JEE Main 29 January 2023 Shift 1 question paper with solutions
JEE Main 29 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
Question 1
Physics · Electromagnetic Induction · Single correct
Match List I with List II : Choose the correct answer from the options given below:
A tennis ball is dropped on to the floor from a height of 9.8 $\mathrm{m}$. It rebounds to a height 5.0 $\mathrm{m}$. Ball comes in contact with the floor for 0.2 $\mathrm{s}$. The average acceleration during contact is _____ $\mathrm{ms^{-2}}$. [Given $g = 10 \mathrm{ms^{-2}}$]
A stone is projected at angle $30^\circ$ to the horizontal. The ratio of kinetic energy of the stone at point of projection to its kinetic energy at the highest point of flight will be:
1 : 2
1 : 4
4 : 1
4 : 3
Answer: (d)
Solution
The ratio of kinetic energies is given by: $$\frac{\mathrm{KE_{POP}}}{\mathrm{KE_{top}}} = \frac{\frac{1}{2} M (u)^2}{\frac{1}{2} M (u \cos 30^\circ)^2} = \frac{4}{3}$$
Question 4
Physics · Laws of Motion · Single correct
A car is moving on a horizontal curved road with radius 50 m. The approximate maximum speed of car will be, if friction between tyres and road is 0.34. [Take g = 10 ms^{-2}]
3.4 ms^{-1}
22.4 ms^{-1}
13 ms^{-1}
17 ms^{-1}
Answer: (c)
Solution
Given $f_s = \frac{mv^2}{r}$. For maximum speed in safe turning, $f_s = f_s max = \mu mg$. $v_{max}$ (for safe turning) $= \sqrt{\mu rg} = \sqrt{0.34 \times 50 \times 10} \approx 13 \, \mathrm{m/s}$.
Question 5
Physics · Laws of Motion · Single correct
A block of mass m slides down the plane inclined at angle 30$^\circ$ with an acceleration $\frac{g}{4}$. The value of coefficient of kinetic friction will be:
$\frac{2\sqrt{3} + 1}{2}$
$\frac{1}{2\sqrt{3}}$
$\frac{\sqrt{3}}{2}$
$\frac{2\sqrt{3} - 1}{2}$
Answer: (b)
Solution
Given $Mg \sin 30^\circ - \mu mg \cos 30^\circ = ma$. $$\frac{g}{2} - \frac{\sqrt{3}}{2} \cdot \mu g = \frac{g}{4}$$ Solving for $\mu$, we have: $$\frac{\sqrt{3}}{2} \mu = \frac{1}{4}$$ Therefore, $$\mu = \frac{1}{2\sqrt{3}}$$
Question 6
Physics · Work, Energy and Power · Numerical
A $0.4 \, \mathrm{kg}$ mass takes $8 \, \mathrm{s}$ to reach ground when dropped from a certain height 'P' above surface of earth. The loss of potential energy in the last second of fall is _______ $\mathrm{J}$. [Take $g = 10 \, \mathrm{m/s^2}$]
Answer: 300
Solution
Displacement is $8^{\mathrm{th}}$ sec. $$S_8 = 0 + \frac{1}{2} \times 10 \times (2 \times 8 - 1)$$ $$S_8 = 5 \times 15$$ $$\Delta U = 0.4 \times 10 \times 5 \times 15$$ $$\Delta U = 20 \times 15 = 300$$
Question 7
Physics · System of Particles and Rotational Motion · Numerical
A solid sphere of mass 2 kg is making pure rolling on a horizontal surface with kinetic energy 2240 \, $\mathrm{J}$. The velocity of centre of mass of the sphere will be _____ \, $\mathrm{ms^{-1}}$.
Answer: 40
Solution
The kinetic energy is given by the equation: $$KE = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2$$ Substituting the given values: $$2240 = \frac{1}{2} \cdot 2 (v)^2 + \frac{1}{2} \cdot \frac{2}{5} R^2 \cdot \left( \frac{v}{R} \right)^2$$ Simplifying the equation: $$2240 = v^2 + \frac{2}{5} v^2$$ Solving for $v$: $$\Rightarrow v = 40 \, \mathrm{m/s}$$
Question 8
Physics · Gravitation · Single correct
Two particles of equal mass 'm' move in a circle of radius 'r' under the action of their mutual gravitational attraction. The speed of each particle will be:
$\sqrt{\frac{GM}{2r}}$
$\sqrt{\frac{4GM}{r}}$
$\sqrt{\frac{GM}{r}}$
$\sqrt{\frac{GM}{4r}}$
Answer: (d)
Solution
Given the equation $$\frac{Gm^2}{4r^2} = \frac{mv^2}{r}$$ for the system shown in the diagram.
Question 9
Physics · Mechanical Properties of Fluids · Single correct
Surface tension of a soap bubble is $2.0 \times 10^{-2} \, \mathrm{Nm}^{-1}$. Work done to increase the radius of soap bubble from $3.5 \, \mathrm{cm}$ to $7 \, \mathrm{cm}$ will be: [Take $\pi = \frac{22}{7}$]
A body cools from $60^\circ \mathrm{C}$ to $40^\circ \mathrm{C}$ in 6 minutes. If, temperature of surroundings is $10^\circ \mathrm{C}$. Then, after the next 6 minutes, its temperature will be _____ $^\circ \mathrm{C}$.
Answer: 28
Solution
By average form of Newton's law of cooling $$\frac{20}{6} = k(50 - 10) \ldots (i)$$ $$\frac{40 - T}{6} = K \left( \frac{40 + T}{2} - 10 \right) \ldots (ii)$$ From equation (i) and (ii) $$\frac{20}{40 - T} = \frac{40}{10 + T/2}$$ $$10 + \frac{T}{2} = 80 - 2T$$
Question 11
Physics · Thermodynamics · Single correct
Given below are two statements. One is labelled as Assertion A and the other is labelled as Reason R. Assertion A : If dQ and dW represent the heat supplied to the system and the work done on the system respectively. Then according to the first law of thermodynamics $dQ = dU - dW$. Reason R : First law of thermodynamics is based on law of conservation of energy. In the light of the above statements, choose the correct answer from the option given below :
A is correct but R is not correct
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
Answer: (c)
Solution
First law of thermodynamics is based on law of conservation of energy and it can be written as $$\mathrm{dQ} = \mathrm{dU} - \mathrm{dW}.$$ where dW is work done on the system.
Question 12
Physics · Thermodynamics · Single correct
A bicycle tyre is filled with air having pressure of $270 \, \mathrm{kPa}$ at $27^\circ \mathrm{C}$. The approximate pressure of the air in the tyre when the temperature increases to $36^\circ \mathrm{C}$ is
Two simple harmonic waves having equal amplitudes of 8 cm and equal frequency of 10 Hz are moving along the same direction. The resultant amplitude is also 8 cm. The phase difference between the individual waves is ____ degree.
Answer: 120
Solution
Given the equation $2A \cos\left(\frac{\Delta \phi}{2}\right) = A$. We simplify to find $\cos\left(\frac{\Delta \phi}{2}\right) = \frac{1}{2}$. This implies $\frac{\Delta \phi}{2} = 60^\circ$.
