JEE Main 25 January 2023 Shift 1 question paper with solutions
JEE Main 25 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 · Mathematics in Physics · Numerical
If $$ \vec{P}=3\hat{i}+\sqrt{3}\hat{j}+2\hat{k} $$ and $$ \vec{Q}=4\hat{i}+\sqrt{3}\hat{j}+2.5\hat{k}, $$ then the unit vector in the direction of $\vec{P}\times\vec{Q}$ is $$ \frac{1}{x}\left(\sqrt{3}\hat{i}+\hat{j}-2\sqrt{3}\hat{k}\right). $$ The value of $x$ is:
Physics · Motion in a Straight Line · Single correct
A car travels a distance of 'x' with speed $V_1$ and then same distance 'x' with speed $V_2$ in the same direction. The average speed of the car is:
$\frac{v_1 v_2}{2(v_1 + v_2)}$
$\frac{v_1 + v_2}{2}$
$\frac{2x}{v_1 + v_2}$
$\frac{2v_1 v_2}{v_1 + v_2}$
Answer: (d)
Solution
Average velocity is given by the formula: $$Average velocity = \frac{Total displacement}{Total time}$$ The total displacement is $x + x$ and the total time is $\frac{x}{v_1} + \frac{x}{v_2}$. Therefore, $$Average velocity = \frac{x + x}{\frac{x}{v_1} + \frac{x}{v_2}} = \frac{2v_1v_2}{v_1 + v_2}$$
Question 4
Physics · Laws of Motion · Single correct
A car is moving with a constant speed of 20 m/s in a circular horizontal track of radius 40 m. A bob is suspended from the roof of the car by a massless string. The angle made by the string with the vertical will be: (Take g = 10 \, $\mathrm{m/s^2}$)
$\frac{\pi}{6}$
$\frac{\pi}{2}$
$\frac{\pi}{4}$
$\frac{\pi}{3}$
Answer: (c)
Solution
Given the forces, we have: $$T \cos \theta = mg$$ $$T \sin \theta = \frac{mv^2}{R}$$ Dividing the second equation by the first, we get: $$\tan \theta = \frac{v^2}{Rg}$$ Substituting the given values: $$\tan \theta = \frac{20^2}{40 \times 10}$$ This simplifies to: $$\tan \theta = 1$$ Therefore, the angle is: $$\theta = \frac{\pi}{4}$$
Question 5
Physics · Work, Energy and Power · Numerical
An object of mass 'm' initially at rest on a smooth horizontal plane starts moving under the action of force $F = 2 \, \mathrm{N}$. In the process of its linear motion, the angle $\theta$ (as shown in figure) between the direction of force and horizontal varies as $\theta = kx$, where $k$ is a constant and $x$ is the distance covered by the object from its initial position. The expression of kinetic energy of the object will be $E = \frac{n}{k} \sin \theta$. The value of $n$ is ___.
Answer: 2
Solution
Given the forces acting on the mass $M$, we have $F \cos \theta = ma$. The equation $2 \cos(kx) = \frac{mv \frac{dv}{dx}}{dx}$ is given. Integrating both sides, we have $$\int_0^v v \, dv = 2 \int_0^x \cos(kx) \, dx.$$ Solving the integrals, we get $$\frac{mv^2}{2} = \frac{2}{k} \sin kx.$$ The kinetic energy is given by $$K.E. = \frac{2}{k} \sin \theta.$$ Therefore, $n = 2$.
Question 6
Physics · System of Particles and Rotational Motion · Single correct
An object of mass 8 kg is hanging from one end of a uniform rod CD of mass 2 kg and length 1 m pivoted at its end C on a vertical wall as shown in figure. It is supported by a cable AB such that the system is in equilibrium. The tension in the cable is: (Take $g = 10 \, \mathrm{m/s^2}$)
240 N
90 N
300 N
30 N
Answer: (c)
Solution
Taking torque about point C $$\frac{T}{2} \times 60 = 20 \times 50 + 80 \times 100$$ $$\Rightarrow 3T = 100 + 800$$ $$\Rightarrow T = 300 \, \mathrm{N}$$
Question 7
Physics · System of Particles and Rotational Motion · Fill in the blank
$I_{CM}$ is moment of inertia of a circular disc about an axis (CM) passing through its center and perpendicular to the plane of disc. $I_{AB}$ is it's moment of inertia about an axis AB perpendicular to plane and parallel to axis CM at a distance $\frac{2}{3}$ R from center. Where R is the radius of the disc.The ratio of $I_{AB}$ and $I_{CM}$ is $x:9$. The value of x is $\_$.
Answer: 17
Solution
Given $$I_{cm} = \frac{mR^2}{2}$$ We have $$I_{AB} = \frac{mR^2}{2} + m \left( \frac{2R}{3} \right)^2 = \frac{17}{18} mR^2$$ Therefore, $$\frac{I_{AB}}{I_{cm}} = \frac{17}{9} \Rightarrow x = 17$$
Question 8
Physics · Gravitation · Single correct
Assume that the earth is a solid sphere of uniform density and a tunnel is dug along its diameter throughout the earth. It is found that when a particle is released in this tunnel, it executes a simple harmonic motion. The mass of the particle is 100 g. The time period of the motion of the particle will be (approximately) (take $g = 10 \, \mathrm{ms^{-2}}$, radius of earth $= 6400 \, \mathrm{km}$)
24 hours
1 hour 24 minutes
1 hour 40 minutes
12 hours
Answer: (b)
Solution
Let at some time particle is at a distance $x$ from centre of Earth, then at that position field $$E = \frac{GM}{R^3} x$$ Therefore, acceleration of particle $$\vec{a} = -\frac{GM}{R^3} \vec{x}$$ Thus, $$\omega = \sqrt{\frac{GM}{R^3}} = \sqrt{\frac{g}{R}}$$ Now $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{R}{g}}$ Therefore, $$T = 2 \times 3.14 \times \sqrt{\frac{6400 \times 10^3}{10}}$$ $$= 2 \times 3.14 \times 800 sec \approx 1 hour 24 minutes$$
Question 9
Physics · Gravitation · Single correct
T is the time period of simple pendulum on the earth's surface. Its time period becomes x T when taken to a height R (equal to earth's radius) above the earth's surface. Then, the value of x will be:
4
2
$\frac{1}{2}$
$\frac{1}{4}$
Answer: (b)
Solution
At surface of earth time period $$T = 2\pi \sqrt{\frac{\ell}{g}}$$ At height $h = R$ $$g' = \frac{g}{\left(1 + \frac{h}{R}\right)^2} = \frac{g}{4}$$ Therefore, $$xT = 2\pi \sqrt{\frac{\ell}{g/4}}$$ $$\Rightarrow xT = 2 \times 2\pi \sqrt{\frac{\ell}{g}}$$ $$\Rightarrow xT = 2T \Rightarrow x = 2$$
Question 10
Physics · Mechanical Properties of Solids · Numerical
As shown in the figure, in an experiment to determine Young's modulus of a wire, the extension-load curve is plotted. The curve is a straight line passing through the origin and makes an angle of $45^\circ$ with the load axis. The length of wire is $62.8 \, \mathrm{cm}$ and its diameter is $4 \, \mathrm{mm}$. The Young's modulus is found to be $x \times 10^4 \, \mathrm{Nm}^{-2}$. The value of $x$ is _____.
A bowl filled with very hot soup cools from $98^{\circ} \mathrm{C}$ to $86^{\circ} \mathrm{C}$ in 2 minutes when the room temperature is $22^{\circ} \mathrm{C}$. How long it will take to cool from $75^{\circ} \mathrm{C}$ to $69^{\circ} \mathrm{C}$?
