JEE Main Question Paper with Solution 2023 January 29th Shift 1 - Morning

JEE Main Physics Question Paper with Solution 2023 January 29th Shift 1 - Morning

Q1. Ratio of thermal energy released in two resistors $R$ and $3 R$ connected in parallel in an electric circuit is :

A

$1: 3$

B

$3: 1$

C

$1: 1$

D

$1: 27$

Solution

$ H =\frac{ V ^2}{ R } \times t$
$ \frac{ H _1}{ H _2}=\frac{\frac{ V ^2 t }{ R }}{\frac{ V ^2 t }{3 R }}=3: 1$

Q2. 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 :

A

$\sqrt{\frac{G m}{r}}$

B

$\sqrt{\frac{G m}{4 r}}$

C

$\sqrt{\frac{4 G m}{r}}$

D

$\sqrt{\frac{G m}{2 r}}$

Q3. 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:

A

$\frac{2 \sqrt{3}+1}{2}$

B

$\frac{2 \sqrt{3}-1}{2}$

C

$\frac{\sqrt{3}}{2}$

D

$\frac{1}{2 \sqrt{3}}$

Solution

$Mg \sin 30^{\circ}-\mu mg \cos 30^{\circ}= ma$
$\frac{g}{2}-\frac{\sqrt{3}}{2} \cdot \mu g=\frac{g}{4}$
image

Q4. 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 -

A

$4: 1$

B

$1: 2$

C

$4: 3$

D

$1: 4$

Solution

$\frac{ KE _{ POP }}{ KE _{\text {top }}}=\frac{\frac{1}{2} M ( u )^2}{\frac{1}{2} M \left( u \cos 30^{\circ}\right)^2}=\frac{4}{3}$

Q5. 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

A

$\frac{1}{8}$

B

$\frac{1}{2}$

C

$\frac{1}{4}$

D

$\frac{1}{16}$

Solution

$\frac{ N }{ N _0}=\left(\frac{1}{2}\right)^{ t /t \frac{1}{2}}=\left(\frac{1}{2}\right)^{\frac{90}{30}}$
$\frac{ N }{ N _0}=\left(\frac{1}{2}\right)^3=\frac{1}{8}$

Q6. In a Young's double slit experiment, two slits are illuminated with a light of wavelength $800\, nm$. The line joining $A_1 P$ is perpendicular to $A_1 A_2$ as shown in the figure. If the first minimum is detected at $P$, the value of slits separation '$a$' will be:
image
The distance of screen from slits $D =5\, cm$

A

$0.4 \,mm$

B

$0.2\, mm$

C

$0.5\, mm$

D

$0.1\, mm$

Solution

$ A _2 P - A _1 P =\frac{\lambda}{2} $ (Condition of minima)
$ \sqrt{ D ^2+ a ^2}- D =\frac{\lambda}{2} $
$ D \left(1+\frac{ a ^2}{ D ^2}\right)^{1 / 2}- D =\frac{\lambda}{2} $
$ D \left(1+\frac{1}{2} \times \frac{ a ^2}{ D ^2}\right)- D =\frac{\lambda}{2} $
$\frac{ a ^2}{2 D }=\frac{\lambda}{2} \Rightarrow a =\sqrt{\lambda \cdot D } $
$ =\sqrt{800 \times 10^{-6} \times 50} $
$ a =0.2\, mm$

Q7. Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: If $d Q$ and $d W$ represent the heat supplied to the system and the work done on the system respectively. Then according to the first law of thermodynamics $d Q=d U-d W$.
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 options given below:

A

Both $A$ and $R$ are correct but $R$ is not the correct explanation of $A$

B

A is correct but $R$ is not correct

C

A is not correct but $R$ is correct

D

Both $A$ and $R$ are correct and $R$ is the correct explanation of $A$

Solution

First law of thermodynamics is based on law of conservation of energy and it can be written as $dQ = dU - dW$.
where $dW$ is work done on the system

Q8. 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 is 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 options given below:

A

B and D only

B

C and D only

C

A and Conly

D

B and C only

Solution

Speed of light does not depend on the motion of source as well as intensity.

Q9. If the height of transmitting and receiving antennas are $80\, m$ each, the maximum line of sight distance will be:
Given: Earth's radius $=6.4 \times 10^6 \, m$

A

$36\, km$

B

$32\, km$

C

$28\, km$

D

$64 \, km$

Solution

Maximum line of sight distance between two antennas, $d _{ M }=\sqrt{2 Rh _{ T }}+\sqrt{2 R \cdot h _{ R }}$
$d _{ M }=2 \times \sqrt{2 \times 6.4 \times 10^6 \times 80}=64 \,km$

Q10. In a cuboid of dimension $2 L \times 2 L \times L$, a charge $q$ is placed at the center of the surface ' $S$ ' having area of $4 L ^2$. The flux through the opposite surface to ' $S$ ' is given by

A

$\frac{q}{12 \epsilon_0}$

B

$\frac{q}{2 \epsilon_0}$

C

$\frac{q}{3 \epsilon_0}$

D

$\frac{q}{6 \in_0}$

Solution

image
Flux passing through shaded face $=\frac{ q }{6 \varepsilon_0}$

Q11. Which one of the following statement is not correct in the case of light emitting diodes?
A. It is a heavily doped $p - n$ junction.
B. It emits light only when it is forward biased.
C. It emits light only when it is reverse biased.
D. The energy of the light emitted is equal to or slightly less than the energy gap of the semiconductor used.
Choose the correct answer from the options given below:

A

B

B

C

C

A

D

C and D

Solution

LED works in forward biasing and light energy maybe slightly less or equal to band gap.

Q12. The magnitude of magnetic induction at mid point $O$ due to current arrangement as shown in Fig will be

A

0

B

$\frac{\mu_0 I}{\pi a}$

C

$\frac{\mu_0 I}{2 \pi a}$

D

$\frac{\mu_0 I}{4 \pi a}$

Solution

Magnetic field due to current in BC and ET are outward at point ' $O$ '
$B _0=\frac{\mu_0 i }{4 \pi r }+\frac{\mu_0 i }{4 \pi r }=\frac{\mu_0 i }{2 \pi r }=\frac{\mu_0 i }{\pi a }$

Q13. A person observes two moving trains, '$A$' reaching the station and ' $B$ ' leaving the station with equal speed of $30 \,m / s$. If both trains emit sounds with frequency $300 \, Hz$, (Speed of sound: $330 \,m / s$ ) approximate difference of frequencies heard by the person will be:

A

$33 \,Hz$

B

$80 \, Hz$

C

$10 \, Hz$

D

$55 \, Hz$

Solution

$ f _1=300\left(\frac{330-0}{330-(-30)}\right)=275$
$ f _2=300\left(\frac{330-0}{330-(30)}\right)=330$
$ \Delta f =330-275=55\, Hz $

Q14. Surface tension of a soap bubble is $2.0 \times 10^{-2} Nm ^{-1}$. Work done to increase the radius of soap bubble from $3.5\, cm$ to $7\, cm$ will be: Take $\left[\pi=\frac{22}{7}\right]$

A

$9.24 \times 10^{-4} J$

B

$18.48 \times 10^{-4} J$

C

$0.72 \times 10^{-4} J$

D

$5.76 \times 10^{-4} J$

Solution

Surface area of soap bubble $=2 \times 4 \pi R ^2$
Work done $=$ change in surface energy $\times T _{ S }$
$= T _{ S } \times 8 \pi \times\left( R _2^2- R _1^2\right) $
$=2 \times 10^{-2} \times 8 \times \frac{22}{7} \times 49 \times \frac{3}{4} \times 10^{-4} $
$=18.48 \times 10^{-4} J$

Q15. A single current carrying loop of wire carrying current I flowing in anticlockwise direction seen from + ve $z$ direction and lying in $x y$ plane is shown in figure. The plot of $\hat{j}$ component of magnetic field (By) at a distance ' $a$ ' (less than radius of the coil) and on $y z$ plane vs $z$ coordinate looks like

A

B

C

D

Solution

image
$B _{ y }=0$ in plane of coil
$B_y$ is opposite of each other in $- z$ and $+ z$ positions.