Question 14
Physics · Waves · Single correct
A person observes two moving trains, 'A' reaching the station and 'B' leaving the station with equal speed of 30 $\mathrm{m/s}$. If both trains emit sounds with frequency 300 $\mathrm{Hz}$, (Speed of sound : 330 $\mathrm{m/s}$) approximate difference of frequencies heard by the person will be :
33 Hz
55 Hz
80 Hz
10 Hz
Answer: (b)
Solution
Given $$f_1 = 300 \left( \frac{330 - 0}{330 - (-30)} \right) = 275$$ $$f_2 = 300 \left( \frac{330 - 0}{330 - (30)} \right) = 330$$ The change in frequency is $$\Delta f = 330 - 275 = 55 \, Hz.$$
Question 15
Physics · Electric Charges and Fields · Single correct
In a cuboid of dimension $2L \times 2L \times L$, a charge $q$ is placed at the centre of the surface ‘S’ having area of $4 \, L^2$. The flux through the opposite surface to ‘S’ is given by
$\frac{q}{12 \varepsilon_0}$
$\frac{q}{3 \varepsilon_0}$
$\frac{q}{2 \varepsilon_0}$
$\frac{q}{6 \varepsilon_0}$
Answer: (d)
Solution
The flux $\phi$ is given by $$\phi = \frac{Q / \varepsilon_0}{6}.$$ The flux passing through the shaded face is $$\frac{q}{6 \varepsilon_0}.$$
Question 16
Physics · Electric Charges and Fields · Numerical
A point charge $q_1 = 4q_0$ is placed at origin. Another point charge $q_2 = -q_0$ is placed at $x = 12 \, \mathrm{cm}$. Charge of proton is $q_0$. The proton is placed on x-axis so that the electrostatic force on the proton in zero. In this situation, the position of the proton from the origin is _____ cm.
Answer: 24
Solution
Given the equation: $$\frac{q_0}{x^2} = \frac{4q_0}{(x+12)^2}$$ Solving for $x$, we have: $$x + 12 = 2x$$ Therefore, $$x = 12$$
Question 17
Physics · Electric Charges and Fields · Single correct
Ratio of thermal energy released in two resistor R and 3R connected in parallel in an electric circuit is :
3 : 1
1 : 1
1 : 3
1 : 27
Answer: (a)
Solution
Given the equation for H as $H = \frac{V^2}{R} \times t$. We have: $$\frac{H_1}{H_2} = \frac{\frac{V^2 t}{R}}{\frac{V^2 t}{3R}} = 3:1$$
Question 18
Physics · Moving Charges and Magnetism · Single correct
A single current carrying loop of wire carrying current I flowing in anticlockwise direction seen from +ve z direction and lying in xy plane in shown in figure. The plot of $\hat{j}$ component of magnetic field ($B_y$) at a distance ‘a’ (less than radius of the coil) and on yz plane vs z coordinate look like
Answer: (c)
Solution
In the plane of the coil, $B_y = 0$. $B_y$ is opposite of each other in $-z$ and $+z$ positions.
Question 19
Physics · Moving Charges and Magnetism · Single correct
The magnitude of magnetic induction at mid-point O due to current arrangement as shown in Fig will be:
$\frac{\mu_0 I}{2 \pi a}$
0
$\frac{\mu_0 I}{4 \pi a}$
$\frac{\mu_0 I}{\pi a}$
Answer: (d)
Solution
Magnetic field due to current in BC and ET are outward at point 'O'. $$B_0 = \frac{\mu_0 i}{4 \pi r} + \frac{\mu_0 i}{4 \pi r} = \frac{\mu_0 i}{2 \pi r} = \frac{\mu_0 i}{\pi a}$$
Question 20
Physics · Alternating Current · Single correct
Find the mutual inductance in the arrangement, when a small circular loop of wire of radius 'R' is placed inside a large square loop of wire of side L (L >> R). The loops are coplanar and their centres coincide:
A certain elastic conducting material is stretched into a circular loop. It is placed with its plane perpendicular to a uniform magnetic field $B = 0.8 \, \mathrm{T}$. When released the radius of the loop starts shrinking at a constant rate of $2 \, \mathrm{cm}^{-1}$. The induced emf in the loop at an instant when the radius of the loop is $10 \, \mathrm{cm}$ will be ____ mV.
Answer: 10
Solution
EMF = $\frac{d}{dt}$ $\left$( B $\pi$ r^2 $\right$) = 2 B $\pi$ r $\frac{dr}{dt}$ = 2 $\times$ $\pi$ $\times$ 0.1 $\times$ 0.8 $\times$ 2 $\times$ 10^{-2} = 2 $\pi$ $\times$ 1.6 = 10.06 [round off 10.06 = 10]
Question 22
Physics · Electromagnetic Waves · Single correct
Which of the following are true? A. Speed of light in vacuum is dependent on the direction of propagation. B. Speed of light in a medium in independent of the wavelength of light. C. The speed of light is independent of the motion of the source. D. The speed of light in a medium is independent of intensity. Choose the correct answer from the option given below :
A and C only
B and D only
B and C only
C and D only
Answer: (d)
Solution
Speed of light does not depend on the motion of source as well as intensity.
Question 23
Physics · Wave Optics · Single correct
In a Young’s double slit experiment, two slits are illuminated with a light of wavelength 800 nm. The line joining $A_1P$ is perpendicular to $A_1A_2$ as shown in the figure. If the first minimum is detected at $P$, the value of slits separation ‘a’ will be:
0.4 mm
0.5 mm
0.2 mm
0.1 mm
Answer: (c)
Solution
Given the condition of minima: $$A_2 P - A_1 P = \frac{\lambda}{2}$$ We have: $$\sqrt{D^2 + a^2} - D = \frac{\lambda}{2}$$ Expanding the square root: $$D \left( 1 + \frac{a^2}{D^2} \right)^{1/2} - D = \frac{\lambda}{2}$$ Approximating for small $a$: $$D \left( 1 + \frac{1}{2} \times \frac{a^2}{D^2} \right) - D = \frac{\lambda}{2}$$ Simplifying gives: $$\frac{a^2}{2D} = \frac{\lambda}{2} \Rightarrow a = \sqrt{\lambda D}$$ Substituting values: $$= \sqrt{800 \times 10^{-6} \times 50}$$ Therefore, $a = 0.2 \, \mathrm{mm}$
Question 24
Physics · Wave Optics · Numerical
As shown in figures, three identical polaroids $P_1$, $P_2$ and $P_3$ are placed one after another. The pass axis of $P_2$ and $P_3$ are inclined at angle of $60^\circ$ and $90^\circ$ with respect to axis of $P_1$. The source $S$ has an intensity of $256 \, \mathrm{W/m^2}$. The intensity of light at point $O$ is $\mathrm{W/m^2}$.
Answer: 24
Solution
By first polaroid P1 intensity will be halved then P2 and P3 will make intensity $\cos^2(60^\circ)$ and $\cos^2(30^\circ)$ times respectively. Intensity out = $$\frac{256}{2} \times \frac{1}{4} \times \left(\frac{\sqrt{3}}{2}\right)^2$$ $$= \frac{256 \times 3}{2 \times 4 \times 4} = 24$$
Question 25
Physics · Dual Nature of Radiation and Matter · Single correct
The threshold wavelength for photoelectric emission from a material is 5500 Å. Photoelectrons will be emitted, when this material is illuminated with monochromatic radiation from a \begin{enumerate} \item[(A)] $75\,\text{W}$ infra-red lamp \item[(B)] $10\,\text{W}$ infra-red lamp \item[(C)] $75\,\text{W}$ ultra-violet lamp \item[(D)] $10\,\text{W}$ ultra-violet lamp \end{enumerate} \textbf{Choose the correct answer from the options given below:}
If a radioactive element having half-life of 30 min is undergoing beta decay, the fraction of radioactive element remains undecayed after 90 min. will be:
Physics · Physical World, Units and Measurements · Numerical
A radioactive element ${}^{242}_{92}\mathrm{X}$ emits two $\alpha$-particles, one electron and two positrons. The product nucleus is represented by ${}^{234}_{P}\mathrm{Y}$. The value of $P$ is
Answer: 87
Solution
P = 92 - 2 - 2 + 1 - 1 - 1 Simplifying, we have: \[ P = 92 - 5 \] Thus, \[ P = 87 \]
Question 28
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
Which of the following statement is not correct in the case of light emitting diodes? \begin{enumerate} \item[(A)] It is a heavily doped p-n junction. \item[(B)] It emits light only when it is forward biased. \item[(C)] It emits light only when it is reverse biased. \item[(D)] The energy of the light emitted is equal to or slightly less than the energy gap of the semiconductor used. \end{enumerate} \textbf{Choose the correct answer from the options given below:}
C and D
A
C
B
Answer: (c)
Solution
LED works in forward biasing and light energy maybe slightly less or equal to band gap.