A Carnot engine with efficiency 50$\%$ takes heat from a source at 600$\mathrm{K}$. In order to increase the efficiency to 70$\%$, keeping the temperature of sink same, the new temperature of the source will be:
360 $\mathrm{K}$
1000 $\mathrm{K}$
900 $\mathrm{K}$
300 $\mathrm{K}$
Answer: (b)
Solution
Initially $\eta = \frac{1}{2}$. But $\eta = 1 - \frac{T_2}{T_1}$. Therefore, $$\frac{1}{2} = 1 - \frac{T_2}{600}$$ $$\Rightarrow \frac{T_2}{600} = \frac{1}{2} \Rightarrow T_2 = 300 \, \mathrm{K}$$ Now efficiency is increased to 70% and $T_2 = 300 \, \mathrm{K}$. Let temperature of source $T_1 = T$. $$\Rightarrow \frac{7}{10} = 1 - \frac{300}{T}$$ $$\Rightarrow \frac{300}{T} = 1 - \frac{7}{10}$$ $$\Rightarrow \frac{300}{T} = \frac{3}{10} \therefore T = 1000 \, \mathrm{K}$$
Question 13
Physics · Kinetic Theory · Single correct
The root mean square velocity of molecules of gas is
Proportional to square of temperature $(T^2)$.
Inversely proportional to square root of temperature $\sqrt{\frac{1}{T}}$.
Proportional to square root of temperature $\sqrt{T}$.
Proportional to temperature $(T)$.
Answer: (c)
Solution
The rms speed of a gas molecule is $$V_{RMS} = \sqrt{\frac{3RT}{M}}$$ $$V_{RMS} \propto \sqrt{T}$$
Question 14
Physics · Waves · Numerical
The distance between two consecutive points with phase difference of $60^\circ$ in a wave of frequency $500 \, \mathrm{Hz}$ is $6.0 \, \mathrm{m}$. The velocity with which wave is traveling is _____ km/s
Answer: 18
Solution
Given the equation for phase difference, $$\Delta \phi = \frac{2\pi}{\lambda} \Delta x$$ we have $$\frac{\pi}{3} = \frac{2\pi}{\lambda} (6 \, \mathrm{m})$$ Solving for $\lambda$, we find $$\lambda = 36 \, \mathrm{m}$$ The velocity $V$ is given by $$V = f \lambda = (500 \, \mathrm{Hz})(36 \, \mathrm{m})$$ which simplifies to $$= 18000 \, \mathrm{m/s} = 18 \, \mathrm{km/s}$$
Question 15
Physics · Electric Charges and Fields · Numerical
A uniform electric field of $10 \, \mathrm{N/C}$ is created between two parallel charged plates (as shown in figure). An electron enters the field symmetrically between the plates with a kinetic energy $0.5 \, \mathrm{eV}$. The length of each plate is $10 \, \mathrm{cm}$. The angle ($\theta$) of deviation of the path of electron as it comes out of the field is ___ (in degree).
Answer: 45
Solution
Given $0.5 e = \frac{1}{2} m v_x^2$, we have $v_x = \sqrt{\frac{e}{m}}$. Along the x-axis, $L = v_x t = \sqrt{\frac{e}{m}} t$. Along the y-axis, $v_y = \frac{eE}{m} t$. Dividing, we get $$\frac{v_y}{L} = E \sqrt{\frac{e}{m}} = E v_x.$$ Therefore, $$\tan \theta = \frac{v_y}{v_x} = E \times L = 10 \times 0.1 = 1.$$ Thus, $\theta = 45^\circ$.
Question 16
Physics · Electrostatic Potential and Capacitance · Single correct
A parallel plate capacitor has plate area $40 \, \mathrm{cm}^2$ and plates separation $2 \, \mathrm{mm}$. The space between the plates is filled with a dielectric medium of a thickness $1 \, \mathrm{mm}$ and dielectric constant $5$. The capacitance of the system is:
$24 \varepsilon_0 \, \mathrm{F}$
$\frac{3}{10} \varepsilon_0 \, \mathrm{F}$
$\frac{10}{3} \varepsilon_0 \, \mathrm{F}$
$10 \varepsilon_0 \, \mathrm{F}$
Answer: (c)
Solution
This can be seen as two capacitors in series combination so $$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$$ $$= \frac{1}{K \varepsilon_0 \frac{A}{t}} + \frac{1}{\varepsilon_0 \frac{A}{d-t}}$$ $$= \frac{t}{K \varepsilon_0 A} + \frac{d-t}{\varepsilon_0 A}$$ $$= \frac{1 \times 10^{-3}}{5 \varepsilon_0 \times 40 \times 10^{-4}} + \frac{1 \times 10^{-3}}{\varepsilon_0 \times 40 \times 10^{-4}}$$ $$\frac{1}{C_{eq}} = \frac{1}{20 \varepsilon_0} + \frac{1}{4 \varepsilon_0}$$ $$C_{eq} = \frac{20 \times 4 \varepsilon_0}{24} = \frac{10}{3} \varepsilon_0 \ \mathrm{F}$$
Question 17
Physics · Current Electricity · Single correct
A uniform metallic wire carries a current 2 A. when 3.4 V battery is connected across it. The mass of uniform metallic wire is $8.92 \times 10^{-3}$ kg, density is $8.92 \times 10^{3}$ kg/m$^{3}$ and resistivity is $1.7 \times 10^{-8}$ $\Omega$-m. The length of wire is:
$l = 6.8$ m
$l = 10$ m
$l = 5$ m
$l = 100$ m
Answer: (b)
Solution
Given $I = 2 \, \mathrm{A}$ and $\Delta V = 3.4 \, \mathrm{V}$. Using Ohm's Law, $$R = \frac{3.4}{2} = 1.7 \, \Omega$$ We have $$1.7 = \frac{\rho L}{A}$$ Solving for $L$, $$L = \frac{1.7 \left( A \right)}{\rho}$$ The mass $M$ is given by the density times volume. The volume is $$Volume = \frac{8.92 \times 10^{-3}}{8.92 \times 10^3} = 10^{-6}$$ Therefore, $$L^2 = \frac{1.7}{\rho} \left( 10^{-6} \right) = \frac{1.7}{1.7} \times 10^2$$ Thus, $L = 10 \, \mathrm{m}$.
Question 18
Physics · Current Electricity · Numerical
In the given circuit, the equivalent resistance between the terminal A and B is _____ $\Omega$.
Answer: 10
Solution
Question 19
Physics · Moving Charges and Magnetism · Single correct
Match List I with List II Choose the correct answer from the option given below:
A-III, B-IV, C-I, D-II
A-I, B-III, C-IV, D-II
A-III, B-I, C-IV, D-II
A-II, B-I, C-IV, D-III
Answer: (c)
Solution
Match List I with List II. A corresponds to III because the magnetic field at point O is given by $$B_0 = \frac{\mu_0 I}{2 \pi l} [\pi - 1]$$. B corresponds to I because the magnetic field at point O is given by $$B_0 = \frac{\mu_0 I}{4 \pi l} [\pi + 2]$$. C corresponds to IV because the magnetic field at point O is given by $$B_0 = \frac{\mu_0 I}{4 \pi l} [\pi + 1]$$. D corresponds to II because the magnetic field at point O is given by $$B_0 = \frac{\mu_0 I}{4 r}$$. Therefore, the correct answer is option (1): A-III, B-IV, C-I, D-II.
