datavorous/entrance-exam-dataset huggingface dataset:
Consider three vectors p→=i^+j^+k^, q→=3i^-j^+k^ and r→=αi^+βj^+λk^, ∀α,β,λ∈R. If p→q→r→ is maximum and r→=26, then the value of α-β-λ is equal to

Options:
A 8
B 4
C 0
D -4

The correct answer is:
4

Solution:
p→×q→=i^j^k^1113-11=2i^+2j^-4k^

For maximum p→ q→ r→, r→ will be parallel to p→×q→
So, r→=2μi^+j^-2k^ ∀μ>0

⇒μ=1 (as r→=26)

⇒r→=2i^+2j^-4k^

i.e. α=2, β=2, λ=-4

Hence, α-β-λ=2-2+4=4

Topic: Mathematics, Vector Algebra, JEE Main
Consider three vectors V→1=sin⁡θi^+cos⁡θj^+a-3k^ , V→2=sin⁡θ+cos⁡θi^+cos⁡θ-sin⁡θj^+b-4k^ and V→3=cos⁡θi^+sin⁡θj^+c-5k^. If the resultant of V→1, V→2 and V→3 is equal to λi^, where θ∈-π,π and a,b,c∈N, then the number of quadruplets a,b,c,θ are

Options:
A 55
B 110
C 91
D 182

The correct answer is:
110

Solution:
Resultant of V→1,V→2,V→3 is V→1+V→2+V→3=λi^
⇒2sin⁡θ+2cos⁡θi^+2cos⁡θ-2sin⁡θj^+a+b+c-12k^=λi^
On comparison, we get,
2sin⁡θ+2cos⁡θ=λ
2cos⁡θ-2sin⁡θ=0⇒tan⁡θ=1⇒ number of values of θ in -π,π is 2
Also, a+b+c=12⇒ number of solutions =11C2=55
Hence, number of quadruplets =55×2=110

Topic: Mathematics, Vector Algebra, JEE Main
Consider titration of NaOH solution versus 1.25 M oxalic acid solution. At the end point following burette readings were obtained.
(i) 4.5 mL
(ii) 4.5 mL
(iii) 4.4 mL
(iv) 4.4 mL
(v) 4.4 mL
If the volume of oxalic acid taken was 10.0 mL then the molarity of the NaOH solution is ___M. (Rounded-off to the nearest integer)

Options:

The correct answer is:
6

Solution:
VNaOH=4.4 ml

eq. of NaOH= eq. of H2C2O4

or, M×4.4×1=1.25×10×2

or, M=5.68M

∴ Nearest integer answer is 6.

Topic: Chemistry, Redox Reactions, JEE Main, JEE Main 2021 (25 Feb Shift 2)
Consider two charged metallic spheres S1 and S2 of radii R1 and R2, respectively. The electric fields E1 (on S1 ) and E2 (on S2 ) on their surfaces are such that E1E2=R1R2. Then the ratio V1 (on S1 )/V2 (on S2 ) of the electrostatic potentials on each sphere is:

Options:
A R1R2
B R1R22
C R2R1
D R1R23

The correct answer is:
R1R22

Solution:
Given, E1E2=R1R2
Then ratio of potential would be,V1V2=E1R1E2R2=R1R2×R1R2=R1R22

Topic: Physics, Electrostatics, JEE Main, JEE Main 2020 (08 Jan Shift 2)
Consider two chemical reactions (A) and (B) that
take place during metallurgical process:
(A) ZnCO3s⟶ΔZnOs+CO2g
(B) 2ZnSs+3O2g⟶Δ2ZnOs+2SO2g
The correct option of names given to them respectively is:

Options:
A (A) is calcination and (B) is roasting
B Both (A) and (B) are producing same product so both are roasting
C Both (A) and (B) are producing same product so both are calcination
D (A) is roasting and (B) is calcination

The correct answer is:
(A) is calcination and (B) is roasting

Solution:
(A) ZnCO3S⟶ΔZnOs+CO2g

Heating in absence of oxygen in calcination.

(B) 2Znss+3O2g→2ZnOg+2SO2g

heating in presence of oxygen in roasting Hence (A) is calcination while (B) in roasting.

Topic: Chemistry, General Principles and Processes of Isolation of Metals, JEE Main, JEE Main 2021 (20 Jul Shift 2)
Consider two coherent, monochromatic (wavelength Î») sources, S1 and S2, separated by a distance d. The ratio of intensities of S1 and that of S2 (which is responsible for interference at point P, where detector is located) is 4. The distance of point P from S1 is (if the resultant intensity at point P is equal to 94 times of intensity of S1) (Given: â S2S1P=90Â°, d>0 and n is a positive integer)

Options:
A d2-n2λ22nλ
B d2+n2λ22nλ
C nλdd2-n2λ2
D 2 n λ dd2-n2λ2

The correct answer is:
d2-n2λ22nλ

Solution:
94 intensity of S1=9 times of intensity S2
Intensity of maxima = I1+I22=9I2
⇒S2P-S1P=nλ for maxima
d2+x2-x=nλ x=d2-n2λ22nλ

Topic: Physics, Waves and Sound, JEE Main
Consider two containers A and B containing monoatomic gases at the same Pressure P, Volume V and Temperature T. The gas in A is compressed isothermally to 18 of its original volume while the gas in B is compressed adiabatically to 18 of its original volume. The ratio of final pressure of gas in B to that of gas in A is

Options:
A 8
B 4
C 832
D 18

The correct answer is:
4

Solution:
In an isothermal process the temperature is constant.