Question 29
Physics · Communication Systems · Single correct
If the height of transmitting and receiving antennas are 80 m each, the maximum line of sight distance will be: Given: Earth's radius = $6.4 \times 10^6 \, \mathrm{m}$.
32 $\mathrm{km}$
28 $\mathrm{km}$
36 $\mathrm{km}$
64 $\mathrm{km}$
Answer: (d)
Solution
Maximum line of sight distance between two antennas, $d_M = \sqrt{2R h_T} + \sqrt{2R h_R}$.
Question 30
Physics · Experimental Physics · Numerical
In a metre bridge experiment the balance point in obtained if the gaps are closed by $2\Omega$ and $3\Omega$. A shunt of $X\Omega$ is added to $3\Omega$ resistor to shift the balancing point by $22.5 \, \mathrm{cm}$. The value of $X$ is
Answer: 2
Solution
The equation is given by: $$\frac{2}{\left(\frac{3x}{3+x}\right)} = \frac{40 + 22.5}{60 - 22.5} = \frac{62.5}{37.5} = \frac{5}{3}$$ Simplifying, we have: $$\frac{6}{5} = \frac{3x}{3+x}$$ Solving for $x$: $$6 + 2x = 5x \implies x = 2$$
Chemistry
Question 31
Chemistry · Structure of Atom · Single correct
The shortest wavelength of hydrogen atom in Lyman series is $\lambda$. The longest wavelength in Balmer series of $\mathrm{He}^{+}$ is
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
The bond dissociation energy is highest for
$\mathrm{Cl}$_2
$\mathrm{I}$_2
$\mathrm{Br}$_2
$\mathrm{F}$_2
Answer: (a)
Solution
Bond energy of $\mathrm{F_2}$ less than $\mathrm{Cl_2}$ due to lone pair lone pair repulsions. Bond energy order $\mathrm{Cl_2} > \mathrm{Br_2} > \mathrm{F_2} > \mathrm{I_2}$
Question 33
Chemistry · Chemical Bonding and Molecular Structure · Numerical
The number of molecules or ions from the following, which do not have odd number of electrons are ________ .
$\mathrm{NO_2}$
$\mathrm{ICl_4^-}$
$\mathrm{BrF_3}$
$\mathrm{ClO_2}$
Answer: (c)
Solution
Q9 (3) $\mathrm{ICl_4^-}$, $\mathrm{BrF_3}$ and $\mathrm{NO_2^+}$ do not have odd number of e
Question 34
Chemistry · States of Matter · Single correct
For 1 mol of gas, the plot of pV vs p is shown below. p is the pressure and V is the volume of the gas. What is the value of compressibility factor at point A?
1 - $\frac{a}{RTV}$
1 + $\frac{b}{V}$
1 - $\frac{b}{V}$
1 + $\frac{a}{RTV}$
Answer: (a)
Solution
For 1 mole of real gas, $PV = ZRT$. From the graph, $PV$ for real gas is less than $PV$ for ideal gas at point A. $Z < 1$. $$Z = 1 - \frac{a}{V_m RT}$$
Question 35
Chemistry · Equilibrium · Numerical
Consider the following reaction approaching equilibrium at 27°C and 1 atm pressure $$A + B \overset{K_f = 10^3}{\underset{K_r = 10^2}{\rightleftharpoons}} C + D$$ The standard Gibb's energy change ($\Delta_r G^\circ$) at 27°C is (−) _______ kJ mol⁻¹ (Nearest integer). \text{(Given: } R = 8.3\,\mathrm{J\,K^{-1}\,mol^{-1}} \text{ and } \ln 10 = 2.3\text{)}
Water decomposes at 2300 K $$\mathrm{H_2O(g) \rightarrow H_2(g) + \frac{1}{2}O_2(g)}$$ The percent of water decomposing at 2300 K and 1 bar is ________ (Nearest integer). Equilibrium constant for the reaction is $2 \times 10^{-3}$ at 2300 K
Answer: 2
Solution
The reaction is given by $\mathrm{H_2O(g)} \rightleftharpoons \mathrm{H_2(g)} + \frac{1}{2} \mathrm{O_2(g)}$. The partial pressures at equilibrium are $P_0[1-\alpha]$, $P_0\alpha$, and $\frac{P_0\alpha}{2}$. The equation for equilibrium is $$P_0 \left[ 1 + \frac{\alpha}{2} \right] = 1 ...(i)$$ The equilibrium constant $K_p$ is given by $$K_p = \frac{(P_{\mathrm{H_2}})(P_{\mathrm{O_2}})^{1/2}}{P_{\mathrm{H_2O}}}$$ Substituting the partial pressures, we have $$\frac{(P_0\alpha) \left( \frac{P_0\alpha}{2} \right)^{1/2}}{P_0[1-\alpha]} = 2 \times 10^{-3}$$ Since $\alpha$ is negligible with respect to 1, we assume $P_0 = 1$ and $1 - \alpha \approx 1$. Therefore, $$\frac{\alpha \sqrt{\alpha}}{\sqrt{2}} = 2 \times 10^{-3}$$ Solving for $\alpha$, we get $$\alpha^{3/2} = 2^{3/2} \times 10^{-3}$$ $$\alpha = 2^{3/2 \times 2/3} \times 10^{-3 \times 2/3}$$ $$\alpha = 2 \times 10^{-2}$$ Thus, the percentage $\alpha = 2\%$.
Question 37
Chemistry · Equilibrium · Numerical
Millimoles of calcium hydroxide required to produce 100 mL of the aqueous solution of pH 12 is $x \times 10^{-1}$. The value of $x$ is ______ (Nearest integer). Assume complete dissociation.
Answer: 5
Solution
Given $pH = 12$. Therefore, $[H^+] = 10^{-12} \, M$. Thus, $[OH^-] = 10^{-2} \, M$. Therefore, $[Ca(OH)_2] = 5 \times 10^{-3} \, M$. $$5 \times 10^{-3} = \frac{milli moles of Ca(OH)_2}{100 \, mL}$$ Milli moles of $Ca(OH)_2 = 5 \times 10^{-1}$. Ans. = 5
Question 38
Chemistry · Hydrogen · Single correct
Which of the given compounds can enhance the efficiency of hydrogen storage tank?
Li/P_4
SiH_4
NaNi_5
Di-isobutylaluminium hydride
Answer: (c)
Solution
Refer NCERT
Question 39
Chemistry · The s-Block Elements · Single correct
The magnetic behaviour of $\mathrm{Li_2O}$, $\mathrm{Na_2O_2}$ and $\mathrm{KO_2}$, respectively, are
The correct order of hydration enthalpies is \begin{enumerate} \item[(A)] $\mathrm{K^+}$ \item[(B)] $\mathrm{Rb^+}$ \item[(C)] $\mathrm{Mg^{2+}}$ \item[(D)] $\mathrm{Cs^+}$ \item[(E)] $\mathrm{Ca^{2+}}$ \end{enumerate}
Chemistry · The d-and f-Block Elements · Single correct
During the borax bead test with $\mathrm{CuSO_4}$, a blue green colour of the bead was observed in oxidising flame due to the formation of
$\mathrm{Cu_3B_2}$
$\mathrm{Cu}$
$\mathrm{Cu(BO_2)_2}$
$\mathrm{CuO}$
Answer: (c)
Solution
Blue green colour is due to formation of $\mathrm{Cu(BO_2)_2}$. $$\mathrm{CuSO_4} \xrightarrow{\Delta} \mathrm{CuO} + \mathrm{SO_3}$$ $$\mathrm{CuO} + \mathrm{B_2O_3} \rightarrow \mathrm{Cu(BO_2)_2}$$
Question 42
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Compound that will give positive Lassaigne’s test for both nitrogen and halogen is
$\mathrm{N_2H_4.HCl}$
$\mathrm{CH_3NH_2.HCl}$
$\mathrm{NH_4Cl}$
$\mathrm{NH_2OH.HCl}$
Answer: (b)
Solution
Given $\mathrm{CH_3NH_2 \cdot HCl} \xrightarrow{Na fusion} \mathrm{NaCN}$ and $\mathrm{NaCl}$. NaCN gives $+\mathrm{ve}$ test for nitrogen and NaCl gives $+\mathrm{ve}$ test for halogen.