Question 20
Physics · Moving Charges and Magnetism · Single correct
A solenoid of 1200 turns is wound uniformly in a single layer on a glass tube 2 m long and 0.2 m in diameter. The magnetic intensity at the center of the solenoid when a current of 2 A flows through it is:
2.4 × 10$^{3}$ A m$^{-1}$
1.2 × 10$^{3}$ A m$^{-1}$
1 A m$^{-1}$
2.4 × 10$^{-3}$ A m$^{-1}$
Answer: (b)
Solution
Magnetic field at centre inside the solenoid is given by $B = \mu_0 n I$. So magnetic intensity at centre $$H = \frac{B}{\mu_0} = n I = \left( \frac{1200}{2} \right)(2)$$ $$H = 1.2 \times 10^3 \, \mathrm{Am^{-1}}$$
Question 21
Physics · Alternating Current · Single correct
In an LC oscillator, if values of inductance and capacitance become twice and eight times, respectively, then the resonant frequency of oscillator becomes $x$ times its initial resonant frequency $\omega_0$. The value of $x$ is:
1/4
16
1/16
4
Answer: (a)
Solution
The resonance frequency of LC oscillations circuit is $$\omega_0 = \frac{1}{\sqrt{LC}}$$ $L \to 2L$ $C \to 8C$ $$\omega = \frac{1}{\sqrt{2L \times 8C}} = \frac{1}{4\sqrt{LC}}$$ $$\omega = \frac{\omega_0}{4}$$ So $x = \frac{1}{4}$
Question 22
Physics · Alternating Current · Numerical
An LCR series circuit of capacitance $62.5 \, \mathrm{nF}$ and resistance of $50 \, \Omega$. is connected to an A.C. source of frequency $2.0 \, \mathrm{kHz}$. For maximum value of amplitude of current in circuit, the value of inductance is _____ mH. (take $\pi^2 = 10$)
Answer: 100
Solution
Given the formula for frequency, $f = \frac{1}{2\pi \sqrt{LC}}$. Substituting $2000 \, \mathrm{Hz}$ into the equation: $$2000 = \frac{1}{2\pi \sqrt{L \times 62.5 \times 10^{-9}}}$$ Solving for $L$: $$L = \frac{1}{4\pi^2 \times 2000^2 \times 62.5 \times 10^{-9}} = 0.1 \, \mathrm{H} = 100 \, \mathrm{mH}$$
Question 23
Physics · Electromagnetic Waves · Single correct
All electromagnetic wave is transporting energy in the negative z direction. At a certain point and certain time the direction of electric field of the wave is along positive y direction. What will be the direction of the magnetic field of the wave at that point and instant?
Positive direction of x
Positive direction of z
Negative direction of x
Negative direction of y
Answer: (a)
Solution
As, poynting vector $\vec{S} = \vec{E} \times \vec{H}$. Given energy transport = negative z direction. Electric field = positive y direction. $$(-\hat{k}) = (+\hat{j}) \times [\hat{i}]$$ Hence according to vector cross product magnetic field should be positive x direction.
Question 24
Physics · Ray Optics and Optical Instruments · Numerical
A ray of light is incident from air on a glass plate having thickness $\sqrt{3} \, \mathrm{cm}$ and refractive index $\sqrt{2}$. The angle of incidence of a ray is equal to the critical angle for glass-air interface. The lateral displacement of the ray when it passes through the plate is ___ $\times 10^{-2} \, \mathrm{cm}$. (given $\sin 15^\circ = 0.26$)
In Young's double slits experiment, the position of 5th bright fringe from the central maximum is 5 cm. The distance between slits and screen is 1 m and wavelength of used monochromatic light is 600 nm. The separation between the slits is:
Physics · Dual Nature of Radiation and Matter · Single correct
Electron beam used in an electron microscope, when accelerated by a voltage of 20 $\mathrm{kV}$, has a de-Broglie wavelength of $\lambda_0$. If the voltage is increased to 40 $\mathrm{kV}$, then the de-Broglie wavelength associated with the electron beam would be:
3$\lambda$_0
9$\lambda$_0
$\frac{\lambda_0}{2}$
$\frac{\lambda_0}{\sqrt{2}}$
Answer: (d)
Solution
When an electron is accelerated through potential difference $V$, then K.E. = $eV$. Therefore, $$\lambda = \frac{h}{\sqrt{2m(KE)}} = \frac{h}{\sqrt{2meV}}$$ Thus, $$\lambda \propto \frac{1}{\sqrt{V}}$$ Therefore, $$\frac{\lambda}{\lambda_0} = \frac{\sqrt{20}}{\sqrt{40}}$$ Thus, $$\lambda = \frac{\lambda_0}{\sqrt{2}}$$
Question 27
Physics · Atoms · Numerical
The wavelength of the radiation emitted is $\lambda_0$ when an electron jumps from the second excited state to the first excited state of the hydrogen atom. If the electron jumps from the third excited state to the second orbit of the hydrogen atom, the wavelength of the radiation emitted will be $\frac{20}{x} \lambda_0$. The value of $x$ is .
Answer: 27
Solution
Second excited state to first excited state $n = 3 \rightarrow n = 2$ $$\frac{hc}{\lambda_0} = 13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) \ldots (i)$$ Third excited state to second orbit $n = 4 \rightarrow n = 2$ $$\frac{hc}{\left( 20 \lambda_0 / x \right)} = 13.6 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) \ldots (ii)$$ $(ii) \div (i)$ $$\frac{x}{20} = \frac{\frac{1}{2^2} - \frac{1}{4^2}}{\frac{1}{2^2} - \frac{1}{3^2}}$$ $$x = 27$$
Question 28
Physics · Nuclei · Single correct
The ratio of the density of oxygen nucleus $\left( ^{16}_{8}\mathrm{O} \right)$ and helium nucleus $\left( ^{4}_{2}\mathrm{He} \right)$ is
4:1
8:1
1:1
2:1
Answer: (c)
Solution
Nuclear density is independent of mass number. As nuclear density = $\($ $\frac{Au}{\frac{4}{3} \pi R^3}$ $\)$. Also, $\($ R = R_0 A^{$\frac{1}{3}$} $\)$. And $\($ R^3 = R_0^3 A $\)$. Therefore, Nuclear density = $\($ $\frac{Au}{\frac{4}{3} \pi R_0^3 A}$ $\)$. Nuclear density = $\($ $\frac{3u}{4 \pi R_0^3}$ $\)$.
Question 29
Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits · Single correct
Given below are two statements : one is labeled as Assertion A and the other is labeled as Reason R Assertion A: Photodiodes are used in forward bias usually for measuring the light intensity. Reason R: For a p-n junction diode, at applied voltage $V$ the current in the forward bias is more than the current in the reverse bias for $|V_z| > \pm V \geq |V_0|$ where $V_0$ is the threshold voltage and $V_z$ is the breakdown voltage. In the light of the above statements, choose the correct answer from the options given below
Both A and R are true and R is correct explanation A
Both A and R are true but R is NOT the correct explanation A
A is false but R is true
A is true but R is false
Answer: (c)
Solution
Theory based. Photodiodes are operated in reverse bias condition. For P-N junction, current in forward bias (for $V \geq V_0$) is always greater than current in reverse bias (for $V \leq V_z$). Hence Assertion is false but Reason is true.
Question 30
Physics · Communication Systems · Single correct
A message signal of frequency $5 \, \mathrm{kHz}$ is used to modulate a carrier signal of frequency $2 \, \mathrm{MHz}$. The bandwidth for amplitude modulation is:
$5 \, \mathrm{kHz}$
$20 \, \mathrm{kHz}$
$10 \, \mathrm{kHz}$
$2.5 \, \mathrm{kHz}$
Answer: (c)
Solution
Given Signal frequency $f_m = 5 \, \mathrm{kHz}$ Carrier wave frequency $f_c = 2 \, \mathrm{MHz}$ $f_c = 2000 \, \mathrm{kHz}$ The resultant signal will have bandwidth of frequency given by $$[(f_c + f_m) - (f_c - f_m)]$$ $$\Rightarrow [(2000 + 5) - (2000 - 5)] \, \mathrm{kHz}$$ $$\Rightarrow 10 \, \mathrm{kHz}$$
Chemistry
Question 31
Chemistry · Structure of Atom · Single correct
The radius of the $2^{\text{nd}}$ orbit of $\mathrm{Li^{2+}}$ is $x$. The expected radius of the $3^{\text{rd}}$ orbit of $\mathrm{Be^{3+}}$ is
$\frac{9}{4} x$
$\frac{4}{9} x$
$\frac{27}{16} x$
$\frac{16}{27} x$
Answer: (c)
Solution
For $\mathrm{Li}^{2+}$, $r_2 = x = k \times \frac{2^2}{3} = \frac{4k}{2}$. For $\mathrm{Be}^{3+}$, $r_3 = y = k \times \frac{3^2}{4}$. The ratio $\frac{y}{x} = \frac{9}{4} \times \frac{3}{4} = \frac{27}{16}$. Thus, $y = \frac{27}{16} x$.
