 2 = m(x - 1)$. Put $(3, 0)$ in the above line to find $m = -1$. The length of the focal chord is $16$.
Question 70
Maths · Conic Sections · Numerical
The line $x = 8$ is the directrix of the ellipse $E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with the corresponding focus $(2, 0)$. If the tangent to $E$ at the point $P$ in the first quadrant passes through the point $(0, 4\sqrt{3})$ and intersects the $x$-axis at $Q$, then $(3PQ)^2$ is equal to _____.
Let $P(x_0, y_0)$ be the point on the hyperbola $3x^2 - 4y^2 = 36$, which is nearest to the line $3x + 2y = 1$. Then $\sqrt{2} (y_0 - x_0)$ is equal to:
-3
9
-9
3
Answer: (c)
Solution
Given the equations $3x^2 - 4y^2 = 36$ and $3x + 2y = 1$. The slope $m$ is given by $m = -\frac{3}{2}$. We have $m = + \frac{\sec \theta \cdot 3}{\sqrt{12} \cdot \tan \theta}$. This implies $$\frac{3}{\sqrt{12}} \times \frac{1}{\sin \theta} = -\frac{3}{2}.$$ Therefore, $$\sin \theta = -\frac{1}{\sqrt{3}}.$$ The coordinates are $\left( \sqrt{12} \cdot \sec \theta, 3 \tan \theta \right)$. Substituting, we get $$\left( \sqrt{12} \cdot \frac{\sqrt{3}}{\sqrt{2}}, -3 \times \frac{1}{\sqrt{2}} \right) \Rightarrow \left( \frac{6}{\sqrt{2}}, -\frac{3}{\sqrt{2}} \right).$$
Question 72
Maths · Mathematical Reasoning · Single correct
Which of the following statements is a tautology?
$p \to (p \land (p \to q))$
$(p \land q) \to (\sim (p) \to q)$
$(p \land (p \to q)) \to \sim q$
$p \lor (p \land q)$
Answer: (b)
Solution
\[ (i)\quad p \rightarrow \bigl(p \wedge (p \rightarrow q)\bigr) \] \[ = \neg p \vee \bigl(p \wedge (\neg p \vee q)\bigr) \] \[ = \neg p \vee \bigl((p \wedge \neg p)\vee (p \wedge q)\bigr) \] \[ = \neg p \vee (p \wedge q) \] \[ = (\neg p \vee p)\wedge (\neg p \vee q) \] \[ = \neg p \vee q \] \[ (ii)\quad (p \wedge q)\rightarrow (p \rightarrow q) \] \[ = \neg (p \wedge q)\vee (\neg p \vee q) \] \[ = (\neg p \vee \neg q)\vee (\neg p \vee q) \] \[ = \neg p \vee (\neg q \vee q) \] \[ = \neg p \vee T \] \[ = T \] \[ \therefore \text{Tautology} \] \[ (iii)\quad (p \wedge (p \rightarrow q))\rightarrow \neg q \] \[ = \neg \bigl(p \wedge (\neg p \vee q)\bigr)\vee \neg q \] \[ = \neg (p \wedge q)\vee \neg q \] \[ = (\neg p \vee \neg q)\vee \neg q \] \[ = \neg p \vee \neg q \] \[ \therefore \text{Not tautology} \] \[ (iv)\quad p \vee (p \wedge q)=p \] \[ \therefore \text{Not tautology.} \]
Question 73
Maths · Sequences and Series · Single correct
Let $9=x_1<x_2<\cdots<x_7$ be in an A.P. with common difference $d$. If the standard deviation of $x_1,x_2,\ldots,x_7$ is $4$ and the mean is $\bar{x}$, then $\bar{x}+x_6$ is equal to:
Maths · Relations and Functions (Advanced) · Single correct
Let $\mathrm{P}$(S) denote the power set of S= \{1,2,3, $\ldots$, 10$\}$. Define the relations R_1 and R_2 on $\mathrm{P}$(S) as AR_1B if (A $\cap$ B^c) $\cup$ (B $\cap$ A^c) = $\emptyset$ and AR_2B if A $\cup$ B^c = B $\cup$ A^c, $\forall$ A, B $\in$ $\mathrm{P}$(S). Then:
Both $R_1$ and $R_2$ are equivalence relations.