Q16. 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 $\left.g =10\, ms ^{-2}\right]$

A

$3.4\, ms ^{-1}$

B

$13\, ms ^{-1}$

C

$17\, ms ^{-1}$

D

$22.4\, ms ^{-1}$

Solution

$f _{ s }=\frac{ mv^{2 } }{ r }$
For maximum speed in safe turning,
$ f _{ s }= f _{ s } \max =\mu mg $
$ v _{\max } \text { (for safe turning }=\sqrt{\mu rg } $
$=\sqrt{0.34 \times 50 \times 10} \approx 13\, m / s$

Q17. 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 $L$ $(L \gg R)$. The loops are coplanar and their centres coincide:

A

$M=\frac{2 \sqrt{2} \mu_0 R^2}{L}$

B

$M=\frac{\sqrt{2} \mu_0 R^2}{L}$

C

$M=\frac{2 \sqrt{2} \mu_0 R}{L^2}$

D

$M=\frac{\sqrt{2} \mu_0 R}{L^2}$

Solution

$ \phi= Mi $
$ \phi=( BA )$
$ \phi=\pi R ^2\left(4 \frac{\mu_0}{4 \pi} \frac{ i }{\left(\frac{ L }{2}\right)} \sqrt{2}\right)$
$ \Rightarrow M =\frac{2 \sqrt{2} \mu_0 R ^2}{ L }$

Q18. The threshold wavelength for photoelectric emission from a material is $5500 \mathring{A}$. Photoelectrons will be emitted, when this material is illuminated with monochromatic radiation from a
A. $75\, W$ infra-red lamp
B. $10 \, W$ infra-red lamp
C. $75\, W$ ultra-violet lamp
D. $10\, W$ ultra-violet lamp
Choose the correct answer from the options given below:

A

C only

B

B and Conly

C

C and D only

D

A and D only

Solution

$ \lambda < 5500 \mathring A $ for photoelectric emission
$\lambda_{ uv } < 5500 \mathring A$

Q19. Match List I with List II:
List I (Physical Quantity) List II (Dimensional Formula)
A Pressure gradient I $\left[ M ^{\circ} L ^2 T ^{-2}\right]$
B Energy density II $\left[ M ^1 L ^{-1} T ^{-2}\right]$
C Electric Field III $\left[ M ^1 L ^{-2} T ^{-2}\right]$
D Latent heat IV $\left[ M ^1 L ^1 T ^{-3} A ^{-1}\right]$
Choose the correct answer from the options given below:

A

A-II, B-III, C-I, D-IV

B

A-III, B-II, C-IV, D-I

C

A-II, B-III, C-IV, D-I

D

A-III, B-II, C-I, D-IV

Solution

Pressure gradient $=\frac{ dp }{ dx }=\frac{\left[ ML ^{-1} T ^{-2}\right]}{[ L ]} $
$ =\left[ M ^1 L ^{-2} T ^{-2}\right] $
Energy density $=\frac{\text { energy }}{\text { volume }}=\frac{\left[ ML ^2 T ^{-2}\right]}{\left[ L ^3\right]} $
$ =\left[ M ^1 L ^{-1} T ^{-2}\right] $
Electric field $=\frac{\text { Force }}{\text { ch arge }}=\frac{\left[ MLT ^{-2}\right]}{[\text { A.T }]} $
$ =\left[ M ^1 L ^1 T ^{-3} A ^{-1}\right] $
Latent heat $=\frac{\text { heat }}{\text { mass }}=\frac{\left[ ML ^2 T ^{-2}\right]}{[ M ]} $
$ =\left[ M ^0 L ^2 T ^{-2}\right] $

Q20. A bicycle tyre is filled with air having pressure of $270 \,k\,Pa$ at $27^{\circ} C$. The approximate pressure of the air in the tyre when the temperature increases to $36^{\circ} C$ is

A

$262\, kPa$

B

$270 \,kPa$

C

$360\, kPa$

D

$278\, kPa$

Solution

Taking volume constant: $\frac{P_1}{T_1}=\frac{P_2}{T_2}$
$ \Rightarrow P _2=\frac{ P _1}{ T _1} \times T _2=\frac{270 \times(309)}{300} $
$ =278\, kPa$

Q21. A body cools from $60^{\circ} C$ to $40^{\circ} C$ in $6$ minutes. If, temperature of surroundings is $10^{\circ} C$. Then, after the next $6$ minutes, its temperature will be ___${ }^{\circ} C$.

Answer: 28

Solution

By average form of Newton's law of cooling
$ \frac{20}{6}=k(50-10) .....$(i)
$ \frac{40- T }{6}= K \left(\frac{40+ T }{2}-10\right)....$(ii)
From equation (i) and (ii)
$\frac{20}{40- T }=\frac{40}{10+ T / 2} $
$10+\frac{ T }{2}=80-2 T$
$ \frac{5 T }{2}=70 \Rightarrow T =28^{\circ} C$

Q22. A tennis ball is dropped on to the floor from a height of $9.8\, m$. It rebounds to a height $5.0\, m$. Ball comes in contact with the floor for $0.2 \,s$. The average acceleration during contact is ___$ms ^{-2}$. $\left(\right.$ Given $g =10\, ms ^{-2}$ )

Answer: 120

Solution

$v _{ i }=\sqrt{2 gh _{ i }} $
$ =\sqrt{2 \times 10 \times 9.8} \downarrow $
$ =14\, m / s \downarrow$
$ v _{ f }=\sqrt{2 gh _{ f }} $
$ =\sqrt{2 \times 10 \times 5} \uparrow$
$ = 1 0 \,m / s \uparrow$
$\left|\vec{ a }_{ avg }\right|=\left|\frac{\Delta \vec{ v }}{\Delta t }\right|=\frac{24}{0.2}=120 \,m / s ^2$

Q23. A $0.4\, kg$ mass takes $8 \,s$ 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 ___$J$.
$\left(\right.$ Take $\left.g =10 \,m / s ^2\right)$

Answer: 300

Solution

Displacement is $8^{\text {th }}$ 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$

Q24. A certain elastic conducting material is stretched into a circular loop. It is placed with its plane perpendicular to a uniform magnetic field $B =0.8 T$. When released the radius of the loop starts shrinking at a constant rate of $2 \,cm ^{-1}$. The induced emf in the loop at an instant when the radius of the loop is $10\, cm$ will be___ $mV$.

Answer: 10

Solution

$ \text { 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= 1 0 . 0 6 \text { [round off } 1 0 . 0 6 = 1 0 \text { ] }$

Q25. As shown in the figure, 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 $256 \frac{ W }{ m ^2}$. The intensity of light at point $O$ is ____ $\frac{W}{m^2}$

Answer: 24

Solution

By first polaroid $P_1$ intensity will be halved then $P_2$ and $P_3$ will make intensity $\cos ^2\left(60^{\circ}\right)$ and $\cos ^2\left(30^{\circ}\right)$ times respectively.
Intensity out $=\frac{256}{2} \times \frac{1}{4} \times\left(\frac{\sqrt{3}}{2}\right)^2=\frac{256 \times 3}{2 \times 4 \times 4}= 2 4$

Q26. A point charge $q_1=4 q_0$ is placed at origin. Another point charge $q_2=-q_0$ is placed at $x=12 \,cm$. Charge of proton is $q_0$. The proton is placed on $x$ axis so that the electrostatic force on the proton is zero. In this situation, the position of the proton from the origin is ____ $cm$.

Answer: 24

Solution

image
$\frac{q_0}{x^2}=\frac{4 q_0}{(x+12)^2} $
$ x+12=2 x$
$ x=12$
Distance from origin $=x+12=24\, cm$.