Question 32
Chemistry · Chemical Bonding and Molecular Structure · Numerical
The total number of lone pairs of electrons on oxygen atoms of ozone is
Answer: 6
Solution
Total number of lone pairs on oxygen atoms is 6.
Question 33
Chemistry · Equilibrium · Numerical
A litre of buffer solution contains 0.1 mole of each of $NH_3$ and $NH_4Cl$. On the addition of 0.02 mole of HCl by dissolving gaseous HCl, the pH of the solution is found to be _____ $\times$ $10^{-3}$ (Nearest integer) [Given : $pK_b(NH_3)$ = 4.745 $\log$ 2 = 0.301 $\log$ 3 = 0.477 T = 298 $\mathrm{K}$]
The density of a monobasic strong acid (Molar mass 24.2 g mol) is 1.21 kg L. The volume of its solution required for the complete neutralization of 25 mL of 0.24 M NaOH is __________ $\times$ 10^{-2} mL (Nearest integer)
Match List I with List II Choose the correct answer from the options given below:
A-II, B-I, C-III, D-IV
A-II, B-IV, C-I, D-III
A-II, B-I, C-IV, D-III
A-IV, B-III, C-II, D-I
Answer: (c)
Solution
The table shows the colors observed in a flame test for different elements. Potassium (K) shows a violet color, calcium (Ca) shows a brick red color, strontium (Sr) shows a crimson red color, and barium (Ba) shows an apple green color.
Question 37
Chemistry · Organic Chemistry — Some Basic Principles & Techniques · Single correct
Which of the following conformations will be the most stable?
Answer: (a)
Solution
Conformation has lowest van der Waals and torsional strain. Hence it must be most stable.
In sulphur estimation, 0.471 g of an organic compound gave 1.4439 g of barium sulphate. The percentage of sulphur in the compound is (Nearest Integer) (Given: Atomic mass Ba: 137 u, S: 32 u, O: 16 u)
The compound which will have the lowest rate towards nucleophilic aromatic substitution on treatment with OH$^-$ is
Answer: (d)
Solution
Electron withdrawing groups are highly ineffective at meta position in nucleophilic aromatic substitution reactions. Hence compound $\mathrm{NO_2}$ $\mathrm{Cl}$ will have lowest rate in nucleophilic aromatic substitution.
Question 40
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
The correct sequence of reagents for the preparation of Q and R is:
The compound $(P)$ undergoes the following reactions: (i) With $\mathrm{Cr_2O_3}$ at $770 \, \mathrm{K}$ and $20 \, \mathrm{atm}$, it forms a compound with a benzene ring and a $\mathrm{CH_3}$ group. (ii) This compound reacts with $\mathrm{CrO_2Cl_2}$ and $\mathrm{H_3O^+}$ to form a compound with a benzene ring and an $\mathrm{CHO}$ group. (iii) The compound undergoes a Cannizzaro reaction with $\mathrm{NaOH}$ to form $\mathrm{COONa}$ and $\mathrm{CH_2OH}$. (iv) Finally, with $\mathrm{H_3O^+}$, it forms $\mathrm{COOH}$ and $\mathrm{CH_2OH}$, which are compounds $(Q)$ and $(R)$ respectively.
Question 41
Chemistry · Environmental Chemistry · Single correct
Some reactions of $\mathrm{NO_2}$ relevant to photochemical smog formation are Identify A, B, X and Y.
X = [O], Y = NO, A = O_2, B = O_3
X = N_2O, Y = [O], A = O_3, B = NO
X = $\frac{1}{2}$ $O_2$, Y = $NO_2$, A = $O_3$, B = $O_2$
X = NO, Y = [O], A = O_2, B = N_2O_3
Answer: (a)
Solution
When $\mathrm{NO_2}$ is exposed to sunlight, it decomposes into $[\mathrm{O}]$ and $\mathrm{NO}$. The atomic oxygen $[\mathrm{O}]$ can react with $\mathrm{O_2}$ to form $\mathrm{O_3}$. Therefore, $X$ is $[\mathrm{O}]$, $Y$ is $\mathrm{NO}$, $A$ is $\mathrm{O_2}$, and $B$ is $\mathrm{O_3}$.
Question 42
Chemistry · The Solid State · Single correct
A cubic solid is made up of two elements X and Y. Atoms of X are present on every alternate corner and one at the center of cube. Y is at $\frac{1}{3}^{\text{rd}}$ of the total faces. The empirical formula of the compound is
The osmotic pressure of solutions of PVC in cyclohexanone at $300 \, \mathrm{K}$ are plotted on the graph. The molar mass of PVC is ________ $\mathrm{g \, mol^{-1}}$ (Nearest integer) (Given: $R = 0.083 \, \mathrm{L \, atm \, K^{-1} \, mol^{-1}}$)
Answer: 41500
Solution
Given $\pi = \mathrm{M'}RT = \left( \frac{W/M}{V} \right) RT$. Therefore, $\pi = \left( \frac{W}{V} \right) \left( \frac{1}{M} \right) RT = C \left( \frac{RT}{M} \right)$. Thus, $\frac{\pi}{C} = \frac{RT}{M} \neq f(c)$. If we assume a graph between $\frac{\pi}{C}$ and $C$, Assuming $\pi$ vs $C$ graph, $$Slope = \frac{RT}{M} = \frac{0.083 \times 300}{M} = 6 \times 10^{-4}$$ Therefore, $M = \frac{0.083 \times 300}{6 \times 10^{-4}} = \frac{830 \times 300}{6} = 41,500 \, \mathrm{gm/mole}$
Question 44
Chemistry · Electrochemistry · Fill in the blank
Consider the cell $\mathrm{Pt(s)|H_2(g)(1\,atm)|H^+(aq,[H^+]=1)||Fe^{3+}(aq),Fe^{2+}(aq)|Pt(s)}$ Given: $E^\circ_{\mathrm{Fe^{3+}/Fe^{2+}}}=0.771\,\mathrm{V}$ and $E^\circ_{\mathrm{H^+/\frac{1}{2}H_2}}=0\,\mathrm{V}$, $T=298\,\mathrm{K}$ If the potential of the cell is $0.712\,\mathrm{V}$ the ratio of concentration of $\mathrm{Fe^{2+}}$ to $\mathrm{Fe^{2+}}$ is $\underline{\hspace{3cm}}$ (Nearest integer)
Answer: 10
Solution
The reaction is given by: $$\frac{1}{2} \mathrm{H_2(g)} + \mathrm{Fe^{3+}(aq.)} \rightarrow \mathrm{H^+(aq)} + \mathrm{Fe^{2+}(aq.)}$$ The Nernst equation is: $$E = E^\circ - \frac{0.059}{1} \log \frac{[\mathrm{Fe^{2+}}]}{[\mathrm{Fe^{3+}}]}$$ Substituting the values, we have: $$0.712 = (0.771 - 0) - \frac{0.059}{1} \log \frac{[\mathrm{Fe^{2+}}]}{[\mathrm{Fe^{3+}}]}$$ Rearranging gives: $$\log \frac{[\mathrm{Fe^{2+}}]}{[\mathrm{Fe^{3+}}]} = \frac{(0.771 - 0.712)}{0.059} = 1$$ Therefore: $$\frac{[\mathrm{Fe^{2+}}]}{[\mathrm{Fe^{3+}}]} = 10$$
Question 45
Chemistry · Chemical Kinetics and Nuclear Chemistry · Numerical
For the first order reaction A $\rightarrow$ B Br the half life is 30 $\,$ mm. The time taken for 75$\%$ completion of the reaction is $\,$ mm. (Nearest integer) $\textbf{Given : }$ $\log$ 2 = 0.3010 $\log$ 3 = 0.4771 $\log$ 5 = 0.6989
$CuFeS_2+O_2\xrightarrow{\text{Partial roasting}}Cu_2S+FeO+SO_2+\text{very small }FeS+\text{very small }Cu_2O$ $Cu_2S+O_2\rightarrow Cu_2O+SO_2$ $FeS+O_2\rightarrow FeO+SO_2$ $FeO+SiO_2\rightarrow FeSiO_3$ No formation of calcium silicate $(CaSiO_3)$ in extraction of $Cu$.