Only $R_1$ is an equivalence relation.
$\text{Only } R_2 \text{ is an equivalence relation}$
Both $R_1$ and $R_2$ are not equivalence relations.
Answer: (a)
Solution
Given $S = \{1, 2, 3, \ldots, 10\}$. $P(S) =$ power set of $S$. $AR, B \Rightarrow (A \cap \bar{B}) \cup (\bar{A} \cap B) = \phi$ $R_1$ is reflexive, symmetric. For transitive: $$(A \cap \bar{B}) \cup (\bar{A} \cap B) = \phi; \{a\} = \phi = \{b\} \Rightarrow A = B$$ $$(B \cap \bar{C}) \cup (\bar{B} \cap C) = \phi \Rightarrow B = C$$ Therefore, $A = C$ equivalence. $R_2 \equiv A \cup \bar{B} = \bar{A} \cup B$ $R_2 \rightarrow$ Reflexive, symmetric. For transitive: $$A \cup \bar{B} = \bar{A} \cup B \Rightarrow \{a, c, d\} = \{b, c, d\}$$ $$\{a\} = \{b\} \Rightarrow A = B$$ $$B \cup \bar{C} = \bar{B} \cup C \Rightarrow B = C$$ Therefore, $A = C$. $A \cup \bar{C} = \bar{A} \cup C$. Therefore, equivalence.
For the system of linear equations $ax + y + z = 1$, $x + ay + z = \beta$, $x + y + az = \beta$, which one of the following statements is NOT correct?
It has infinitely many solutions if $\alpha = 2$ and $\beta = -1$
It has no solution if $\alpha = -2$ and $\beta = 1$
$x + y + z = \frac{3}{4}$ if $\alpha = 2$ and $\beta = 1$
It has infinitely many solutions if $\alpha = 1$ and $\beta = 1$
Answer: (a)
Solution
Given the determinant equation: $$\begin{vmatrix} \alpha & 1 & 1 \\ 1 & \alpha & 1 \\ 1 & 1 & \alpha \end{vmatrix} = 0$$ Expanding the determinant, we have: $$\alpha (\alpha^2 - 1) - 1(\alpha - 1) + 1(1 - \alpha) = 0$$ Simplifying, we get: $$\alpha^3 - 3\alpha + 2 = 0$$ Factoring gives: $$\alpha^2 (\alpha - 1) + \alpha (\alpha - 1) - 2(\alpha - 1) = 0$$ $$(\alpha - 1)(\alpha^2 + \alpha - 2) = 0$$ Thus, $\alpha = 1, \alpha = -2, 1$ For $\alpha = 1$, $\beta = 1$ The system of equations becomes: $$\begin{cases} x + y + z = 1 \\ x + y + z = b \end{cases}$$ This results in an infinite solution. For $\alpha = 2$, $\beta = 1$ The determinant $\Delta = 4$ Calculating $\Delta_1$: $$\begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{vmatrix} = 3 - 1 - 1 = 1 \implies x = \frac{1}{4}$$ Calculating $\Delta_2$: $$\begin{vmatrix} 2 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 2 \end{vmatrix} = 2 - 1 = 1 \implies y = \frac{1}{4}$$ Calculating $\Delta_3$: $$\begin{vmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 1 \end{vmatrix} = 2 - 1 = 1 \implies z = \frac{1}{4}$$ For $\alpha = 2$, there is a unique solution.
Question 77
Maths · Inverse Trigonometric Functions · Single correct
Let $S=\left\{x\in\mathbb{R}:0<x<1\text{ and }2\tan^{-1}\left(\frac{1-x}{1+x}\right)=\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right\}$. If $n(S)$ denotes the number of elements in $S$, then:
$n(S) = 2$ and only one element in $S$ is less than $\frac{1}{2}$.