Q27. A solid sphere of mass $2 \,kg$ is making pure rolling on a horizontal surface with kinetic energy $2240\, J$. The velocity of centre of mass of the sphere will be ____$ms ^{-1}$.

Answer: 40

Solution

$ KE =\frac{1}{2} mv ^2+\frac{1}{2} I \omega^2 $
$ 2240=\frac{1}{2} 2( v )^2+\frac{1}{2} \frac{2}{5}(2) R ^2 \cdot\left(\frac{ v }{ R }\right)^2 $
$2240= v ^2+\frac{2}{5} v ^2$
$ \Rightarrow v =40 \,m / s $

Q28. In a metre bridge experiment the balance point is 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 $22.5 \,cm$. The value of $X$ is _______

Answer: 2

Solution

$ \frac{2}{\left(\frac{3 x}{3+x}\right)}=\frac{40+22.5}{60-22.5}=\frac{62.5}{37.5}=\frac{5}{3} $
$ \frac{6}{5}=\frac{3 x}{3+x} $
$ 6+2 x=5 x \Rightarrow x=2$

Q29. 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

$ 2 A \cos \left(\frac{\Delta \phi}{2}\right)= A$
$ \cos \left(\frac{\Delta \phi}{2}\right)=\frac{1}{2} $
$ \frac{\Delta \phi}{2}=60^{\circ}$

Q30. A radioactive element ${ }_{92}^{222} X$ emits two $\alpha$-particles, one electron and two positrons. The product nucleus is represented by ${ }_p^{2 34} Y$. The value of $P$ is ___

Answer: 87

Solution

$ P=92-2-2+1-1-1 $
$ P=92-5 $
$ P = 8 7$

JEE Main Chemistry Question Paper with Solution 2023 January 29th Shift 1 - Morning

Q1. The increasing order of $pK _{ a }$ for the following phenols is
(A) 2,4-Dinitrophenol
(B) 4- Nitrophenol
(C) 2,4,5-Trimethylphenol
(D) Phenol
(E) 3-Chlorophenol
Choose the correct answer from the option given below:

A

(A), (E), (B), (D), (C)

B

(C), (E), (D), (B), (A)

C

(C), (D), (E), (B), (A)

D

(A), (B), (E), (D), (C)

Solution

Order of acidity for following phenol is
image
- $M$ and $- I$ increases acidity
$+ M$ and $+ I$ decreases acidity

Q2. During the borax bead test with $CuSO _4$, a blue green colour of the bead was observed in oxidising flame due to the formation of

A

$Cu_3B_2$

B

$Cu \left( BO _2\right)_2$

C

$Cu$

D

$CuO$

Solution

Blue green colour is due to formation of $Cu \left( BO _2\right)_2$
$ CuSO _4 \xrightarrow{\Delta} CuO + SO _3 $
$CuO + B _2 O _3 \rightarrow Cu \left( BO _2\right)_2$

Q3. The reaction representing the Mond process for metal refining is

A

$Ni +4 CO \stackrel{\Delta}{\longrightarrow} Ni ( CO )_4$

B

$2 K \left[ Au ( CN )_2\right]+ Zn \stackrel{\Delta}{\longrightarrow} K _2\left[ Zn ( CN )_4\right]+2 Au$

C

$Zr +2 I _2 \stackrel{\Delta}{\longrightarrow} Zr I _4$

D

$ZnO + C \stackrel{\Delta}{\longrightarrow} Zn + CO$

Solution

Mond's process uses:
$Ni +4 CO \rightarrow\left[ Ni ( CO )_4\right]$

Q4. Match List I with List II.
List I Antimicrobials List II Names
A Narrow Spectrum Antibiotic 1 Furacin
B Antiseptic 2 Sulphur dioxide
C Disinfectants 3 Penicillin G
D Broad spectrum antibiotic 4 Chloramphenicol
Choose the correct answer from the options given below:

A

(A) - III, (B) - I, (C) - II, (D) - IV

B

(A) - II, (B) - I, (C) - IV, (D) - III

C

(A) - III, (B) - I, (C) - IV, (D) - II

D

(A) - I, (B) - II, (C) - IV, (D) - III

Solution

(A) Narrow spectrum antibiotic - penicillin-G
(B) Antiseptic - Furacine
(C) Disinfectants - sulphur dioxide
(D) Broad spectrum antisiotics - chloramphenicol

Q5. The bond dissociation energy is highest for

A

$F_2$

B

$Cl_2$

C

$I_2$

D

$Br_2$

Solution

Bond energy of $F _2$ less than $Cl _2$ due to lone pair lone pair repulsions. Bond energy order $Cl _2> Br _2> F _2> I _2$

Q6. Correct statement about smog is:

A

Both $NO _2$ and $SO _2$ are present in classical smog

B

Classical smog also has high concentration of oxidizing agents

C

$NO _2$ is present in classical smog

D

Photochemical smog has high concentration of oxidizing agents

Solution

Photochemical smog has high concentration of oxidising agents
$NO _2$ is produced from $NO$ and $O _3$ in the presence of sunlight
Classical smog contain smoke, fog and $SO _2$ and it is known as reducing smog, as chemically it is reducing mixture

Q7. Identify the correct order for the given property for following compounds. image Choose the correct answer from the option given below:

A

(B), (C) and (D) only

B

( A ),( C ) and (E) only

C

(A), (C) and (D) only

D

(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

Q8. 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
image
What is the value of compressibility factor at point $A$ ?

A

$1-\frac{b}{v}$

B

$1-\frac{a}{R T V}$

C

$1+\frac{b}{V}$

D

$1+\frac{a}{R T V}$

Solution

For 1 mole of real gas
$PV = ZRT$
from graph PV for real gas is less than PV for ideal gas at point $A$
$ Z < 1 $
$Z=1-\frac{a}{ V _{ m } RT }$

Q9. Which of the given compounds can enhance the efficiency of hydrogen storage tank?

A

$Li / P _4$

B

Di-isobutylaluminium hydride

C

$SiH _4$

D

$NaNi _5$

Q10. Number of cyclic tripeptides formed with 2 amino acids $A$ and $B$ is:

A

2

B

4

C

3

D

5

Q11. The magnetic behavior of $Li _2 O , Na _2 O _2$ and $KO _2$, respectively, are

A

diamagnetic, paramagnetic and diamagnetic

B

diamagnetic, diamagnetic and paramagnetic

C

paramagnetic, paramagnetic and diamagnetic

D

paramagnetic, diamagnetic and paramagnetic

Solution

$Li _2 O \rightarrow O ^{2-} \rightarrow$ diamagnetic
$Na _2 O _2 \rightarrow O _2{ }^{2-} \rightarrow$ diamagnetic
$KO _2 \rightarrow O _2^{-} \rightarrow$ paramagnetic

Q12. The major product ' $P$ ' for the following sequence of reactions is:

A

B

C

D

Q13. Chiral complex from the following is: Here en = ethylene diamine

A

$\operatorname{cis}-\left[ PtCl _2\left( NH _3\right)_2\right]$

B

trans $-[ PtCl 2( en ) 2]^{2+}$

C

trans - $\left[ Co \left( NH _3\right)_4 Cl _2\right]^{+}$

D

$cis -\left[ PtCl _2( en )_2\right]^{2+}$

Q14. The correct order of hydration enthalpies is
(A) $K ^{+}$
(B) $Rb ^{+}$
(C) $Mg ^{2+}$
(D) $Cs ^{+}$
(E) $Ca ^{2+}$
Choose the correct answer from the options given below:

A

$C > E > A > B > D$

B

$C > E > A > D > B$

C

$C > A > E > B > D$

D

$E > C > A > B > D$

Solution

Hydration enthalpies:
(i) $K ^{\prime}> Rb ^{\prime}> Cs ^{\prime}$ : (A) $>$ (B) $>$ (D)
(ii) $Mg ^{+2}> Ca ^{+2} : (C) > (E) $
Option (D)
$( C )>( E )>( A )>( B )>( D )$