Question 47
Chemistry · The p-Block Elements (Group 15, 16, 17 & 18) · Single correct
Reaction of thionyl chloride with white phosphorus forms a compound [A], which on hydrolysis gives [B], a dibasic acid. [A] and [B] are respectively
P_4O_6 and H_3PO_3
PCl_3 and H_3PO_3
PCl_5 and H_3PO_4
POCl_3 and H_3PO_4
Answer: (b)
Solution
The reaction given in [A] is: $$\mathrm{P_4 + 8SOCl_2 \rightarrow 4PCl_3 + 4SO_2 + 2S_2Cl_2}$$ The reaction given in [B] is: $$\mathrm{PCl_3 + 3H_2O \rightarrow H_3PO_3 + 3HCl}$$
Question 48
Chemistry · The s-Block Elements · Single correct
Compound A reacts with $\mathrm{NH_4Cl}$ and forms a compound B. Compound B reacts with $\mathrm{H_2O}$ and excess of $\mathrm{CO_2}$ to form compound C, which on passing through or reacting with saturated $\mathrm{NaCl}$ solution forms sodium hydrogen carbonate. Compound A, B and C are respectively:
Chemistry · Classification of Elements and Periodicity in Properties · Single correct
Inert gases have positive electron gain enthalpy. Its correct order is
Xe < Kr < Ne < He
He < Ne < Kr < Xe
He < Xe < Kr < Ne
He < Kr < Xe < Ne
Answer: (c)
Solution
The table shows the electron gain enthalpy $\Delta_{eg}H$ in $\mathrm{kJ/mol}$ for different elements: Element: He, $\Delta_{eg}H$: $+48$ Element: Ne, $\Delta_{eg}H$: $+116$ Element: Kr, $\Delta_{eg}H$: $+96$ Element: Xe, $\Delta_{eg}H$: $+77$ From NCERT So, order is Ne $>$ Kr $>$ Xe $>$ He
Question 50
Chemistry · Analytical Chemistry · Single correct
Match the List-I with List-II : \begin{tabular}{|l|l|} \hline \textbf{Cations} & \textbf{Group reaction} \\ \hline (P) $\rightarrow$ $Pb^{2+}$, $Cu^{2+}$ & (i) $H_2S$ gas in presence of dilute HCl \\ \hline (Q) $\rightarrow$ $Al^{3+}$, $Fe^{3+}$ & (ii) $(NH_4)_2CO_3$ in presence of $NH_4OH$ \\ \hline (R) $\rightarrow$ $Co^{2+}$, $Ni^{2+}$ & (iii) $NH_4OH$ in presence of $NH_4Cl$ \\ \hline (S) $\rightarrow$ $Ba^{2+}$, $Ca^{2+}$ & (iv) $H_2S$ in presence of $NH_4OH$ \\ \hline \end{tabular} Choose the correct answer from the options given below :-
P→i, Q→iii, R→ii, S→iv
P→iv, Q→ii, R→iii, S→i
P→iii, Q→i, R→iv, S→ii
P→i, Q→iii, R→iv, S→ii
Answer: (d)
Solution
Question 51
Chemistry · Co-ordination Compounds · Numerical
The number of paramagnetic species from the following is_____. $[Ni(CN)_4]^{2-}$, $[Ni(CO)_4]$, $[NiCl_4]^{2-}$ $[Fe(CN)_6]^{4-}$, $[Cu(NH_3)_4]^{2+}$ $[Fe(CN)_6]^{3-}$ and $[Fe(H_2O)_6]^{2+}$
Answer: 4
Solution
For $[\mathrm{Ni(CN)_4}]^{2-}$: $\mathrm{Ni}^{2+} = 3d^8$: diamagnetic. $\mathrm{CN}^-$ is a strong field ligand. For $[\mathrm{Ni(CO)_4}]$: $\mathrm{Ni} = 3d^{10}$: diamagnetic. For $[\mathrm{NiCl_4}]^{2-}$: $\mathrm{Ni}^{2+} = 3d^8$: paramagnetic. $\mathrm{Cl}^-$ is a weak field ligand. For $[\mathrm{Fe(CN)_6}]^{4-}$: $\mathrm{Fe}^{2+} = 3d^6$: diamagnetic. $\mathrm{CN}^-$ is a strong field ligand. For $[\mathrm{Cu(NH_3)_4}]^{2+}$: $\mathrm{Cu}^{2+}$ has one unpaired electron: paramagnetic. For $[\mathrm{Fe(CN)_6}]^{3-}$: $\mathrm{Fe}^{3+} = 3d^5$: paramagnetic. $\mathrm{CN}^-$ is a strong field ligand. For $[\mathrm{Fe(H_2O)_6}]^{2+}$: $\mathrm{Fe}^{2+} = 3d^6$: paramagnetic. $\mathrm{H_2O}$ is a weak field ligand.
Question 52
Chemistry · Structure of Atom · Numerical
How many of the following metal ions have similar value of spin only magnetic moment in gaseous state ? (Given: Atomic number : V, 23 ; Cr, 24 ; Fe, 26 ; Ni, 28) $\mathrm{V}^{3+}, \mathrm{Cr}^{3+}, \mathrm{Fe}^{2+}, \mathrm{Ni}^{3+}$
Answer: 2
Solution
The magnetic moment $\mu_s$ is given by $\mu_s = \sqrt{n(n+2)} BM$, where $n$ is the number of unpaired electrons. For $\mathrm{V^{3+}}$: $[\mathrm{Ar}] \, 3d^2 4s^0$, $n = 2$. For $\mathrm{Cr^{3+}}$: $[\mathrm{Ar}] \, 3d^3 4s^0$, $n = 3$. For $\mathrm{Fe^{2+}}$: $[\mathrm{Ar}] \, 3d^6 4s^0$, $n = 4$. For $\mathrm{Ni^{3+}}$: $[\mathrm{Ar}] \, 3d^7 4s^0$, $n = 3$. $\mathrm{Cr^{3+}}$ and $\mathrm{Ni^{3+}}$ have the same value of $\mu_s$.
Question 53
Chemistry · Alcohols, Phenols and Ethers · Single correct
In the cumene to phenol preparation in presence of air, the intermediate is
Answer: (d)
Solution
Question 54
Chemistry · Aldehydes, Ketones and Carboxylic Acids · Single correct
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R: Assertion A: Acetal/Ketal is stable in basic medium. Reason R: The high leaving tendency of alkoxide ion gives the stability to acetal/ketal in basic medium. In the light of the above statements, choose the correct answer from the options given below:
A is true but R is false
A is false but R is true
Both A and R are true and R is the correct explanation of A
Both A and R are true but R is NOT the correct explanation of A
Answer: (a)
Solution
For Assertion: Acetal and ketals are basically ethers hence they must be stable in basic medium but should break down in acidic medium. Hence assertion is correct. For reason: Alkoxide ion (RO^-) is not considered a good leaving group hence reason must be false.