Following chromatogram was developed by adsorption of compound 'A' on a 6 cm TLC glass plate. Retardation factor of the compound 'A' is _____ $\times 10^{-1}$.
Answer: 6
Solution
The formula for the retention factor $R_f$ is given by: $$R_f = \frac{Distance moved by the substance from base line}{Distance moved by the solvent from base line}$$ Substituting the given values: $$R_f = \frac{3.0 \, cm}{5.0 \, cm} = 0.6 or 6 \times 10^{-1}$$
Question 44
Chemistry · Hydrocarbons · Numerical
17 $\mathrm{mg}$ of a hydrocarbon (M.F. $C_{10}H_{16}$) takes up 8.40 $\mathrm{mL}$ of the $\mathrm{H_2}$ gas measured at $0^\circ$ $\mathrm{C}$ and 760 $\mathrm{mm}$ of Hg. Ozonolysis of the same hydrocarbon yields The number of double bond/s present in the hydrocarbon is ________.
Answer: 3
Solution
Moles of hydrocarbon $= \frac{17 \times 10^{-3}}{136} = 1.25 \times 10^{-4}$ Mole of $\mathrm{H_2}$ gas $$\Rightarrow 1 \times \frac{8.40}{1000} = n \times 0.0821 \times 273$$ $$\Rightarrow n = 3.75 \times 10^{-4}$$ Hydrogen molecule used for 1 molecule of hydrocarbon is 3 $$= \frac{3.75 \times 10^{-4}}{1.25 \times 10^{-4}} = 3$$
Question 45
Chemistry · Environmental Chemistry · Single correct
Correct statement about smog is
$\mathrm{NO}_2$ is present in classical smog
Both $\mathrm{NO}_2$ and $\mathrm{SO}_2$ are present in classical smog
Photochemical smog has high concentration of oxidizing agents
Classical smog also has high concentration of oxidizing agents
Answer: (c)
Solution
Photochemical smog has high concentration of oxidising agents. $\mathrm{NO_2}$ is produced from $\mathrm{NO}$ and $\mathrm{O_3}$ in the presence of sunlight. Classical smog contains smoke, fog and $\mathrm{SO_2}$ and it is known as reducing smog, as chemically it is a reducing mixture.
Question 46
Chemistry · Solutions · Numerical
Solid Lead nitrate is dissolved in 1 litre of water. The solution was found to boil at $100.15^\circ \mathrm{C}$. When $0.2$ mol of NaCl is added to the resulting solution, it was observed that the solution froze at $-0.8^\circ \mathrm{C}$. The solubility product of $\mathrm{PbCl}_2$ formed is ______ $\times 10^{-6}$ at $298 \, \mathrm{K}$. (Nearest integer) Given : $K_b = 0.5 \, \mathrm{K \, kg \, mol^{-1}}$ and $K_f = 1.8 \, \mathrm{kg \, mol^{-1}}$. Assume molality to be equal to molarity in all cases.
Answer: 13
Solution
Let a mole $\mathrm{Pb(NO_3)_2}$ be added. $$\mathrm{Pb(NO_3)_2 \rightarrow Pb^{2+} + 2NO_3^-}$$ $a a 2a$ $$\Delta T_b = 0.15 = 0.5 [3a] \Rightarrow a = 0.1$$ $$\mathrm{Pb^{2+}_{(aq)} + 2Cl^-_{(aq)} \rightarrow PbCl_2(s)}$$ $t = 0 0.1 0.2$ $t = \infty (0.1 - x) (0.2 - 2x)$ In final solution $$\Delta T_f = 0.8 = 1.8 \left[ \frac{0.3 - 3x + 0.2 + 0.2}{1} \right]$$ $$\Rightarrow x = \frac{2.3}{27}$$ $$\Rightarrow K_{sp} = \left(0.1 - \frac{2.3}{27}\right) \left(0.2 - \frac{4.6}{27}\right)^2 = 13 \times 10^{-6}$$
Question 47
Chemistry · Electrochemistry · Single correct
The standard electrode potential $(M^{3+}/M^{2+})$ for V, Cr, Mn $\&$ Co are $-0.26 \, \mathrm{V}$, $-0.41 \, \mathrm{V}$, $+1.57 \, \mathrm{V}$ and $+1.97 \, \mathrm{V}$, respectively. The metal ions which can liberate $\mathrm{H}_2$ from a dilute acid are
$\mathrm{V}^{2+}$ and $\mathrm{Mn}^{2+}$
$\mathrm{Cr}^{2+}$ and $\mathrm{Co}^{2+}$
$\mathrm{V}^{2+}$ and $\mathrm{Cr}^{2+}$
$\mathrm{Mn}^{2+}$ and $\mathrm{Co}^{2+}$
Answer: (c)
Solution
Metal cation with (−) value of reduction potential $(\mathrm{M^{+3}/M^{+2}})$ or with (+) value of oxidation potential $(\mathrm{M^{+2}/M^{+3}})$ will liberate $\mathrm{H_2}$. Therefore they will reduce $\mathrm{H^+}$. i.e. $\mathrm{V^{+2}}$ and $\mathrm{Cr^{+2}}$.
Question 48
Chemistry · Electrochemistry · Numerical
Following figure shows dependence of molar conductance of two electrolytes on concentration. $\Lambda_m^0$ is the limiting molar conductivity. The number of \underline{\textbf{Incorrect}} statement(s) from the following is ____ \begin{enumerate} \item[(A)] $\Lambda_m^{0}$ for electrolyte A is obtained by extrapolation. \item[(B)] For electrolyte B, $\nu_x \Lambda_m$ vs. $\sqrt{c}$ graph is a straight line with intercept equal to $\Lambda_m^{0}$. \item[(C)] At infinite dilution, the value of degree of dissociation approaches zero for electrolyte B. \item[(D)] $\Lambda_m^{0}$ for any electrolyte A or B can be calculated using $\lambda^{0}$ for individual ions. \end{enumerate}
$\Lambda_m^0$ for electrolyte A is obtained by extrapolation
For electrolyte B, vx $\Lambda_m$ vs $\sqrt{c}$ graph is a straight line with intercept equal to $\Lambda_m^0$
At infinite dilution, the value of degree of dissociation approach zero for electrolyte B.
$\Lambda_m^0$ for any electrolyte A or B can be calculated using $\lambda^0$ for individual ions.
Answer: (b)
Solution
Statement (A) and Statement (C) are incorrect
Question 49
Chemistry · Chemical Kinetics and Nuclear Chemistry · Single correct
For certain chemical reaction $X \rightarrow Y$, the rate of formation of product is plotted against the time as shown in the figure. The number of Correct statement/s from the following is _______
Over all order of this reaction is one
Order of this reaction can’t be determined
In region-I and III, the reaction is of first and zero order respectively
In region-II, the reaction is of first order
Answer: (b)
Solution
Only option (B) is correct as order cannot be determined.