Q15. Match List I with List II.
List I Reaction List II Reagents
A Hoffmann Degradation I Cone.KOH, $\Delta$
B Clemenson reduction II $CHCl _3 NaOH / H _3 O ^+$
C Cannizaro reaction III $Br_2 , NaOH$
D Reimer-Tiemann Reaction IV $Zn \cdot Hg / HCl$
Choose the correct answer from the options given below:

A

(A)−II,(B)−IV,(C)−I,(D)−I

B

(A) −III,(B)−IV,(C)−I,(D)−II

C

(A)−II,(B)−IV,(C)−I,(D)−III

D

(A) −II,(B)−I,(C)−III,(D)−IV

Solution

Reactions Reagent used
A Hoffmann Degradation $Br_2 , NaOH$
B Clemenson reduction $Zn - Hg / HCl$
C Cannizaro reaction Conc.KOH, $\Delta$
D Reimer-Tiemann Reaction $CHCl _3, NaOH / H _3 O ^+$

Q16. "$A$" obtained by Ostwald's method involving air oxidation of $NH _3$, 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. Identify $A$ and $B$, respectively.

A

$NO , NO _2$

B

$NO _2, N _2 O _4$

C

$N _2 O _3, NO _2$

D

$NO _2, N _2 O _5$

Solution

$4 NH _3+5 O _2 \xrightarrow{\Delta} \underset{(A)}{4 NO} +6 H _2 O$
$2 NO + O _2 \longrightarrow \underset{(B)}{2 NO _2}$

Q17. The shortest wavelength of hydrogen atom in Lyman series is $\lambda$. The longest wavelength in Balmer series of $He ^{+}$is

A

$\frac{9 \lambda}{5}$

B

$\frac{36 \lambda}{5}$

C

$\frac{5 \lambda}{9}$

D

$\frac{5}{9 \lambda}$

Solution

For $H : \frac{1}{\lambda}= R _{ H } \times 1^2\left(\frac{1}{1^2}-\frac{1}{\infty^2}\right) .....$(1)
$\frac{1}{\lambda_{ He ^{+}}}= R _{ H } \times 2^2 \times\left(\frac{1}{4}-\frac{1}{9}\right).....$(2)
From (1) & (2) $\frac{\lambda_{ He ^{+}}}{\lambda}=\frac{9}{5}$
$ \lambda_{ He ^{+}}=\lambda \times \frac{9}{5}$
$ \lambda_{ He ^{+}}-\frac{9 \lambda}{5}$

Q18. Compound that will give positive Lassaigne's test for both nitrogen and halogen is:

A

$NH _2 OH . HCl$

B

$CH _3 NH _2 \cdot HCl$

C

$NH _4 Cl$

D

$N _2 H _4 \cdot HCl$

Solution

$CH _3 NH _2 . HCl \xrightarrow[\text { fusion }]{ Na } NaCN$ and $NaCl$
$NaCN$ gives $+ ve$ test for nitrogen and
$NaCl$ gives +ve test for halogen

Q19. The standard electrode potential $\left( M ^{3+} / M ^{2+}\right)$ for $V , Cr , Mn$ & Co are $-0.26 V$, $-0.41 V ,+1.57 V$ and $+1.97 V$, respectively. The metal ions which can liberate $H _2$ from a dilute acid are

A

$V ^{2+}$ and $Mn ^{2+}$

B

$Mn ^{2+}$ and $Co ^{2+}$

C

$Cr ^{2+}$ and $Co ^{2+}$

D

$V ^{2+}$ and $Cr ^{2+}$

Solution

Metal cation with $(-)$ value of reduction potential $\left( M ^{+3} / M ^{+2}\right)$ or with $(+)$ value of oxidation potential $\left( M ^{+2} / M ^{+3}\right)$ will liberate $H _2$
Therefore they will reduce $H ^{+}$
i.eV ${ }^{+2}$ and $Cr ^{+2}$

Q20. Which of the following salt solutions would coagulate the colloid solution formed when $FeCl _3$ is added to $NaOH$ solution, at the fastest rate?

A

$10\, mL$ of $0.1 \,mol\, dm ^{-3} Ca _3\left( PO _4\right)_2$

B

$10\, mL$ of $0.2 \,mol \,dm ^{-3} AlCl _3$

C

$10 \,mL$ of $0.15\, mol \,dm ^{-3} CaCl _2$

D

$10 \,mL$ of $0.1 \,mol \,dm ^{-3} Na _2 SO _4$

Solution

Formed is negatively charged solution, therefore $Al ^{3+}$ has highest coagulating power

Q21. The sum of bridging carbonyls in $W ( CO )_6$ and $Mn _2( CO )_{10}$ is _____.

Answer: 0

Q22. Following figure shows dependence of molar conductance of two electrolytes on concentration. $\Lambda \overset{\circ}{m}$ is the limiting molar conductivity.
image
The number of incorrect statement(s) from the following is _______
(A) $\Lambda \overset{\circ}{m}$ for electrolyte $A$ is obtained by extrapolation
(B) For electrolyte $B, \Lambda m$ vs $\sqrt{c}$ graph is a straight line with intercept equal to $\Lambda \overset{\circ}{m}$
(C) At infinite dilution, the value of degree of dissociation approaches zero for electrolyte $B$.
(D) $\Lambda \overset{\circ}{m}$ for any electrolyte $A$ or $B$ can be calculated using $\lambda^{\circ}$ for individual ions

Answer: 2

Solution

Statement (A) and Statement (C) are incorrect

Q23. The number of molecules or ions from the following, which do not have odd number of electrons are _____
(A) $NO _2$
(B) $ICl _4^{-}$
(C) $ BrF _3$
(D) $ClO _2$
(E) $NO _2^{+}$
(F) $NO$

Answer: 3

Solution

$ICl _4^{-}, BrF _3$ and $NO _2^{+}$do not have odd number of $e ^{-}$

Q24. $17 \,mg$ of a hydrocarbon (M.F. $C _{10} H _{16}$ ) takes up $8.40\, mL$ of the $H _2$ gas measured at $0^{\circ} C$ and $760 \,mm$ of $Hg$. Ozonolysis of the same hydrocarbon yields
image
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 $H _2$ 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
$=\frac{3.75 \times 10^{-4}}{1.25 \times 10^{-4}}=3$

Q25. Water decomposes at $2300\, K$
$H _2 O ( g ) \rightarrow H _2( g )+\frac{1}{2} O _2( g )$
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

image
$ P _0\left[1+\frac{\alpha}{2}\right]=1 $...(i)
$ K _{ p }=\frac{\left( P _{ H _2}\right)\left( P _{ O _2}\right)^{1 / 2}}{ P _{ H _2 O }}$
$ \frac{\left( P _0 \alpha\right)\left(\frac{ P _0 \alpha}{2}\right)^{1 / 2}}{ P _0[1-\alpha]}=2 \times 10^{-3} $
since $ \alpha $ is negligible w.r.t $ 1 \text { so } P _0=1 \text { and } 1-\alpha \approx 1 $
$ \frac{\alpha \sqrt{\alpha}}{\sqrt{2}}=2 \times 10^{-3} $
$ \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} \% \alpha=2 \%$

Q26. 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

$ R _{ f }=\frac{\text { Distance moved by the substance from base line }}{\text { Distance moved by the solvent from base line }} $
$ =\frac{3.0\, cm }{5.0 \, cm }=0.6 \text { or } 6 \times 10^{-1}$

Q27. Solid Lead nitrate is dissolved in 1 litre of water. The solution was found to boil at $100.15^{\circ} C$. When $0.2\, mol$ of $NaCl$ is added to the resulting solution, it was observed that the solution froze at $-0.8^0 C$. The solubility product of $PbCl _2$ formed is ___ $\times 10^{-6}$ at $298 \,K$. (Nearest integer)
Given : $K _{ b }=0.5\, K \,kg\, mol ^{-1}$ and $K _{ f }=1.8 \,K \,kg\, mol ^{-1}$. Assume molality to be equal to molarity in all cases.