Question 55
Chemistry · Amines · Single correct
The correct order in aqueous medium of basic strength in case of methyl substituted amines is :
Me_2NH > MeNH_2 > Me_3N > NH_3
Me_2NH > Me_3N > MeNH_2 > NH_3
NH_3 > Me_3N > MeNH_2 > Me_2NH
Me_3N > Me_2NH > MeNH_2 > NH_3
Answer: (a)
Solution
In aqueous medium basic strength is dependent on electron density on nitrogen as well as solvation of cation formed after accepting $\mathrm{H}^+$. After considering all these factors overall basic strength order is $\mathrm{Me_2NH} > \mathrm{MeNH_2} > \mathrm{Me_3N} > \mathrm{NH_3}$.
Question 56
Chemistry · Amines · Single correct
Identify the product formed (A and E)
Answer: (b)
Solution
The reaction sequence starts with the nitration of toluene to form compound (A) with a nitro group. Bromination of (A) gives a bromo derivative. Reduction of the nitro group in (A) using $\mathrm{Sn/HCl}$ leads to the formation of an amine (B). Diazotization of (B) with $\mathrm{NaNO_2/HCl}$ at $273-278 \, \mathrm{K}$ forms the diazonium salt (C). Hydrolysis of (C) results in the formation of compound (D). Oxidation of (D) using $\mathrm{KMnO_4/KOH}$ followed by acidification gives compound (E) with a carboxylic acid group.
Question 57
Chemistry · Chemistry in Everyday Life · Single correct
Which of the following statements is incorrect for antibiotics?
An antibiotic must be a product of metabolism.
An antibiotic is a synthetic substance produced as a structural analogue of naturally occurring antibiotic.
An antibiotic should promote the growth or survival of microorganisms.
An antibiotic should be effective in low concentrations.
Answer: (c)
Solution
An antibiotic should not promote growth or survival of microorganisms. Antibiotics should inhibit growth of microbes.
Question 58
Chemistry · Chemical Kinetics and Nuclear Chemistry · Single correct
The variation of the rate of an enzyme catalyzed reaction with substrate concentration is correctly represented by graph
Answer: (b)
Solution
Fact base.
Question 59
Chemistry · General Principles and Processes of Isolation of Elements · Single correct
Match items of Row I with those of Row II. \textbf{Row II:} (i) $\alpha$-D-($-$) Fructofuranose (ii) $\beta$-D-($-$) Fructofuranose (iii) $\alpha$-D-($-$) Glucopyranose (iv) $\beta$-D-($-$) Glucopyranose Correct Match is:
P \to iv,\quad Q \to iii,\quad R \to i,\quad S \to ii
P \to i,\quad Q \to ii,\quad R \to iii,\quad S \to iv
P \to iii,\quad Q \to iv,\quad R \to ii,\quad S \to i
P \to iii,\quad Q \to iv,\quad R \to i,\quad S \to ii
Answer: (d)
Question 60
Chemistry · Thermodynamics · Numerical
An athlete is given 100 g of glucose ($C_6H_{12}O_6$) for energy. This is equivalent to 1800 kJ of energy. The 50\% of this energy gained is utilized by the athlete for sports activities at the event. In order to avoid storage of energy, the weight of extra water he would need to perspire is ________ g (Nearest integer) Assume that there is no other way of consuming stored energy. Given: The enthalpy of evaporation of water is 45 kJ mol$^{-1}$ Molar mass of C, H \& O are 12, 1 and 16 g mol$^{-1}$.
Answer: 360
Solution
$\mathrm{C_6H_{12}O_6}(s) + 6\mathrm{O_2} \rightarrow 6\mathrm{CO_2}(g) + 6\mathrm{H_2O}(l)$ Extra energy used to convert $\mathrm{H_2O}(l)$ into $\mathrm{H_2O}(g)$ $\displaystyle = \frac{1800}{2} = 900\,\mathrm{kJ}$ $\Rightarrow 900 = n_{\mathrm{H_2O}} \times 45$ $\displaystyle n_{\mathrm{H_2O}} = \frac{900}{45} = 20\ \text{mol}$ $W_{\mathrm{H_2O}} = 20 \times 18 = 360\ \mathrm{g}$
Maths
Question 61
Maths · Complex Numbers and Quadratic Equations · Numerical
Let $$S = \left\{ \alpha : \log_2 \left( 9^{2\alpha - 4} + 13 \right) - \log_2 \left( \frac{5}{2} \cdot 3^{2\alpha - 4} + 1 \right) = 2 \right\}.$$ Then the maximum value of $\beta$ for which the equation $$x^2 - 2 \left( \sum_{\alpha \in S} \alpha \right) x + \sum_{\alpha \in S} (\alpha + 1)^2 \beta = 0$$ has real roots, is ____.
Answer: 25
Solution
Given $\log_2(9^{2\alpha-4} + 13) - \log_2\left(\frac{5}{2} \cdot 3^{2\alpha-4} + 1\right) = 2$. This implies $$\frac{9^{2\alpha-4} + 13}{\frac{5}{2} \cdot 3^{2\alpha-4} + 1} = 4$$ Therefore, $\alpha = 2$ or $3$. Given $\sum_{\alpha \in S} \alpha = 5$ and $\sum_{\alpha \in S} (\alpha + 1)^2 = 25$. This implies $x^2 - 50x + 25\beta = 0$ has real roots. Therefore, $\beta \leq 25$. Thus, $\beta_{\max} = 25$.
Question 62
Maths · Conic Sections · Single correct
Let $( z_1 = 2 + 3i )$ and $( z_2 = 3 + 4i )$. The set $$S = \left\{ z \in \mathbb{C} : |z - z_1|^2 - |z - z_2|^2 = |z_1 - z_2|^2 \right\}$$ represents a
straight line with sum of its intercepts on the coordinate axes equals 14
hyperbola with the length of the transverse axis 7
straight line with the sum of its intercepts on the coordinate axes equals -18
hyperbola with eccentricity 2
Answer: (a)
Solution
Given the equation $$((x-2)^2 + (y-3)^2) - ((x-3)^2 - (y-4)^2) = 1 + 1$$ which simplifies to $$x + y = 7$$
Question 63
Maths · Permutations and Combinations · Numerical
Let x and y be distinct integers where $1 \leq x \leq 25$ and $1 \leq y \leq 25$. Then, the number of ways of choosing x and y, such that $x + y$ is divisible by 5, is _____.
Answer: 120
Solution
Given $x + y = 5\lambda$. Cases: For $x = 5\lambda$, $y = 5\lambda$, the number of ways is 20. For $x = 5\lambda + 1$, $y = 5\lambda + 4$, the number of ways is 25. For $x = 5\lambda + 2$, $y = 5\lambda + 3$, the number of ways is 25. For $x = 5\lambda + 3$, $y = 5\lambda + 2$, the number of ways is 25. For $x = 5\lambda + 4$, $y = 5\lambda + 1$, the number of ways is 25.
Question 64
Maths · Permutations and Combinations · Numerical
Let $S = \{1, 2, 3, 5, 7, 10, 11\}$. The number of non-empty subsets of $S$ that have the sum of all elements a multiple of 3, is ___.
Answer: 43
Solution
Elements of the type $3k = 3$. Elements of the type $3k + 1 = 1, 7, 9$. Elements of the type $3k + 2 = 2, 5, 11$. Subsets containing one element $S_1 = 1$. Subsets containing two elements $$S_2 = {^3C_1} \times {^3C_1} = 9$$ Subsets containing three elements $$S_3 = {^3C_1} \times {^3C_1} + 1 + 1 = 11$$ Subsets containing four elements $$S_4 = {^3C_3} + {^3C_3} + {^3C_2} \times {^3C_2} = 11$$ Subsets containing five elements $$S_5 = {^3C_2} \times {^3C_2} \times 1 = 9$$ Subsets containing six elements $S_6 = 1$. Subsets containing seven elements $S_7 = 1$. Therefore, the sum is $43$.