$n(S) = 1$ and the element in $S$ is more than $\frac{1}{2}$.
$n(S) = 1$ and the element in $S$ is less than $\frac{1}{2}$.
Let $f : \mathbb{R} - \{0, 1\} \rightarrow \mathbb{R}$ be a function such that $f(x) + f\left(\frac{1}{1-x}\right) = 1 + x$. Then $f(2)$ is equal to:
$\frac{9}{2}$
$\frac{9}{4}$
$\frac{7}{4}$
$\frac{7}{3}$
Answer: (b)
Solution
Given $f(x) + f\left(\frac{1}{1-x}\right) = 1 + x$. For $x = 2$, we have $f(2) + f(-1) = 3$ (1) For $x = -1$, we have $f(-1) + f\left(\frac{1}{2}\right) = 0$ (2) For $x = \frac{1}{2}$, we have $f\left(\frac{1}{2}\right) + f(2) = \frac{3}{2}$ (3) Adding (1) and (3) and subtracting (2), we get $2f(2) = \frac{9}{2}$. Therefore, $f(2) = \frac{9}{4}$.
Question 79
Maths · Continuity and Differentiability · Single correct
If $y(x) = x^x$, $x > 0$, then $y''(2) - 2y'(2)$ is equal to
Maths · Applications of Derivatives · Single correct
The sum of the abosolute maximum and minimum values of the function $f(x) = |x^2 - 5x + 6| - 3x + 2$ in the interval $[-1, 3]$ is equal to:
10
12
13
24
Answer: (a)
Solution
Given $f(x) = |x^2 - 5x + 6| - 3x + 2$. The piecewise function is defined as follows: $$f(x) = \begin{cases} x^2 - 8x + 8, & x \in [-1, 2] \\ -x^2 + 2x - 4, & x \in [2, 3] \end{cases}$$ The maximum value is $17$ and the minimum value is $-17$.
Question 81
Maths · Integrals · Single correct
The value of the integral $$\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x + \frac{\pi}{4}}{2 - \cos 2x} \, dx$$ is
If $$ \int_{0}^{\pi} \frac{5^{\cos x} \left( 1 + \cos x \cos 3x + \cos^2 x + \cos^3 x \cos 3x \right) \mathrm{d}x}{1 + 5^{\cos x}} = \frac{k \pi}{16}, $$ then $k$ is equal to .
Answer: 26
Solution
Given $$I = \int_0^{\pi} \frac{5^{\cos x} (1 + \cos x \cos 3x + \cos^2 x + \cos^3 x \cos 3x)}{1 + 5^{\cos x}} \, dx$$ We have $$I = \int_0^{\pi} \frac{5^{-\cos x} (1 + \cos x \cos 3x + \cos^2 x + \cos^3 x \cos 3x)}{1 + 5^{-\cos x}} \, dx$$ Adding these two equations, we get $$2I = \int_0^{\pi} (1 + \cos x \cos 3x + \cos^2 x + \cos^3 x \cos 3x) \, dx$$ This implies $$2I = 2 \int_0^{\pi/2} (1 + \cos x \cos 3x + \cos^2 x + \cos^3 x \cos 3x) \, dx$$ Now, $$I = \int_0^{\pi/2} (1 + \sin x (-\sin 3x) + \sin^2 x - \sin^3 x \sin 3x) \, dx$$ Thus, $$2I = \int_0^{\pi/2} (3 + \cos 4x + \cos^3 x \cos 3x - \sin^3 x \sin 3x) \, dx$$ Simplifying further, $$2I = \int_0^{\pi/2} \left( 3 + \cos 4x + \frac{\cos 3x + \cos 3x}{4} \cos 3x - \frac{3 \sin x \sin 3x}{4} \right) \, dx$$ This gives $$2I = \int_0^{\pi/2} \left( 3 + \cos 4x + \frac{1}{4} + \frac{3}{4} \cos 4x \right) \, dx$$ Finally, $$2I = \frac{13}{4} \times \frac{\pi}{2} + \frac{7}{4} \left( \frac{\sin 4x}{4} \right)_0^{\pi/2} \implies I = \frac{13\pi}{16}$$
Question 83
Maths · Applications of Integrals · Single correct
The area of the region given by $\{(x, y) : xy \leq 8, 1 \leq y \leq x^2\}$ is:
8 $\log_e$ 2 - $\frac{13}{3}$
16 $\log_e$ 2 - $\frac{14}{3}$
8 $\log_e$ 2 + $\frac{7}{6}$
16 $\log_e$ 2 + $\frac{7}{3}$
Answer: (b)
Solution
Question 84
Maths · Differential Equations · Single correct
Let $\alpha x = \exp(x^\beta y^\gamma)$ be the solution of the differential equation $2x^2y \, dy - (1 - xy^2) \, dx = 0$, $x > 0$, $y(2)=\sqrt{\log_e 2}$. Then $\alpha + \beta - \gamma$ equals :
1
-1
0
3
Answer: (a)
Solution
Given $\alpha x = e^{x^\beta \cdot y^\gamma}$. $2x^2 y \frac{dy}{dx} = 1 - x \cdot y^2$ Let $y^2 = t$. Then $x^2 \frac{dt}{dx} = 1 - xt$. Rearranging gives $\frac{dt}{dx} + \frac{t}{x} = \frac{1}{x^2}$. The integrating factor (I.F.) is $e^{\int \frac{1}{x} dx} = x$. Thus, $t(x) = \int \frac{1}{x^2} \cdot x \, dx$. This simplifies to $y^2 \cdot x = \ln x + C$. Therefore, $2 \cdot \ln 2 = \ln 2 + C$. So, $C = \ln 2$. Hence, $xy^2 = \ln 2x$. Thus, $2x = e^{x \cdot y^2}$. Hence $\alpha = 2$, $\beta = 1$, $\gamma = 2$.
Question 85
Maths · Vector Algebra · Single correct
Let \[ \vec{a}=5\hat{i}-\hat{j}-3\hat{k} \] and \[ \vec{b}=\hat{i}+3\hat{j}+5\hat{k} \] be two vectors. Then which one of the following statements is TRUE?
Projection of $\vec{a}$ on $\vec{b}$ is $\dfrac{17}{\sqrt{35}}$ and the direction of the projection vector is same as that of $\vec{b}$.
Projection of $\vec{a}$ on $\vec{b}$ is $\dfrac{17}{\sqrt{35}}$ and the direction of the projection vector is opposite to the direction of $\vec{b}$.
Projection of $\vec{a}$ on $\vec{b}$ is $-\dfrac{17}{\sqrt{35}}$ and the direction of the projection vector is same as that of $\vec{b}$.
Projection of $\vec{a}$ on $\vec{b}$ is $-\dfrac{17}{\sqrt{35}}$ and the direction of the projection vector is opposite to the direction of $\vec{b}$.
Answer: (a)
Solution
Given $\vec{a} = 5\hat{i} - \hat{j} - 3\hat{k}$ and $\vec{b} = \hat{i} - 3\hat{j} + 5\hat{k}$.
Question 86
Maths · Vector Algebra · Single correct
Let $\mathbf{a} = 2\mathbf{i} - 7\mathbf{j} + 5\mathbf{k}$, $\mathbf{b} = \mathbf{i} + \mathbf{k}$ and $\mathbf{c} = \mathbf{i} + 2\mathbf{j} - 3\mathbf{k}$ be three given vectors. If $\mathbf{r}$ is a vector such that $\mathbf{r} \times \mathbf{a} = \mathbf{c} \times \mathbf{a}$ and $\mathbf{r} \cdot \mathbf{b} = 0$, then $|\mathbf{r}|$ is equal to:
Maths · Three Dimensional Geometry · Single correct
Let the plane P pass through the intersection of the planes $2x + 3y - z = 2$ and $x + 2y + 3z = 6$, and be perpendicular to the plane $2x + y - z + 1 = 0$. If $d$ is the distance of P from the point $(-7, 1, 1)$, then $d^2$ is equal to:
Maths · Three Dimensional Geometry · Fill in the blank
Let $\alpha x + \beta y + \gamma z = 1$ be the equation of a plane passing through the point $(3, -2, 5)$ and perpendicular to the line joining the points $(1, 2, 3)$ and $(-2, 3, 5)$. Then the value of $\alpha \beta \gamma$ is equal to _____.