Answer: 13

Solution

image
In final solution
$\Delta T _{ f }=0.8=1.8\left[\frac{0.3-3 x +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}$

Q28. 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 _______
image
(A) Over all order of this reaction is one
(B) Order of this reaction can't be determined
(C) In region I and III, the reaction is of first and zero order respectively
(D) In region-II, the reaction is of first order
(E) In region-II, the order of reaction is in the range of $0.1$ to $0.9$.

Answer: 2

Solution

Only option (B) is correct as order cannot be determined

Q29. Consider the following reaction approaching equilibrium at $27^{\circ} C$ and $1 atm$ pressure
$A + B \xrightleftharpoons[k _r=10^2]{k_{f}=1 0^3}C + D$
The standard Gibb's energy change $\left(\Delta_r G^{\circ}\right)$ at $27^{\circ} C$ is $(-)$___ $kJ mol ^{-1}$
(Nearest integer).
(Given: $R =8.3 J K ^{-1} mol ^{-1}$ and $\ln 10=2.3$ )

Answer: 6

Solution

$ \because \Delta G ^{ o }=- RT \ln K _{ eq } $
$ \text { and } K _{ eq }=\frac{ K _{ f }}{ K _{ b }} $
$ \therefore K _{ eq }=\frac{10^3}{10^2}=10$
$ \therefore \Delta G =- RT \ln 10 $
$\Rightarrow-(8.3 \times 300 \times 2.3)=-5.7 \,kJ \,mole ^{-1} \approx 6 kJ$
$ \text { mole }^{-1}(\text { nearest integer }) $
$ \text { Ans }=6$

Q30. 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

$ \because pH =12 $
$ \therefore\left[ H ^{+}\right]=10^{-12} M$
$ \therefore[ OH ]=10^{-2} M $
$ \therefore\left[ Ca ( OH )_2\right]=5 \times 10^{-3} M $
$5 \times 10^{-3}=\frac{\text { milli moles of } Ca ( OH )_2}{100 mL } $
$\text { milli moles of } Ca ( OH )_2=5 \times 10^{-1}$
$ \text { Ans. }=5$

JEE Main Mathematics Question Paper with Solution 2023 January 29th Shift 1 - Morning

Q1. Let the tangents at the points $A(4,-11)$ and $B(8,-5)$ on the circle $x^2+y^2-3 x+10 y-15=0$, 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

A

$2 \sqrt{13}$

B

$\frac{2 \sqrt{13}}{3}$

C

$\sqrt{13}$

D

$\frac{3 \sqrt{3}}{4}$

Solution

Equation of tangent at $A (4, -11)$ on circle is
$ \Rightarrow 4 x-11 y-3\left(\frac{x+4}{2}\right)+10\left(\frac{y-11}{2}\right)-15=0$
$\Rightarrow 5 x-12 y-152=0 \ldots \ldots(1)$
Equation of tangent at B $(8,-5)$ on circle is
$ \Rightarrow 8 x-5 y-3\left(\frac{x+8}{2}\right)+10\left(\frac{y-5}{2}\right)-15=0 $
$\Rightarrow 13 x-104=0 \Rightarrow x=8$
put in (1) $\Rightarrow y =\frac{28}{3}$
$r =\left|\frac{3.8+\frac{2.28}{3}-34}{\sqrt{13}}\right|=\frac{2 \sqrt{13}}{3}$

Q2. Three 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

A

250

B

20

C

25

D

30

Solution

$X$ $P(X)$ $XP(X)$ $X^2P(X)$
0 1/6 0 0
1 1/2 1/2 1/2
2 3/10 6/10 12/10
4 1/30 1/10 9/30
$ \sum xP ( x )=\frac{6}{2}=\mu $
$ \sigma^2=\sum x ^2 P ( x )-\mu^2 $
$ \sigma^2+\mu^2=0+\frac{1}{2}+\frac{12}{10}+\frac{9}{30}=2$
$ 10\left(\sigma^2+\mu^2\right)=20 \text { Ans. }$

Q3. Let $A=\left\{(x, y) \in R ^2: y \geq 0,2 x \leq y \leq \sqrt{4-(x-1)^2}\right\}$ and
$B=\left\{(x, y) \in R \times R : 0 \leq y \leq \min \left\{2 x, \sqrt{4-(x-1)^2}\right\}\right\} .$
Then the ratio of the area of $A$ to the area of $B$ is

A

$\frac{\pi}{\pi+1}$

B

$\frac{\pi-1}{\pi+1}$

C

$\frac{\pi+1}{\pi-1}$

D

$\frac{\pi}{\pi-1}$

Solution

$y^2+(x-1)^2=4$
image
shaded portion $=\text { circular }( OABC )$
$ -\operatorname{Ar}(\triangle OAB ) $
$=\frac{\pi(4)}{4}-\frac{1}{2}(2)(1)$
$ A =(\pi-1)$
image
Area $B =\operatorname{Ar}(\triangle AOB )+$ 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}$

Q4. Let $\lambda \neq 0$ be a real number. Let $\alpha, \beta$ be the roots of the equation $14 x^2-31 x+3 \lambda=0$ and $\alpha, \gamma$ be the roots of the equation $35 x^2-53 x+4 \lambda=0$. Then $\frac{3 \alpha}{\beta}$ and $\frac{4 \alpha}{\gamma}$ are the roots of the equation

A

$7 x^2+245 x-250=0$

B

$49 x^2+245 x+250=0$

C

$49 x^2-245 x+250=0$

D

$7 x^2-245 x+250=0$

Solution

$ 14 x^2-31 x+3 \lambda=0$
$ \alpha+\beta=\frac{31}{14} \ldots . .(1) $
and $ \alpha \beta=\frac{3 \lambda}{14} \ldots .(2) $
$ 35 x^2-53 x+4 \lambda=0$
$ \alpha+\gamma=\frac{53}{35} \ldots .(3) $
and $ \alpha \gamma=\frac{4 \lambda}{35} \ldots .(4) $
$ \frac{(2)}{(4)} \Rightarrow \frac{\beta}{\gamma}=\frac{3 \times 35}{4 \times 14}=\frac{15}{8} \Rightarrow \beta=\frac{15}{8} \gamma$
$(1)-(3) \Rightarrow \beta-\gamma=\frac{31}{14}-\frac{53}{35}=\frac{155-106}{70}=\frac{7}{10}$
$ \frac{15}{8} \gamma-\gamma=\frac{7}{10} \Rightarrow \gamma=\frac{4}{5} $
$\Rightarrow \beta=\frac{15}{8} \times \frac{4}{5}=\frac{3}{2} $
$ \Rightarrow \alpha=\frac{31}{14}-\beta=\frac{31}{14}-\frac{3}{2}=\frac{5}{7} $
$ \Rightarrow \lambda=\frac{14}{3} \alpha \beta=\frac{14}{3} \times \frac{5}{7} \times \frac{3}{2}=5$
so, sum of roots $\frac{3 \alpha}{\beta}+\frac{4 \alpha}{\gamma}=\left(\frac{3 \alpha \gamma+4 \alpha \beta}{\beta \gamma}\right)$
$=\frac{\left(3 \times \frac{4 \lambda}{35}+4 \times \frac{3 \lambda}{14}\right)}{\beta \gamma}=\frac{12 \lambda(14+35)}{14 \times 35 \beta \gamma}$
$=\frac{49 \times 12 \times 5}{490 \times \frac{3}{2} \times \frac{4}{5}}=5$
Product of roots
$=\frac{3 \alpha}{\beta} \times \frac{4 \alpha}{\gamma}=\frac{12 \alpha^2}{\beta \gamma}=\frac{12 \times \frac{25}{49}}{\frac{3}{2} \times \frac{4}{5}}=\frac{250}{49}$
So, required equation is $x^2-5 x+\frac{250}{49}=0$
$\Rightarrow 49 x^2-245 x+250=0$