Question 50
Chemistry · Surface Chemistry · Single correct
Which of the following salt solutions would coagulate the colloid solution formed when $\mathrm{FeCl_3}$ is added to NaOH solution, at the fastest rate?
$10 \, \mathrm{mL}$ of $0.2 \, \mathrm{mol \, dm^{-3}}$ $\mathrm{AlCl_3}$
$10 \, \mathrm{mL}$ of $0.15 \, \mathrm{mol \, dm^{-3}}$ $\mathrm{Na_2SO_4}$
$10 \, \mathrm{mL}$ of $0.1 \, \mathrm{mol \, dm^{-3}}$ $\mathrm{Ca_3(PO_4)_2}$
$10 \, \mathrm{mL}$ of $0.5 \, \mathrm{mol \, dm^{-3}}$ $\mathrm{Na_2SO_4}$
Answer: (a)
Solution
Formed is negatively charged solution, therefore $\mathrm{Al^{3+}}$ has highest coagulating power.
Question 51
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
The reaction representing the Mond process for metal refining is
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
“A” obtained by Ostwald’s method involving air oxidation of $\mathrm{NH_3}$, upon further air oxidation produces “B”. “B” on hydration forms an oxoacid of Nitrogen along with evolution of “A”. The oxoacid also produces “A” and gives positive brown ring test
NO_2, N_2O_5
NO_2, N_2O_4
NO, NO_2
N_2O_3, NO_2
Answer: (c)
Solution
The reaction given is: $$4\mathrm{NH_3} + 5\mathrm{O_2} \xrightarrow{\Delta} 4\mathrm{NO} + 6\mathrm{H_2O}$$ (A) $$2\mathrm{NO} + \mathrm{O_2} \rightarrow 2\mathrm{NO_2}$$ (B)
Question 53
Chemistry · Co-ordination Compounds · Single correct
Chiral complex from the following is Here en = ethylene diamine
cis – $[\mathrm{PtCl}_2(\mathrm{en})_2]^{2+}$
trans – $[\mathrm{PtCl}_2(\mathrm{en})_2]^{2+}$
cis – $[\mathrm{PtCl}_2(\mathrm{NH}_3)_2]$
trans – $[\mathrm{Co}(\mathrm{NH}_3)_4 \mathrm{Cl}_2]^+$
Answer: (a)
Solution
This is a chiral complex form.
Question 54
Chemistry · Co-ordination Compounds · Fill in the blank
The sum of bridging carbonyl ligands in $\mathrm{W(CO)_6}$ and $\mathrm{Mn_2(CO)_{10}}$ is:
Answer: 0
Solution
The structure shown is a metal carbonyl complex with a central tungsten (W) atom surrounded by carbon monoxide (CO) ligands. The formula for the complex is $[\mathrm{(CO)_5Mn-Mn(CO)_5}]$.
Question 55
Chemistry · Haloalkanes and Haloarenes · Single correct
Identify the correct order for the given property for each of the following compounds Choose the correct answer from the option given below :-
(B) , $(C)$ and (D) only
(A) , $(C)$ and (E) only
(A) , $(C)$ and (D) only
(A) , (B) and (E) only
Solution
Boiling point of alkyl halide increases with increase in size, mass of halogen atom and size of alkyl group. Boiling point of isomeric alkyl halide decreases with increase in branching. Density increases with increase in atomic mass of halogen atom.
Question 56
Chemistry · Alcohols, Phenols and Ethers · Single correct
\textbf{Q56.} The increasing order of $pK_a$ for the following phenols is \begin{enumerate} \item[(A)] 2,4-Dinitrophenol \item[(B)] 4-Nitrophenol \item[(C)] 2,4,5-Trimethylphenol \item[(D)] Phenol \item[(E)] 3-Chlorophenol \end{enumerate} Choose the correct answer from the options given below:
(A), (E), (B), (D), (C)
(A), (B), (E), (D), (C)
(C), (D), (E), (B), (A)
(C), (E), (D), (B), (A)
Answer: (b)
Solution
Order of acidity for following phenol is $-M$ and $-I$ increases acidity
Reactions and their corresponding reagents are listed as follows: (A) Hoffmann degradation uses $\mathrm{Br_2/NaOH}$. (B) Clemenson reduction uses $\mathrm{Zn-Hg/HCl}$. (C) Cannizaro reaction uses $\mathrm{conc.\ KOH/\Delta}$. (D) Reimer-Tiemann reaction uses $\mathrm{CHCl_3,\ NaOH/H_3O^+}$.
Question 58
Chemistry · Hydrocarbons · Single correct
The major product 'P' for the following sequence of reactions is:
Answer: (c)
Solution
The given reaction involves two steps. First, the Clemmensen reduction is performed using Zn/Hg and HCl, which reduces the carbonyl groups to methylene groups. The intermediate product is then subjected to reduction with LiAlH4 followed by hydrolysis with $\mathrm{H_3O^+}$ to yield the final product.
Question 59
Chemistry · Chemistry in Everyday Life · Single correct
Number of cyclic tripeptides formed with 2 amino acids A and B is:
2
3
5
4
Answer: (d)
Solution
Two amino acids are Tripeptides are formed from three amino acids
Maths
Question 61
Maths · Complex Numbers and Quadratic Equations · Single correct
Let $\lambda \neq 0$ be a real number. Let $\alpha, \beta$ be the roots of the equation $14x^2 - 31x + 3\lambda = 0$ and $\alpha, \gamma$ be the roots of the equation $35x^2 - 53x + 4\lambda = 0$. Then $\frac{3\alpha}{\beta}$ and $\frac{4\alpha}{\gamma}$ are the roots of the equation:
$7x^2 + 245x - 250 = 0$
$7x^2 - 245x + 250 = 0$
$49x^2 - 245x + 250 = 0$
$49x^2 + 245x + 250 = 0$
Answer: (c)
Solution
Given the equation $14x^2 - 31x + 3\lambda = 0$, we have $\alpha + \beta = \frac{31}{14}$ (1) and $\alpha \beta = \frac{3\lambda}{14}$ (2). For the equation $35x^2 - 53x + 4\lambda = 0$, we have $\alpha + \gamma = \frac{53}{35}$ (3) and $\alpha \gamma = \frac{4\lambda}{35}$ (4). From (2) and (4), we get: $$\frac{\beta}{\gamma} = \frac{3 \times 35}{4 \times 14} = \frac{15}{8} \implies \beta = \frac{15}{8} \gamma$$ From (1) and (3), we have: $$\beta - \gamma = \frac{31}{14} - \frac{53}{35} = \frac{155 - 106}{70} = \frac{7}{10}$$ Thus, $\frac{15}{8} \gamma - \gamma = \frac{7}{10}$, which implies $\gamma = \frac{4}{5}$. Therefore, $\beta = \frac{15}{8} \times \frac{4}{5} = \frac{3}{2}$. Now, $\alpha = \frac{31}{14} - \beta = \frac{31}{14} - \frac{3}{2} = \frac{5}{7}$. Then, $\lambda = \frac{14}{3} \alpha \beta = \frac{14}{3} \times \frac{5}{7} \times \frac{3}{2} = 5$. So, the sum of roots $\frac{3\alpha}{\beta} + \frac{4\alpha}{\gamma} = \frac{(3\alpha \gamma + 4\alpha \beta)}{\beta \gamma}$. This equals: $$\frac{\left(\frac{3\times \frac{4\lambda}{35} + 4 \times \frac{3\lambda}{14}}{\beta \gamma}\right)} = \frac{12\lambda (14 + 35)}{14 \times 35 \beta \gamma}$$
Question 62
Maths · Complex Numbers and Quadratic Equations · Multiple correct
For two non-zero complex numbers $z_1$ and $z_2$, if $\operatorname{Re}(z_1z_2)=0$ and $\operatorname{Re}(z_1+z_2)=0$, then which of the following are possible? \begin{enumerate} \item[(A)] $\operatorname{Im}(z_1)>0$ and $\operatorname{Im}(z_2)>0$ \item[(B)] $\operatorname{Im}(z_1) 0$ \item[(C)] $\operatorname{Im}(z_1)>0$ and $\operatorname{Im}(z_2)<0$ \item[(D)] $\operatorname{Im}(z_1)<0$ and $\operatorname{Im}(z_2)<0$ \end{enumerate} Choose the correct answer from the options given below:
B and D
B and C
A and B
A and C
Answer: (b)
Solution
Given $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$. The real part of the product is given by $$\mathrm{Re}(z_1 z_2) = x_1 x_2 - y_1 y_2 = 0.$$ The real part of the sum is given by $$\mathrm{Re}(z_1 + z_2) = x_1 + x_2 = 0.$$ Therefore, $x_1$ and $x_2$ are of opposite sign, and $y_1$ and $y_2$ are of opposite sign.