Question 65
Maths · Sequences and Series · Numerical
Let $A_1, A_2, A_3$ be the three A.P. with the same common difference d and having their first terms as $A, A + 1, A + 2,$ respectively. Let a, b, c be the $7^{th}, 9^{th}, 17^{th}$ terms of $A_1, A_2, A_3,$ respectively such that \[ \left| \begin{array}{ccc} a & 7 & 1\\ 2b & 17 & 1\\ c & 17 & 1 \end{array} \right| +70=0 \] If a = 29, then the sum of first 20 terms of an AP whose first term is c - a - b and common difference is $\frac{d}{12}$, is equal to .
Answer: 495
Solution
Given the equations: $$\begin{vmatrix} A + 6d & 7 & 1 \\ 2(A + 1 + 8d) & 17 & 1 \\ A + 2 + 16d & 17 & 1 \end{vmatrix} + 70 = 0$$ We find that $A = -7$ and $d = 6$. Therefore, $c - a - b = 20$. The sum $S_{20} = 495$.
Question 66
Maths · Binomial Theorem · Single correct
If $a_r$ is the coefficient of $x^{10-r}$ in the Binomial expansion of $(1 + x)^{10}$, then $$\sum_{r=1}^{10} r^3 \left( \frac{a_r}{a_{r-1}} \right)^2$$ is equal to
The constant term in the expansion of $$\left(2x + \frac{1}{x^7} + 3x^2\right)^5$$ is .
Answer: 1080
Solution
General term is $$\sum \frac{5!(2x)^{n_1}(x^{-7})^{n_2}(3x^2)^{n_3}}{n_1! \, n_2! \, n_3!}$$ For constant term, $$n_1 + 2n_3 = 7n_2$$ and $$n_1 + n_2 + n_3 = 5$$ Only possibility $$n_1 = 1, \, n_2 = 1, \, n_3 = 3$$ Therefore, constant term = 1080
Question 68
Maths · Conic Sections · Single correct
The points of intersection of the line $ax + by = 0$, $(a \neq b)$ and the circle $x^2 + y^2 - 2x = 0$ are $A(\alpha, 0)$ and $B(1, \beta)$. The image of the circle with $AB$ as a diameter in the line $x + y + 2 = 0$ is:
$x^2 + y^2 + 5x + 5y + 12 = 0$
$x^2 + y^2 + 3x + 5y + 8 = 0$
$x^2 + y^2 + 3x + 3y + 4 = 0$
$x^2 + y^2 - 5x - 5y + 12 = 0$
Answer: (a)
Solution
Only possibility $\alpha = 0$, $\beta = 1$. Therefore, the equation of the circle $x^2 + y^2 - x - y = 0$. Image of circle in $x + y + 2 = 0$ is $x^2 + y^2 + 5x + 5y + 12 = 0$.
Question 69
Maths · Conic Sections · Single correct
The distance of the point $(6, -2\sqrt{2})$ from the common tangent $y = mx + c$, $m > 0$, of the curves $x = 2y^2$ and $x = 1 + y^2$ is
$\frac{1}{3}$
5
$\frac{14}{3}$
$5\sqrt{3}$
Answer: (b)
Solution
For $y^2 = \frac{x}{2}$, $T: y = mx + \frac{1}{8m}$. For tangent to $y^2 + 1 = x$, $$\Rightarrow \left(mx + \frac{1}{8m}\right)^2 + 1 = x$$ $D = 0 \Rightarrow m = \frac{1}{2\sqrt{2}}$. Therefore, $T: x - 2\sqrt{2}y + 1 = 0$. $$d = \frac{6 + 8 + 1}{\sqrt{9}} = 5$$
Question 70
Maths · Conic Sections · Numerical
The vertices of a hyperbola $H$ are $(\pm 6, 0)$ and its eccentricity is $\frac{\sqrt{5}}{2}$. Let $N$ be the normal to $H$ at a point in the first quadrant and parallel to the line $\sqrt{2}x + y = 2\sqrt{2}$. If $d$ is the length of the line segment of $N$ between $H$ and the $y$-axis then $d^2$ is equal to $\ldots$
The mean and variance of the marks obtained by the students in a test are 10 and 4 respectively. Later, the marks of one of the students is increased from 8 to 12. If the new mean of the marks is 10.2, then their new variance is equal to:
Let $x,y,z > 1$ and $$A = \begin{bmatrix} 1 & \log_x y & \log_x z \\ \log_y x & 2 & \log_y z \\ \log_z x & \log_z y & 3 \end{bmatrix}$$ Then $|adj(adj \, A^2)|$ is equal to
$6^4$
$2^8$
$4^8$
$2^4$
Answer: (b)
Solution
Given $$|A| = \frac{1}{\log x \cdot \log y \cdot \log z} \begin{vmatrix} \log x & \log y & \log z \\ \log x & 2 \log y & \log z \\ \log x & \log y & 3 \log z \end{vmatrix} = 2$$ Therefore, $$|adj(adj A^2)| = |A^2|^4 = 2^8$$
Question 75
Maths · Determinants · Single correct
Let $S_1$ and $S_2$ be respectively the sets of all $a \in \mathbb{R} - \{0\}$ for which the system of linear equations $$ax + 2ay - 3az = 1$$ $$(2a + 1)x + (2a + 3)y + (a + 1)z = 2$$ $$(3a + 5)x + (a + 5)y + (a + 2)z = 3$$ has unique solution and infinitely many solutions. Then
$n(S_1) = 2$ and $S_2$ is an infinite set
$S_1$ is an infinite set and $n(S_2) = 2$
$S_1 = \Phi$ and $S_2 = \mathbb{R} - \{0\}$
$S_1 = \mathbb{R} - \{0\}$ and $S_2 = \Phi$
Answer: (d)
Solution
Given the determinant $$\Delta = \begin{vmatrix} a & 2a & -3a \\ 2a+1 & 2a+3 & a+1 \\ 3a+5 & a+5 & a+2 \end{vmatrix}$$ we have $$= a(15a^2 + 31a + 36) = 0 \Rightarrow a = 0$$ $$\Delta \neq 0 for all a \in \mathbb{R} - \{0\}$$ Hence, $$S_1 = \mathbb{R} - \{0\}$$ and $$S_2 = \emptyset$$
If the sum of all the solutions of $\tan^{-1}\left(\frac{2x}{1-x^2}\right) + \cot^{-1}\left(\frac{1-x^2}{2x}\right) = \frac{\pi}{3}$, $-1 < x < 1, x \neq 0$, is $\alpha - \frac{4}{\sqrt{3}}$, then $\alpha$ is equal to
Maths · Applications of Derivatives · Single correct
Let $f : (0,1) \rightarrow \mathbb{R}$ be a function defined by $f(x) = \frac{1}{1-e^{-x}}$, and $g(x) = (f(-x) - f(x))$. Consider two statements (I) $g$ is an increasing function in $(0, 1)$ (II) $g$ is one-one in $(0, 1)$ Then,
Only (I) is true
Only (II) is true
Neither (I) nor (II) is true
Both (I) and (II) are true
Answer: (d)
Solution
Given $g(x) = f(-x) - f(x) = \frac{1 + e^x}{1 - e^x}$. Therefore, $g'(x) = \frac{2e^x}{(1 - e^x)^2} > 0$. Thus, $g$ is increasing in $(0, 1)$. Therefore, $g$ is one-one in $(0, 1)$.
Question 78
Maths · Relations and Functions · Numerical
For some $a, b, c \in \mathbb{N}$, let $f(x) = ax - 3$ and $g(x) = x^b + c, x \in \mathbb{R}$. If $(f \circ g)^{-1}(x) = \left( \frac{x - 7}{2} \right)^{1/3}$ then $(f \circ g)(ac) + (g \circ f)(b)$ is equal to ____.