Q5. Let $y=f(x)$ be the solution of the differential equation $y(x+1) d x-x^2 d y=0, y(1)=e$. Then $\displaystyle\lim _{x \rightarrow 0^{+}} f(x)$ is equal to

A

$\frac{1}{e^2}$

B

$\frac{1}{e}$

C

0

D

$e^2$

Solution

$ \frac{x+1}{x^2} d x=\frac{d y}{y} $
$ \ln x-\frac{1}{x}=\ln y+c$
$ (1, e ) $
$ c =-2 $
$ \ln x-\frac{1}{x}=\ln y-2 $
$ y=e^{\ln x}-\frac{1}{x}+2 $
$ \displaystyle\lim _{x \rightarrow 0^{+}} e^{\ln x-1}-\frac{1}{x}+2$
$ =e^{-\infty} $
$ =0$

Q6. Let $[x]$ denote the greatest integer $\leq x$. Consider the function $f(x)=\max \left\{x^2, 1+[x]\right\}$. Then the value of the integral $\int\limits_0^2 f(x) d x$ is

A

$\frac{5+4 \sqrt{2}}{3}$

B

$\frac{8+4 \sqrt{2}}{3}$

C

$\frac{1+5 \sqrt{2}}{3}$

D

$\frac{4+5 \sqrt{2}}{3}$

Solution

image
$ A=\int\limits_0^1 1 \cdot d x+\int\limits_1^{\sqrt{2}} 2 d x+\int\limits_{\sqrt{2}}^2 x^2 d x $
$ =1+2 \sqrt{2}-2+\frac{8}{3}-\frac{2 \sqrt{2}}{3} $
$ =\frac{5}{3}+\frac{4 \sqrt{2}}{3}$

Q7. Let $\Delta$ be the area of the region $\left\{(x, y) \in R ^2: x^2+y^2 \leq 21, y^2 \leq 4 x, x \geq 1\right\}$. Then $\frac{1}{2}\left(\Delta-21 \sin ^{-1} \frac{2}{\sqrt{7}}\right)$ is equal to

A

$\sqrt{3}-\frac{2}{3}$

B

$2 \sqrt{3}-\frac{2}{3}$

C

$2 \sqrt{3}-\frac{1}{3}$

D

$\sqrt{3}-\frac{4}{3}$

Solution

image
Area $2 \int\limits_1^3 2 \sqrt{x} d x+2 \int\limits_3^{\sqrt{21}} \sqrt{21-x^2 d x}$
$ \Delta=\frac{8}{3}(3 \sqrt{3}-1)+21 \sin ^{-1}\left(\frac{2}{\sqrt{7}}\right)-6 \sqrt{3} $
$ \frac{1}{2}\left(\Delta-21 \sin ^{-1}\left(\frac{2}{\sqrt{7}}\right)\right)=\frac{2 \sqrt{3}-\frac{8}{3}}{2} $
$ =\sqrt{3}-\frac{4}{3}$

Q8. Let $x=2$ be a root of the equation $x^2+p x+q=0$ and $f(x)=\begin{cases}\frac{1-\cos \left(x^2-4 p x+q^2+8 q+16\right)}{(x-2 p)^4}, & x \neq 2 p \\ 0, & , x=2 p\end{cases}$ Then $\displaystyle\lim _{x \rightarrow 2 p^{+}}[f(x)]$, where [.] denotes greatest integer function, is

A

2

B

1

C

$-1$

D

0

Solution

$\displaystyle \lim _{x \rightarrow p ^{+}}\left(\frac{1-\cos \left(x^2-4 p x+q^2+8 q+16\right)}{\left(x^2-4 p x+q^2+8 q+16\right)^2}\right)\left(\frac{\left(x^2-4 p x+q^2+8 q+16\right)^2}{(x-2 p)^2}\right) $
$\displaystyle\lim _{h \rightarrow 0} \frac{1}{2}\left(\frac{(2 p+h)^2-4 p(2 p+h)+q^2+82+16}{h^2}\right)^2=\frac{1}{2}$
Using L'Hospital's
$\displaystyle\lim _{x \rightarrow 2 p^{+}}[f(x)]=0$

Q9. 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

A

$\frac{2}{(\sqrt{3}-1)}$

B

$\frac{\sqrt{3}}{2(\sqrt{3}+1)}$

C

$\frac{2}{3+\sqrt{3}}$

D

$\frac{2}{3-\sqrt{3}}$

Solution

Slope of reflected ray $=\tan 60^{\circ}=\sqrt{3}$
Line $y=\frac{x}{\sqrt{3}}$ intersect $y+x=1$ 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 $y =0 \Rightarrow x =\frac{2}{3+\sqrt{3}}$

Q10. 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 $y-2 x=2$ such that $\triangle A B C$ is an equilateral triangle. Then, the area of the $\triangle A B C$ is

A

$2 \sqrt{3}$

B

$\frac{8}{\sqrt{3}}$

C

$3 \sqrt{3}$

D

$\frac{10}{\sqrt{3}}$

Solution

image
At $A x=y$
$Y-2 x=2$
$(-2,-2)$
Height from line $x + y =0$
$h=\frac{4}{\sqrt{2}}$
Area of $\Delta=\frac{\sqrt{3}}{4} \frac{ h ^2}{\sin ^2 60}=\frac{8}{\sqrt{3}}$

Q11. 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 $\{(p \vee q) \wedge((\sim p) \vee r)\} \rightarrow((\sim q) \vee r)$ false?

A

$p=T, q=T, r=F$

B

$p=T, q=F, r=T$

C

$p=F, q=T, r=F$

D

$p=T, q=F, r=F$

Q12. Consider the following system of equations
$ \alpha x+2 y+z=1 $
$ 2 \alpha x+3 y+z=1 $
$ 3 x+\alpha y+2 z=\beta$
for some $\alpha, \beta \in R$. Then which of the following is NOT correct.

A

It has no solution if $\alpha=-1$ and $\beta \neq 2$

B

It has no solution for $\alpha=-1$ and for all $\beta \in R$

C

It has a solution for all $\alpha \neq-1$ and $\beta=2$

D

It has no solution for $\alpha=3$ and for all $\beta \neq 2$

Solution

$ D=\begin{vmatrix}\alpha & 2 & 1 \\ 2 \alpha & 3 & 1 \\ 3 & \alpha & 2\end{vmatrix}=0 \Rightarrow \alpha=-1,3 $
$ D_x=\begin{vmatrix}2 & 1 & 1 \\ 3 & 1 & 1 \\ \alpha & 2 & \beta\end{vmatrix}=0 \Rightarrow \beta=2 $
$ D_y= \begin{vmatrix}\alpha & 1 & 1 \\ 2 \alpha & 1 & 1 \\ 3 & 2 & \beta\end{vmatrix}=0$
$ D_z= \begin{vmatrix} \alpha & 2 & 1 \\ 2 \alpha & 3 & 1 \\ 3 & \alpha & \beta\end{vmatrix}=0 $
$ \beta=2, \alpha=-1 $
$\alpha=-1, \beta=2 $ Infinite solution

Q13. Let $f(\theta)=3\left(\sin ^4\left(\frac{3 \pi}{2}-\theta\right)+\sin ^4(3 \pi+\theta)\right)-2\left(1-\sin ^2 2 \theta\right)$ and $S=\left\{\theta \in[0, \pi]: f^{\prime}(\theta)=-\frac{\sqrt{3}}{2}\right\}$. If $4 \beta=\displaystyle\sum_{\theta \in S} \theta$, then $f(\beta)$ is equal to