Answer: 6
Solution
Given Equation is not equation of plane as $yz$ is present. If we consider $y$ is $\gamma$ then answer would be $6$. Normal vector of plane $= 3\hat{i} - \hat{j} - 2\hat{k}$. Plane: $3x - y - 2z + \lambda = 0$. Point $(3, -2, 5)$ satisfies the plane. $\lambda = -1$. $3x - y - 2z = 1$. $\alpha \beta y = 6$.
Question 89
Maths · Three Dimensional Geometry · Numerical
The point of intersection $C$ of the plane $8x + y + 2z = 0$ and the line joining the points $A(-3, -6, 1)$ and $B(2, 4, -3)$ divides the line segment $AB$ internally in the ratio $k : 1$. If $a, b, c$ $(|a|, |b|, |c|$ are coprime) are the direction ratios of the perpendicular from the point $C$ on the line $$\frac{1-x}{1} = \frac{y+4}{2} = \frac{z+2}{3},$$ then $|a + b + c|$ is equal to
Answer: 10
Solution
Plane: $8x + y + 2z = 0$ Given line $AB$: $\frac{x-2}{5} = \frac{y-4}{10} = \frac{z+3}{-4} = \lambda$ Any point on line $(5\lambda + 2, 10\lambda + 4, -4\lambda - 3)$ Point of intersection of line and plane: $$8(5\lambda + 2) + 10\lambda + 4 - 8\lambda - 6 = 0$$ $$\lambda = -\frac{1}{3}$$ $C \left( \frac{1}{3}, \frac{2}{3}, -\frac{5}{3} \right)$ $L: \frac{x-1}{-1} = \frac{y+4}{2} = \frac{z+2}{3} = \mu$ $\overline{CD} = \left( -\mu + \frac{2}{3} \right) \hat{i} + \left( 2\mu - \frac{14}{3} \right) \hat{j} + \left( 3\mu - \frac{1}{3} \right) \hat{k}$ $$\left( -\mu + \frac{2}{3} \right)(-1) + \left( 2\mu - \frac{14}{3} \right)2 + \left( 3\mu - \frac{1}{3} \right)3 = 0$$ $$\mu = \frac{11}{14}$$ $$\overline{CD} = \frac{-5}{42}, \frac{-130}{42}, \frac{85}{42}$$ Direction ratios $\rightarrow (-1, -26, 17)$ $|a + b + c| = 10$
Question 90
Maths · Probability · Single correct
Two dice are thrown independently. Let $A$ be the event that the number appeared on the $1^{st}$ die is less than the number appeared on the $2^{nd}$ die, $B$ be the event that the number appeared on the $1^{st}$ die is even and that on the second die is odd, and $C$ be the event that the number appeared on the $1^{st}$ die is odd and that on the $2^{nd}$ is even. Then
the number of favourable cases of the event $(A \cup B) \cap C$ is 6
$A$ and $B$ are mutually exclusive
The number of favourable cases of the events $A$, $B$ and $C$ are 15, 6 and 6 respectively
$B$ and $C$ are independent
Answer: (a)
Solution
A: number on 1st die < number on 2nd die B: number on 1st die = even and number on 2nd die = odd C: number on 1st die = odd and number on 2nd die = even $n(A) = 5 + 4 + 3 + 2 + 1 = 15$ $n(B) = 9$ $n(C) = 9$ $n((A \cup B) \cap C) = (A \cap C) \cup (B \cap C)$ $= (3 + 2 + 1) + 0 = 6.$