Maths · Continuity and Differentiability · Single correct
Let $$y(x) = (1+x)(1+x^2)(1+x^4)(1+x^8)(1+x^{16})$$ Then $y' - y''$ at $x = -1$ is equal to
976
464
496
944
Answer: (c)
Solution
Given $$y = \frac{1-x^{32}}{1-x} \implies y - xy = 1 - x^{32}$$ Differentiating, we have $$y' - xy' - y = -32x^{31}$$ Differentiating again, $$y'' - xy'' - y' - y' = -(32)(31)x^{30}$$ At $x = -1$, $$y' - y'' = 496$$
Question 80
Maths · Applications of Derivatives · Single correct
Let x = 2 be a local minima of the function f(x) = 2$x^4$ - 18$x^2$ + 8x + 12, x $\in$ (-4, 4). If M is local maximum value of the function f in (-4, 4), then M =
The minimum value of the function $$f(x) = \int_{0}^{2} e^{|x-t|} \, dt$$ is
2(e-1)
2e-1
2
e(e-1)
Answer: (a)
Solution
For $x \leq 0$ $$f(x) = \int_0^2 e^{t-x} \, dt = e^{-x} \left( e^2 - 1 \right)$$ For $0 < x < 2$ $$f(x) = \int_0^x e^{x-t} \, dt + \int_x^2 e^{t-x} \, dt = e^x + e^{2-x} - 2$$ For $x \geq 2$ $$f(x) = \int_0^2 e^{x-t} \, dt = e^{x-2} \left( e^2 - 1 \right)$$ For $x \leq 0$, $f(x)$ is decreasing and $x \geq 2$, $f(x)$ is increasing. Therefore, the minimum value of $f(x)$ lies in $x \in (0, 2)$. Applying A.M $\geq$ G.M, minimum value of $f(x)$ is $2(e - 1)$
Question 83
Maths · Applications of Integrals · Numerical
If the area enclosed by the parabolas $P_1: 2y = 5x^2$ and $P_2: x^2 - y + 6 = 0$ is equal to the area enclosed by $P_1$ and $y = \alpha x, \alpha > 0$, then $\alpha^3$ is equal to _____.
Answer: 600
Solution
Abscissa of point of intersection of $2y = 5x^2$ and $y = x^2 + 6$ is $\pm 2$. $$Area = 2 \int_0^2 \left( x^2 + 6 - \frac{5x^2}{2} \right) \, dx = \int_0^{\frac{2\alpha}{5}} \left( \alpha x - \frac{5x^2}{2} \right) \, dx$$ $$\Rightarrow \int_0^{\frac{2\alpha}{5}} \left( \alpha x - \frac{5x^2}{2} \right) \, dx = 16$$ $$\Rightarrow \alpha^3 = 600$$
Question 84
Maths · Differential Equations · Single correct
Let $y = y(x)$ be the solution curve of the differential equation $\frac{dy}{dx} = \frac{y}{x} (1 + xy^2 (1 + \log_e x))$, $x > 0, y(1) = 3$. Then $\frac{y^2(x)}{9}$ is equal to:
Let $\vec{a}$, $\vec{b}$ and $\vec{c}$ be three non zero vectors such that $\vec{b} \cdot \vec{c} = 0$ and $\vec{a} \times (\vec{b} \times \vec{c}) = \frac{\vec{b} - \vec{c}}{2}$. If $\vec{d}$ be a vector such that $\vec{b} \cdot \vec{d} = \vec{a} \cdot \vec{b}$, then $(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d})$ is equal to
The vector $\vec{a} = -\hat{i} + 2\hat{j} + \hat{k}$ is rotated through a right angle, passing through the $y$-axis in its way and the resulting vector is $\vec{b}$. Then the projection of $3\vec{a} + \sqrt{2}\vec{b}$ on $\vec{c} = 5\hat{i} + 4\hat{j} + 3\hat{k}$ is
$3\sqrt{2}$
$1$
$\sqrt{6}$
$2\sqrt{3}$
Answer: (a)
Solution
Given $\mathbf{b} = \lambda \mathbf{a} \times (\mathbf{a} \times \mathbf{j})$. Therefore, $\mathbf{b} = \lambda (-2\hat{\imath} - 2\hat{\jmath} + 2\hat{k})$. The magnitude $|\mathbf{b}| = |\mathbf{a}|$. Thus, $\sqrt{6} = \sqrt{12} |\lambda| \Rightarrow \lambda = \pm \frac{1}{\sqrt{2}}$. The value $\lambda = \frac{1}{\sqrt{2}}$ is rejected because $\mathbf{b}$ makes an acute angle with the $y$ axis. Therefore, $\mathbf{b} = -\sqrt{2} (-\hat{\imath} - \hat{\jmath} + \hat{k})$. Finally, $\($ $\frac{(3\mathbf{a} + \sqrt{2} \mathbf{b}) \cdot \mathbf{c}}{|\mathbf{c}|}$ = 3$\sqrt{2}$ $\)$.
Question 87
Maths · Three Dimensional Geometry · Single correct
The distance of the point P(4, 6, -2) from the line passing through the point (-3, 2, 3) and parallel to a line with direction ratios 3, 3, -1 is equal to:
3
$\sqrt{6}$
$2\sqrt{3}$
$\sqrt{14}$
Answer: (d)
Solution
Equation of line is $\frac{x+3}{3} = \frac{y-2}{3} = \frac{z-3}{-1} = \lambda$. M is $(3\lambda - 3, 3\lambda + 2, 3 - \lambda)$. D.R of PM is $(3\lambda - 7, 3\lambda - 4, 5 - \lambda)$. Since PM is perpendicular to line, $3(3\lambda - 7) + 3(3\lambda - 4) - 1(5 - \lambda) = 0$. Solving gives $\lambda = 2$. Therefore, M is $(3, 8, 1)$ and PM is $\sqrt{14}$.
Question 88
Maths · Three Dimensional Geometry · Single correct
Consider the lines $L_1$ and $L_2$ given by $$L_1: \frac{x-1}{2} = \frac{y-3}{1} = \frac{z-2}{2}$$ $$L_2: \frac{x-2}{1} = \frac{y-2}{2} = \frac{z-3}{3}$$ A line $L_3$ having direction ratios $1, -1, -2$, intersects $L_1$ and $L_2$ at the points $P$ and $Q$ respectively. Then the length of line segment $PQ$ is
Let the equation of the plane passing through the line $x - 2y - z - 5 = 0 = x + y + 3z - 5$ and parallel to the line $x + y + 2z - 7 = 0 = 2x + 3y + z - 2$ be $ax + by + cz = 65$. Then the distance of the point $(a, b, c)$ from the plane $2x + 2y - z + 16 = 0$ is ____.
Answer: 9
Solution
Equation of plane is (x - 2y - z - 5) + b(x + y + 3z - 5) = 0$$ \begin{vmatrix} 1 + b & -2 + b & -1 + 3b \\ 1 & 1 & 2 \\ 2 & 3 & 1 \end{vmatrix} = 0 $$ Therefore, b = 12 . Thus, the plane is 13x + 10y + 35z = 65 .
Question 90
Maths · Probability · Single correct
Let M be the maximum value of the product of two positive integers when their sum is 66. Let the sample space $S = \left\{ x \in \mathbb{Z} : x(66-x) \geq \frac{5}{9}M \right\}$ and the event $A=\{x\in S:x\text{ is a multiple of }3\}$. Then $P(A)$ is equal to
$\frac{15}{44}$
$\frac{1}{3}$
$\frac{1}{5}$
$\frac{7}{22}$
Answer: (b)
Solution
Given $M = 33 \times 33$. We have $x(66 - x) \geq \frac{5}{9} \times 33 \times 33$. The range is $11 \leq x \leq 55$. Set $A : \{12, 15, 18, \ldots, 54\}$. The probability is $\mathrm{P}(A) = \frac{15}{45} = \frac{1}{3}$.