A

$\frac{11}{8}$

B

$\frac{3}{2}$

C

$\frac{9}{8}$

D

$\frac{5}{4}$

Solution

$f (\theta)=3\left(\sin ^4\left(\frac{3 \pi}{2}-\theta\right)+\sin ^4(3 x+\theta)\right)-2\left(1-\sin ^2 2 \theta\right) $
$ S =\left\{\theta \in[0, \pi]: f^{\prime}(\theta)=-\frac{\sqrt{3}}{2}\right\} $
$ \Rightarrow f(\theta)=3\left(\cos ^4 \theta+\sin ^4 \theta\right)-2 \cos ^2 2 \theta $
$ \Rightarrow f(\theta)=3\left(1-\frac{1}{2} \sin ^2 2 \theta\right)-2 \cos ^2 2 \theta $
$ \Rightarrow f(\theta)=3-\frac{3}{2} \sin ^2 2 \theta-2 \cos ^2 \theta$
$=\frac{3}{2}-\frac{1}{2} \cos ^2 2 \theta=\frac{3}{2}-\frac{1}{2}\left(\frac{1+\cos 4 \theta}{2}\right)$
$f(\theta)=\frac{5}{4}-\frac{\cos 4 \theta}{4}$
$ f^{\prime}(\theta)=\sin 4 \theta$
$ \Rightarrow f^{\prime}(\theta)=\sin 4 \theta=-\frac{\sqrt{3}}{2} $
$\Rightarrow 4 \theta= n \pi+(-1)^{ n } \frac{\pi}{3}$
$ \Rightarrow \theta=\frac{ n \pi}{4}+(-1)^{ n } \frac{\pi}{12}$
$ \Rightarrow \theta=\frac{\pi}{12},\left(\frac{\pi}{4}-\frac{\pi}{12}\right),\left(\frac{\pi}{2}+\frac{\pi}{12}\right),\left(\frac{3 \pi}{4}-\frac{\pi}{12}\right) $
$ \Rightarrow 4 \beta=\frac{\pi}{4}+\frac{\pi}{2}+\frac{3 \pi}{4}=\frac{3 \pi}{2} $
$ \Rightarrow \beta=\frac{3 \pi}{8} \Rightarrow f (\beta)=\frac{5}{4}-\frac{\cos \frac{3 \pi}{2}}{4}=\frac{5}{4}$

Q14. For two non-zero complex numbers $z_1$ and $z_2$, if $\text{Re}\left(z_1 z_2\right)=0$ and $\text{Re}\left(z_1+z_2\right)=0$, then which of the following are possible?
A. $\text{Im}\left(z_1\right)>0$ and $\text{Im}\left(z_2\right)>0$
B. $\text{Im}\left(z_1\right)<0$ and $\text{Im}\left(z_2\right)>0$
C. $\text{Im}\left(z_1\right)>0$ and $\text{Im}\left(z_2\right)<0$
D. $\text{Im}\left(z_1\right)<0$ and $\text{Im}\left(z_2\right)<0$
Choose the correct answer from the options given below:

A

A and B

B

$B$ and $C$

C

B and D

D

A and C

Solution

$z_1=x_1+i y_1$
$z _2= x _2+ i y _2$
$\operatorname{Re}\left( z _1 z _2\right)= x _1 x _2- y _1 y _2=0$
$\operatorname{Re}\left( z _1+ z _2\right)= x _1+ x _2=0$
$x _1 \& x _2$ are of opposite sign
$y_1 \& y_2$ are of opposite sign

Q15. Let $\alpha$ and $\beta$ be real numbers. Consider a $3 \times 3$ matrix $A$ such that $A^2=3 A+\alpha I$. If $A^4=21 A+\beta I$, then

A

$\beta=-8$

B

$\beta=8$

C

$a=4$

D

$\alpha=1$

Solution

$ A ^2=3 A +\alpha I$
$ A ^3=3 A ^2+\alpha A$
$ A ^3=3(3 A +\alpha I )+\alpha A$
$ A ^3=9 A +\alpha A +3 \alpha I$
$ A ^4=(9+\alpha) A ^2+3 \alpha A $
$=(9+\alpha)(3 A +\alpha I )+3 \alpha A$
$ = A (27+6 \alpha)+\alpha(9+\alpha) $
$ \Rightarrow 27+6 \alpha=21 \Rightarrow \alpha=-1 $
$ \Rightarrow \beta=\alpha(9+\alpha)=-8$

Q16. Fifteen football players of a elub-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

A

$\frac{5}{36}$

B

$\frac{2}{15}$

C

$\frac{5}{24}$

D

$\frac{1}{6}$

Solution

Required probability $=1-\frac{D_{(15)}+{ }^{15} C_1 \cdot D_{(14)}+{ }^{15} C_2 D_{(13)}}{15 !}$
Taking $D _{(15)}$ as $\frac{15 \text { ! }}{e}$
$D _{(14)}$ as $\frac{14 \text { ! }}{e}$
$D _{(13)}$ as $\frac{13 !}{e}$
We get, $1-\left(\frac{\frac{15 !}{e}+15 \cdot \frac{14 !}{e}+\frac{15 \times 14}{2} \times \frac{13 !}{e}}{15 !}\right)$
$=1-\left(\frac{1}{e}+\frac{1}{e}+\frac{1}{2 e}\right)=1-\frac{5}{2 e} \approx .08$

Q17. Let $f: R \rightarrow R$ be a function such that $f(x)=\frac{x^2+2 x+1}{x^2+1}$. Then

A

$f(x)$ is many-one in $(-\infty,-1)$

B

$f(x)$ is one-one in $(-\infty, \infty)$

C

$f(x)$ is many-one in $(1, \infty)$

D

$f(x)$ is one-one in $[1, \infty)$ but not in $(-\infty, \infty)$

Solution

image
$ f(x)=\frac{(x+1)^2}{x^2+1}=1+\frac{2 x}{x^2+1} $
$ f(x)=1+\frac{2}{x+\frac{1}{x}}$

Q18. 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

A

18

B

6

C

24

D

0

Solution

$ \begin{vmatrix} \lambda & \mu & 4 \\ -2 & 4 & -2 \\ 2 & 3 & 1 \end{vmatrix}=0$
$ \lambda(10)=\mu(2)+4(-14)=0 $
$ 10 \lambda-2 \mu=56$
$ 5 \lambda-\mu=28 .....$(1)
$ \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}=\sqrt{54}$
$\frac{-2 \lambda+4 \mu-8}{\sqrt{24}}=\sqrt{54} $
$ -2 \lambda+4 \mu-8=\sqrt{54 \times 24} .....$(2)
By solving equation (1) & (2)
$\Rightarrow \lambda+\mu=24$

Q19. Let $f(x)=x+\frac{a}{\pi^2-4} \sin x+\frac{b}{\pi^2-4} \cos x, x \in R$ be a function which satisfies $f(x)=x+\int\limits_0^{\pi / 2} \sin (x+y) f(y) d y$. Then $(a+b)$ is equal to