Question 63
Maths · Permutations and Combinations · Numerical
If all the six digit numbers $x_1, x_2, x_3, x_4, x_5, x_6$ with $0 < x_1 < x_2 < x_3 < x_4 < x_5 < x_6$ are arranged in the increasing order, then the sum of the digits in the $72^{th}$ number is .
Five digit numbers are formed using the digits 1, 2, 3, 5, 7 with repetitions and are written in descending order with serial numbers. For example, the number 77777 has serial number 1. Then the serial number of 35337 is.
Answer: 1436
Solution
No of 5 digit numbers starting with digit 1 $$= 5 \times 5 \times 5 \times 5 = 625$$ No of 5 digit numbers starting with digit 2 $$= 5 \times 5 \times 5 \times 5 = 625$$ No of 5 digit numbers starting with 31 $$= 5 \times 5 \times 5 = 125$$ No of 5 digit numbers starting with 32 $$= 5 \times 5 \times 5 = 125$$ No of 5 digit numbers starting with 33 $$= 5 \times 5 \times 5 = 125$$ No of 5 digit numbers starting with 351 $$= 5 \times 5 = 25$$ No of 5 digit numbers starting with 352 $$= 5 \times 5 = 25$$ No of 5 digit numbers starting with 3531 $$= 5$$ No of 5 digit numbers starting with 3532 $$= 5$$ Before 35337 will be 4 numbers, So rank of 35337 will be 1690 So, in descending order serial number will be $$3125 - 1690 + 1 = 1436$$
Question 65
Maths · Sequences and Series · Fill in the blank
Let $a_1, a_2, a_3, \ldots$ be a GP of increasing positive numbers. If the product of fourth and sixth terms is 9 and the sum of fifth and seventh terms is 24, then $a_1 a_9 + a_2 a_4 a_9 + a_5 + a_7$ is equal to ____.
Let the coefficients of three consecutive terms in the binomial expansion of $(1 + 2x)^n$ be in the ratio $2 : 5 : 8$. Then the coefficient of the term, which is in the middle of these three terms, is_____.
Answer: 1120
Solution
Given $t_{r+1} = {}^{n}C_{r} (2x)^r$. $$\Rightarrow \frac{{}^{n}C_{r-1} (2)^{r-1}}{{}^{n}C_{r} (2)^r} = \frac{2}{5}$$ $$\Rightarrow \frac{\frac{n!}{(r-1)!(n-r+1)!}}{\frac{n!}{r!(n-r)!}} = \frac{2}{5}$$ $$\Rightarrow \frac{r}{n-r+1} = \frac{4}{5} \Rightarrow 5r = 4n - 4r + 4$$ $$\Rightarrow 9r = 4(n+1) \ldots (1)$$ $$\Rightarrow \frac{{}^{n}C_{r} (2)^r}{{}^{n}C_{r+1} (2)^{r+1}} = \frac{5}{8}$$ $$\Rightarrow \frac{\frac{n!}{r!(n-r)!}}{\frac{n!}{(r+1)!(n-r-1)!}} = \frac{5}{4} \Rightarrow \frac{r+1}{n-r} = \frac{5}{4}$$ $$\Rightarrow 4r + 4 = 5n - 5r \Rightarrow 5n - 4 = 9r \ldots (2)$$ From (1) and (2) $$\Rightarrow 4n + 4 = 5n - 4 \Rightarrow n = 8$$ (1) $\Rightarrow r = 4$ So, coefficient of middle term is
Question 67
Maths · Binomial Theorem · Numerical
If the co-efficient of $x^9$ in $$\left( \alpha x^3 + \frac{1}{\beta x} \right)^{11}$$ and the co-efficient of $x^{-9}$ in $$\left( \alpha x - \frac{1}{\beta x^3} \right)^{11}$$ are equal, then $(\alpha \beta)^2$ is equal to .
Answer: 1
Solution
Coefficient of $x^9$ in $$\left(\alpha x^3 + \frac{1}{\beta x}\right) = {^{11}C_6} \cdot \frac{\alpha^5}{\beta^6}$$ Therefore, both are equal $$\frac{11}{C_6} \cdot \frac{\alpha^5}{\beta^6} = -\frac{11}{C_5} \cdot \frac{\alpha^6}{\beta^5}$$ $$\Rightarrow \frac{1}{\beta} = -\alpha$$ $$\Rightarrow \alpha \beta = -1$$ $$\Rightarrow (\alpha \beta)^2 = 1$$
Question 68
Maths · Trigonometric Functions · Single correct
Let $f(\theta) = 3 \left( \sin^4 \left( \frac{3\pi}{2} - \theta \right) + \sin^4 (3\pi + \theta) \right) - 2(1 - \sin^2 2\theta)$ and $$ S = \left\{ \theta \in [0, \pi] : f'(\theta) = -\frac{\sqrt{3}}{2} \right\} $$. If $4\beta = \sum_{\theta \in S} \theta$, then $f(\beta)$ is equal to
Maths · Straight Lines and Pair of Straight Lines · Single correct
A light ray emits from the origin making an angle $30^\circ$ with the positive $x$-axis. After getting reflected by the line $x + y = 1$, if this ray intersects $x$-axis at $Q$, then the abscissa of $Q$ is
$\frac{2}{\sqrt{3} - 1}$
$\frac{2}{3 + \sqrt{3}}$
$\frac{2}{3 - \sqrt{3}}$
$\frac{\sqrt{3}}{2(\sqrt{3} + 1)}$
Answer: (b)
Solution
Slope of reflected ray is $\tan 60^\circ = \sqrt{3}$. Line $y = \frac{x}{\sqrt{3}}$ intersects $y + x = 1$ at $$\left( \frac{\sqrt{3}}{\sqrt{3} + 1}, \frac{1}{\sqrt{3} + 1} \right).$$ Equation of reflected ray is $$y - \frac{1}{\sqrt{3} + 1} = \sqrt{3} \left( x - \frac{\sqrt{3}}{\sqrt{3} + 1} \right).$$ Put $y = 0$ to get $$x = \frac{2}{3 + \sqrt{3}}.$$
Question 70
Maths · Straight Lines and Pair of Straight Lines · Single correct
Let B and C be the two points on the line $y + x = 0$ such that B and C are symmetric with respect to the origin. Suppose A is a point on $y - 2x = 2$ such that $\Delta ABC$ is an equilateral triangle. Then, the area of the $\Delta ABC$ is
$3\sqrt{3}$
$2\sqrt{3}$
$\frac{8}{\sqrt{3}}$
$\frac{10}{\sqrt{3}}$
Answer: (c)
Solution
At point A, $x = y$. The equation $y - 2x = 2$ gives the point $(-2, -2)$. The height from the line $x + y = 0$ is given by $$h = \frac{4}{\sqrt{2}}.$$ The area of $\Delta$ is $$\frac{\sqrt{3}}{4} \frac{h^2}{\sin^2 60} = \frac{8}{\sqrt{3}}.$$
Question 71
Maths · Conic Sections · Single correct
Let the tangents at the points A (4, -11) and B(8, -5) on the circle $x^2 + y^2 - 3x + 10y - 15 = 0$, intersect at the point C. Then the radius of the circle, whose centre is C and the line joining A and B is its tangent, is equal to
$\frac{3\sqrt{3}}{4}$
$2\sqrt{13}$
$\sqrt{13}$
$\frac{2\sqrt{13}}{3}$
Answer: (d)
Solution
Equation of tangent at A (4, -11) on circle is $$\Rightarrow 4x - 11y - 3 \left( \frac{x + 4}{2} \right) + 10 \left( \frac{y - 11}{2} \right) - 15 = 0$$ $$\Rightarrow 5x - 12y - 152 = 0 \ldots (1)$$ Equation of tangent at B (8, -5) on circle is $$\Rightarrow 8x - 5y - 3 \left( \frac{x + 8}{2} \right) + 10 \left( \frac{y - 5}{2} \right) - 15 = 0$$ $$\Rightarrow 13x - 104 = 0 \Rightarrow x = 8$$ Put in (1) $\($ $\Rightarrow$ y = $\frac{28}{3}$ $\)$ $$r = \left| \frac{3.8 + \frac{2.28}{3} - 34}{\sqrt{13}} \right| = \frac{2\sqrt{13}}{3}$$
Question 72
Maths · Limits and Derivatives · Single correct
Let x = 2 be a root of the equation $x^2 + px + q = 0$ and $f(x) = \begin{cases} \frac{1 - \cos(x^2 - 4px + q^2 + 8q + 16)}{(x - 2p)^4}, & x \neq 2p \\ 0, & x = 2p \end{cases}$ Then $\lim_{x \to 2p^+} [f(x)]$ where $[.]$ denotes greatest integer function, is
If p, q and r are three propositions, then which of the following combination of truth values of p, q and r makes the logical expression $$\{(p \lor q) \land ((\sim p) \lor r)\} \rightarrow ((\sim q) \lor r)$$ false?