A

$-\pi(\pi+2)$

B

$-\pi(\pi-2)$

C

$-2 \pi(\pi+2)$

D

$-2 \pi(\pi-2)$

Solution

$f(x)=x+\int\limits_0^{\pi / 2}(\sin x \cos y+\cos x \sin y) f(y) d y$
$f(x)=x+\int\limits_0^{\pi / 2}((\cos y f(y) d y) \sin x+(\sin y f(y) d y) \cos x) .......$(1)
On comparing with
$f(x)=x+\frac{a}{\pi^2-4} \sin x+\frac{b}{\pi^2-4} \cos x, x \in R$ then
$\Rightarrow \frac{a}{\pi^2-4}=\int\limits_0^{\pi / 2} \cos y f(y) d y .....$(2)
$\Rightarrow \frac{b}{\pi^2-4}=\int\limits_0^{\pi / 2} \sin y f(y) d y .....$(3)
Add (2) and (3)
$ \frac{a+b}{\pi^2-4}=\int\limits_0^{\pi / 2}(\sin y+\cos y) f(y) d y \ldots(4) $
$\frac{a+b}{\pi^2-4}=\int\limits_0^{\pi / 2}(\sin y+\cos y) f\left(\frac{\pi}{2}-y\right) d y \ldots(5) $
Add (4) and (5)
$ \frac{2(a+b)}{\pi^2-4} =\int\limits_0^{\pi / 2}(\sin y+\cos y)\left(\frac{\pi}{2}+\frac{(a+b)}{\pi^2-4}(\sin y+\cos y)\right) d y $
$ =\pi+\frac{a+b}{\pi^2-4}\left(\frac{\pi}{2}+1\right) $
$(a+b) =-2 \pi(\pi+2)$

Q20. The domain of $f(x)=\frac{\log _{(x+1)}(x-2)}{e^{2 \log _e x}-(2 x+3)}, x \in R$ is

A

$(2, \infty)-\{3\}$

B

$R -\{3\}$

C

$R -\{-1,3\}$

D

$(-1, \infty)-\{3\}$

Solution

$x-2>0 \Rightarrow x>2$
$x+1>0 \Rightarrow x>-1 $
$ x+1 \neq 1 \Rightarrow x \neq 0 \text { and } x>0$
Denominator
$ x^2-2 x-3 \neq 0 $
$ (x-3)(x+1) \neq 0$
$x \neq-1,3$
So Ans $(2, \infty)-\{3\}$

Q21. Let $a_1, a_2, a_3, \ldots$ be a $G P$ 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 $a_1 a_9+a_2 a_4 \alpha_9+a_5+a_7$ is equal to ____

Answer: 60

Solution

$a _4 \cdot a _6=9 \Rightarrow\left( a _5\right)^2=9 \Rightarrow a _5=3 $
$ \& a _5+ a _7=24 \Rightarrow a _5+ a _5 r ^2=24 \Rightarrow\left(1+ r ^2\right)=8 \Rightarrow r =\sqrt{7} $
$ \Rightarrow a =\frac{3}{49} $
$ \Rightarrow a _1 a _9+ a _2 a _4 a _9+ a _5+ a _7=9+27+3+21=60$

Q22. 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 $\left(\alpha x^3+\frac{1}{\beta x}\right)={ }^{11} C_6 \cdot \frac{\alpha^5}{\beta^6}$
$\because$ Both are equal
$ \therefore \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 $
$\Rightarrow(\alpha \beta)^2=1$

Q23. Suppose $f$ is a function satisfying $f(x+y)=f(x)+f(y)$ for all $x, y \in N$ and $f(1)=\frac{1}{5}$. If $\displaystyle\sum_{n=1}^{\text {m }} \frac{f(n)}{n(n+1)(n+2)}=\frac{1}{12}$, then $m$ is equal to_____

Answer: 10

Solution

$ \because f(1)=\frac{1}{5} \therefore f(2)=f(1)+f(1)=\frac{2}{5} $
$ f(2)=\frac{2}{5} f(3)=f(2)+f(1)=\frac{3}{5}$
$f(3)=\frac{3}{5} $
$ \therefore \displaystyle\sum_{n=1}^m \frac{f(n)}{n(n+1)(n+2)} $
$=\frac{1}{5} \displaystyle\sum_{n=1}^m\left(\frac{1}{n+1}-\frac{1}{n+2}\right) $
$=\frac{1}{5}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\ldots .+\frac{1}{m+1}-\frac{1}{m+2}\right)$
$ =\frac{1}{5}\left(\frac{1}{2}-\frac{1}{m+2}\right)=\frac{m}{10(m+2)}=\frac{1}{12} $
$ \therefore m=10$

Q24. Let the coefficients of three consecutive terms in the binomial expansion of $(1+2 x)^n$ 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

$ t_{r+1}={ }^n C_r(2 x)^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 !(2)}{r !(n-r) !}}=\frac{2}{5} $
$ \Rightarrow \frac{r}{n-r+1}=\frac{4}{5} \Rightarrow 5 r=4 n-4 r+4$
$ \Rightarrow 9 r=4(n+1) .....$(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 4 r+4=5 n-5 r \Rightarrow 5 n-4=9 r \ldots$(2)
From (1) and (2)
$\Rightarrow 4 n+4=5 n-4 \Rightarrow n=8$
(1) $\Rightarrow r =4$
so, coefficient of middle term is
${ }^8 C _4 2^4=16 \times \frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1}=16 \times 70=1120$

Q25. Let $f: R \rightarrow R$ be a differentiable function that satisfies the relation $f(x+y)=f(x)+f(y)-1, \forall x, y \in R$. If $f^{\prime}(0)=2$, then $|f(-2)|$ is equal to _____

Answer: 3

Solution

$ f ( x + y )= f ( x )+ f ( y )-1 $
$ f^{\prime}(x)=\displaystyle\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} $
$ f^{\prime}(x)=\displaystyle\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}=f^{\prime}(0)=2$
$f^{\prime}(x)=2 \Rightarrow d y=2 d x $
$ y =2 x + C $
$ x =0, y =1, c =1 $
$ y =2 x +1 $
$ |f(-2)|=|-4+1|=|-3|=3$

Q26. Let the equation of the plane $P$ containing the line $x+10=\frac{8-y}{2}=z$ be $a x+b y+3 z=2(a+b)$ 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 d. r. $(1,-2,1)$
$ \because \text { the plane } a x+b y+3 z=2(a+b)$
$ \Rightarrow b=2 a$
$\&$ dot product of d.r.'s is zero
$\therefore a -2 b +3=0$
$\therefore a =1 \& b =2$
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$

Q27. 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 $3125-1690+1=1436$

Q28. Let the co-ordinates of one vertex of $\triangle A B C$ 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 $\alpha \in Z$, if the area of $\triangle A B C$ is $21$ sq. units and the line segment $B C$ has length $2 \sqrt{21}$ units, then $\alpha^2$ is equal to______

Answer: 9

Solution

image
$ \left|\frac{1}{2} \cdot 2 \sqrt{21} \begin{vmatrix} i & j & k \\ \alpha & 1 & \alpha+4 \\ 5 & 2 & 3\end{vmatrix} \frac{1}{\sqrt{25+4+9}}\right|=21 \sqrt{21} $
$ \sqrt{(2 \alpha+5)^2+(2 \alpha+20)^2+(2 \alpha-5)^2}=\sqrt{21} \sqrt{38} $
$\Rightarrow 12 \alpha^2+80 \alpha+450=798 $
$ \Rightarrow 12 \alpha^2+80 \alpha-348=0$
$ \Rightarrow \alpha=3 \Rightarrow \alpha^2=9$

Q29. 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 $72^{\text {th }}$ number is _____

Answer: 32

Solution

image
$ 245678 \rightarrow 72^{\text {th }} \text { word } $
$ 2+4+5+6+7+8=32$

Q30. 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 $\overrightarrow{A B}, \overrightarrow{A C}$ and $\overrightarrow{A D}$ are coplanar, then $\lambda$ is equal to

Answer: 2

Solution

$ \overline{A B}=(\lambda-1) \bar{a}-2 \bar{b}+3 \bar{c}$
$\overline{A C}=2 \bar{a}+3 \bar{b}-4 \bar{c} $
$ \overline{A D}=\bar{a}-3 \bar{b}+5 \bar{c} $
$\begin{vmatrix}\lambda-1 & -2 & 3 \\ -2 & 3 & -4 \\ 1 & -3 & 5\end{vmatrix}=0 $
$ \Rightarrow(\lambda-1)(15-12)+2(-10+4)+3(6-3)=0$
$\Rightarrow(\lambda-1)=1 \Rightarrow \lambda=2$