p = T, q = F, r = T
p = T, q = T, r = F
p = F, q = T, r = F
p = T, q = F, r = F
Answer: (c)
Solution
Option (3) $(p \lor q) \land (\sim q \lor r) \to (\sim p \lor r)$ will be False.
Question 74
Maths · Matrices · Single correct
Let $\alpha$ and $\beta$ be real numbers. Consider a $3 \times 3$ matrix $A$ such that $A^2 = 3A + \alpha I$. If $A^4 = 21A + \beta I$, then
Consider the following system of questions $$\alpha x + 2y + z = 1$$ $$2\alpha x + 3y + z = 1$$ $$3x + \alpha y + 2z = \beta$$ For some $\alpha, \beta \in \mathbb{R}$. Then which of the following is NOT correct.
It has no solution if $\alpha = -1$ and $\beta \neq 2$
It has no solution for $\alpha = -1$ and for all $\beta \in \mathbb{R}$
It has no solution for $\alpha = 3$ and for all $\beta \neq 2$
It has a solution for all $\alpha \neq -1$ and $\beta = 2$
The domain of $f(x) = \frac{\log_{e(x+1)}(x-2)}{e^{2\log_e x} - (2x+3)}, x \in \mathbb{R}$ is
$\mathbb{R} - \{1-3\}$
$(2, \infty) - \{3\}$
$(-1, \infty) - \{3\}$
$\mathbb{R} - \{3\}$
Answer: (b)
Solution
Given the inequality $x - 2 > 0$, we have $x > 2$. For $x + 1 > 0$, we get $x > -1$. The condition $x + 1 \neq 1$ implies $x \neq 0$ and $x > 0$. For the denominator, $x^2 - 2x - 3 \neq 0$, we factor it as $(x - 3)(x + 1) \neq 0$, giving $x \neq -1, 3$. Therefore, the solution is $(2, \infty) - \{3\}$.
Question 77
Maths · Relations and Functions · Single correct
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function such that $$ f(x) = \frac{x^2 + 2x + 1}{x^2 + 1}. $$ Then
$f(x)$ is many-one in $(-\infty, -1)$
$f(x)$ is many-one in $(1, \infty)$
$f(x)$ is one-one in $[1, \infty)$ but not in $(-\infty, \infty)$
$f(x)$ is one-one in $(-\infty, \infty)$
Answer: (c)
Solution
Given the function $$f(x) = \frac{(x+1)^2}{x^2+1} = 1 + \frac{2x}{x^2+1}$$ we can rewrite it as $$f(x) = 1 + \frac{2}{x + \frac{1}{x}}.$$
Question 78
Maths · Continuity and Differentiability · Numerical
Let $f : \mathbb{R} \to \mathbb{R}$ be a differentiable function that satisfies the relation $f(x+y) = f(x) + f(y) - 1$, $\forall x, y \in \mathbb{R}$. If $f'(0) = 2$, then $|f(-2)|$ is equal to _____.
Answer: 3
Solution
Given $f(x + y) = f(x) + f(y) - 1$. The derivative $f'(x)$ is given by $$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$$ $$f'(x) = \lim_{h \to 0} \frac{f(h) - f(0)}{h} = f'(0) = 2$$ Since $f'(x) = 2$, we have $dy = 2dx$. Therefore, $y = 2x + C$. Given $x = 0$, $y = 1$, $c = 1$, we find $y = 2x + 1$. Finally, $$|f(-2)| = |-4 + 1| = |-3| = 3$$
Question 79
Maths · Sequences and Series · Numerical
Suppose f is a function satisfying f(x+y) = f(x) + f(y) for all x, y $\in \mathbb{N}$ and f(1) = $\frac{1}{5}$. If $\sum_{n=1}^{m} \frac{f(n)}{n(n+1)(n+2)}$ = $\frac{1}{12}$ then m is equal to $\ldots$
Answer: 10
Solution
Question 80
Maths · Integrals · Single correct
Let $f(x) = x + \frac{a}{\pi^2 - 4} \sin x + \frac{b}{\pi^2 - 4} \cos x$, $x \in \mathbb{R}$ be a function which satisfies $f(x) = x + \int_{0}^{\pi/2} \sin(x+y) f(y) \, dy$. Then $(a+b)$ is equal to
$-\pi(\pi+2)$
$-2\pi(\pi+2)$
$-2\pi(\pi-2)$
$-\pi(\pi-2)$
Answer: (b)
Solution
Given $$f(x) = x + \int_0^{\pi/2} (\sin x \cos y + \cos x \sin y) f(y) \, dy$$ $$f(x) = x + \int_0^{\pi/2} ((\cos y \, f(y) \, dy) \sin x + (\sin y \, f(y) \, dy) \cos x) \ldots (1)$$ On comparing with $$f(x) = x + \frac{a}{\pi^2 - 4} \sin x + \frac{b}{\pi^2 - 4} \cos x, x \in \mathbb{R}$$ then $$\Rightarrow \frac{a}{\pi^2 - 4} = \int_0^{\pi/2} \cos y \, f(y) \, dy \ldots (2)$$ $$\Rightarrow \frac{b}{\pi^2 - 4} = \int_0^{\pi/2} \sin y \, f(y) \, dy \ldots (3)$$ Add (2) and (3) $$\frac{a+b}{\pi^2 - 4} = \int_0^{\pi/2} (\sin y + \cos y) f(y) \, dy \ldots (4)$$ $$\frac{a+b}{\pi^2 - 4} = \int_0^{\pi/2} (\sin y + \cos y) f\left(\frac{\pi}{2} - y\right) \, dy \ldots (5)$$ Add (4) and (5) $$2 \frac{(a+b)}{\pi^2 - 4} = \int_0^{\pi/2} (\sin y + \cos y) \left(\frac{\pi}{2} + \frac{(a+b)}{\pi^2 - 4} (\sin y + \cos y)\right) \, dy$$ $$= \pi + \frac{a+b}{\pi^2 - 4} \left(\frac{\pi}{2} + 1\right)$$ $$(a+b) = -2\pi(\pi + 2)$$
Question 81
Maths · Integrals · Single correct
Let [x] denote the greatest integer $\leq x$. Consider the function $f(x) = \max \{x^2, 1 + [x]\}$. Then the value of the integral $$\int_{0}^{2} f(x) \, dx$$ is:
$\frac{5 + 4\sqrt{2}}{3}$
$\frac{8 + 4\sqrt{2}}{3}$
$\frac{1 + 5\sqrt{2}}{3}$
$\frac{4 + 5\sqrt{2}}{3}$
Answer: (a)
Solution
The area $A$ is given by the sum of the integrals: $$A = \int_0^1 1 \, dx + \int_1^{\sqrt{2}} 2 \, dx + \int_{\sqrt{2}}^2 x^2 \, dx$$ Calculating each integral, we have: $$= 1 + 2\sqrt{2} - 2 + \frac{8}{3} - \frac{2\sqrt{2}}{3}$$ Simplifying, we get: $$= \frac{5}{3} + \frac{4\sqrt{2}}{3}$$
Question 82
Maths · Applications of Integrals · Single correct
Let A = $\{ (x, y) \in \mathbb{R}^2 : y \geq 0,\; 2x \leq y \leq \sqrt{4 - (x-1)^2} \}$ and B = $\{ (x, y) \in \mathbb{R} \times \mathbb{R} : 0 \leq y \leq \min \{ 2x,\; \sqrt{4 - (x-1)^2} \} \}$ Then the ratio of the area of A to the area of B is
$\frac{\pi - 1}{\pi + 1}$
$\frac{\pi}{\pi - 1}$
$\frac{\pi}{\pi + 1}$
$\frac{\pi + 1}{\pi - 1}$
Answer: (a)
Solution
Given $y^2 + (x-1)^2 = 4$. The shaded portion is circular (OABC) minus the area of $\triangle OAB$. $$-Ar(\triangle OAB)$$ $$= \frac{\pi(4)}{4} - \frac{1}{2}(2)(1)$$ $$A = (\pi - 1)$$ Area $B = Ar(\triangle OAB) + Area of arc of circle (ABC)$ $$= \frac{1}{2}(1)(2) + \frac{\pi(2)^2}{4} = \pi + 1$$ $$\frac{A}{B} = \frac{\pi - 1}{\pi + 1}$$
Question 83
Maths · Applications of Integrals · Single correct
Let $\Delta$ be the area of the region $\{(x,y) \in \mathbb{R}^2 : x^2 + y^2 \leq 21, y^2 \leq 4x, x \geq 1\}$. Then $$\frac{1}{2}\left(\Delta - 21 \sin^{-1} \frac{2}{\sqrt{7}}\right)$$ is equal to
Let $y = f(x)$ be the solution of the differential equation $y(x + 1) \, dx - x^2 \, dy = 0$, $y(1) = e$. Then $\lim_{x \to 0^+} f(x)$ is equal to
0
$\frac{1}{e}$
$e^2$
$\frac{1}{e^2}$
Answer: (a)
Solution
Given $\frac{x+1}{x^2} \, dx = \frac{dy}{y}$. \[ \ln x - \frac{1}{x} = \ln y + c \] At $(1, e)$, $c = -2$ \[ \ln x - \frac{1}{x} = \ln y - 2 \] \[ y = e^{\ln x - \frac{1}{x} + 2} \] \[ \lim_{x \to 0^+} e^{\ln x - 1 - \frac{1}{x} + 2} \] \[ = e^{-\infty} \] \[ = 0 \]
Question 85
Maths · Vector Algebra · Single correct
If the vectors $\vec{a} = \lambda \hat{i} + \mu \hat{j} + 4 \hat{k}$, $\vec{b} = - 2 \hat{i} + 4 \hat{j} - 2 \hat{k}$ and $\vec{c} = 2 \hat{i} + 3 \hat{j} + \hat{k}$ are coplanar and the projection of $\vec{a}$ on the vector $\vec{b}$ is $\sqrt{54}$ units, then the sum of all possible values of $\lambda + \mu$ is equal to
Let $\vec{a}$, $\vec{b}$ and $\vec{c}$ be three non-zero non-coplanar vectors. Let the position vectors of four points $A$, $B$, $C$ and $D$ be $\vec{a} - \vec{b} + \vec{c}$, $\lambda \vec{a} - 3\vec{b} + 4\vec{c}$, $-\vec{a} + 2\vec{b} - 3\vec{c}$ and $2\vec{a} - 4\vec{b} + 6\vec{c}$ respectively. If $\overrightarrow{AB}$, $\overrightarrow{AC}$ and $\overrightarrow{AD}$ are coplanar, then $\lambda$ is :
Let the equation of the plane $P$ containing the line $x + 10 = \frac{8 - y}{2} = z$ be $ax + by + 3z = 2(a+b)$ and the distance of the plane $P$ from the point $(1, 27, 7)$ be $c$. Then $a^2 + b^2 + c^2$ is equal to ____.
Answer: 355
Solution
The line $\frac{x+10}{1} = \frac{y-8}{-2} = \frac{z}{1}$ have a point $(-10, 8, 0)$ with direction ratios $(1, -2, 1)$. Therefore, the plane $ax + by + 3z = 2(a + b)$. This implies $b = 2a$. The dot product of direction ratios is zero. Therefore, $a - 2b + 3 = 0$. Thus, $a = 1$ and $b = 2$. The distance from $(1, 27, 7)$ is $$c = \frac{1 + 54 + 21 - 6}{\sqrt{14}} = \frac{70}{\sqrt{14}} = 5\sqrt{14}.$$ Therefore, $a^2 + b^2 + c^2 = 1 + 4 + 350 = 355$.
Question 88
Maths · Three Dimensional Geometry · Numerical
Let the co-ordinates of one vertex of $\Delta ABC$ be $A(0, 2, \alpha)$ and the other two vertices lie on the line $\frac{x + \alpha}{5}$ = $\frac{y - 1}{2}$ = $\frac{z + 4}{3}$. For $\alpha \in \mathbb{Z}$, if the area of $\Delta ABC$ is 21 sq. units and the line segment BC has length $2\sqrt{21}$ units, then ( $\alpha^2$ ) is equal to .
Fifteen football players of a club-team are given 15 T-shirts with their names written on the backside. If the players pick up the T-shirts randomly, then the probability that at least 3 players pick the correct T-shirt is
There rotten apples are mixed accidently with seven good apples and four apples are drawn one by one without replacement. Let the random variable $X$ denote the number of rotten apples. If $\mu$ and $\sigma^2$ represent mean and variance of $X$, respectively, then $10 \left( \mu^2 + \sigma^2 \right)$ is equal to
20
250
25
30
Answer: (a)
Solution
The sum of $xP(x)$ is given by $$\sum xP(x) = \frac{6}{2} = \mu$$ The variance $\sigma^2$ is calculated as $$\sigma^2 = \sum x^2P(x) - \mu^2$$ Calculating $\sigma^2 + \mu^2$: $$\sigma^2 + \mu^2 = 0 + \frac{1}{2} + \frac{12}{10} + \frac{9}{30} = 2$$ Therefore, $$10(\sigma^2 + \mu^2) = 20$$ Ans.