PV=nRT

So, it can be written that

PV=PAV8⇒PA=8P

In an adiabatic process,

PVγ=constant

γ for monoatomic gas is 53

So, it can be written

PBP=VV853⇒PB=32P

Thus, the ratio of the pressures is

PBPA=32P8P=4

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2023 (10 Apr Shift 1)
Consider two ideal diatomic gases A and B at some temperature T . Molecules of the gas A are rigid, and have a mass m . Molecules of the gas B have an additional vibrational mode and have a mass m4 . The ratio of the specific heats CV Aand CVB of gas A and B, respectively is:

Options:
A 7:9
B 5:9
C 3:5
D 5:7

The correct answer is:
5:7

Solution:
Molar heat capacity of A at constant volume =5R2
Molar heat capacity of B at constant volume =7R2
Dividing both CvACvB=57

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2020 (09 Jan Shift 1)
Consider two identical, coaxial circular loops carrying a current i each, circulating in the same direction. If the loops approach each other

Options:
A the current in each loop will decrease
B the current in each loop will increase
C The current in each loop will remain the same
D the current in one loop will increase and in the other loop will decrease

The correct answer is:
the current in each loop will decrease

Solution:
According to Lenz's law, induced current will be developed in both the loops in opposite direction. Therefore, the net current in each loop will decrease.

Topic: Physics, Electromagnetic Induction, NEET
Consider two identical rings of mass M and radius R fixed on a horizontal axis in a gravity-free space. A particle of mass m is held at rest at the centre of 1st ring and projected along the axis towards 2nd ring. The minimum speed of projection so that the particle reaches the centre of 2nd ring is

Options:
A GMR
B 2GMR
C 2GMR1+15-2
D GMR1+5-12

The correct answer is:
2GMR1+15-2

Solution:
The particle should be projected such that it can reach halfway because after midpoint it will move under the attraction of 2nd ring and will reach the center of the ring.
PEi + KEi = KEf + PEf
-GMmR-GMm5R+12mv2=-GMm2R×2
v=2GMR1+15-2

Topic: Physics, Gravitation, JEE Main
Consider two infinitely long wires parallel to Z-axis carrying same current I in the positive Z direction. One wire passes through the point $L$ at coordinates $(-1,+1)$ and the other wire passes through the point $M$ at coordinates $(-1,-1)$. The resultant magnetic field at the origin $\mathrm{O}$ will be

Options:
A $\frac{\mu_{0} I}{2 \sqrt{2 \pi}} \hat{\mathrm{j}}$
B $\frac{\mu_{0} I}{2 \pi} \hat{j}$
C $\frac{\mu_{0} \mathrm{I}}{2 \sqrt{2 \pi}} \hat{\mathrm{i}}$
D $\frac{\mu_{0} I}{4 \pi} \hat{j}$

The correct answer is:
$\frac{\mu_{0} I}{2 \pi} \hat{j}$

Solution:
$\mathrm{B}_{\text {net }}=\sqrt{2} \mathrm{~B}$ where $\mathrm{B}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}}$
$\mathrm{B}_{\text {net }}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{r}} \sqrt{2} \hat{\mathrm{j}}$
$=\frac{\mu_{0} \mathrm{I}}{2 \pi \sqrt{2}} \times \sqrt{2} \hat{\mathrm{j}}$ and $\mathrm{r}=\sqrt{2}$
$\mathrm{B}_{\text {net }}=\frac{\mu_{0} \mathrm{I}}{2 \pi} \hat{\mathrm{j}}$

Topic: Physics, Magnetic Effects of Current, JEE Main
Consider two masses with $\mathrm{m}_{1}>\mathrm{m}_{2}$ connected by a light inextensible string that passes over a pulley of radius $\mathrm{R}$ and moment of inertia I about its axis of rotation. The string does not slip on the pulley and the pulley turns without friction. The two masses are released from rest separated by a vertical distance $2 \mathrm{~h}$. When the two masses pass each other, the speed of the masses is proportional to

Options:
A $\sqrt{\frac{m_{1}-m_{2}}{m_{1}+m_{2}+\frac{1}{R^{2}}}}$
B $\sqrt{\frac{\left(m_{1}+m_{2}\right)\left(m_{1}-m_{2}\right)}{m_{1}+m_{2}+\frac{1}{R^{2}}}}$
C $\sqrt{\frac{\mathrm{m}_{1}+\mathrm{m}_{2}+\frac{1}{\mathrm{R}^{2}}}{\mathrm{~m}_{1}-\mathrm{m}_{2}}}$
D $\sqrt{\frac{\frac{1}{\mathrm{R}^{2}}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}}$

The correct answer is:
$\sqrt{\frac{\mathrm{m}_{1}+\mathrm{m}_{2}+\frac{1}{\mathrm{R}^{2}}}{\mathrm{~m}_{1}-\mathrm{m}_{2}}}$

Solution:
The total mechanical energy of system $=$ conserved Hence
$$
\begin{array}{l}
\mathrm{KE}_{\mathrm{i}}+\mathrm{PE}_{\mathrm{i}}=\mathrm{KE}_{\mathrm{f}}+\mathrm{PE}_{\mathrm{f}} \\
0-\mathrm{m}_{2} \mathrm{~g} \times 2 \mathrm{~h}=\frac{1}{2} \mathrm{~m}_{2} \mathrm{v}^{2}+\frac{1}{2} \mathrm{~m}_{1} \mathrm{v}^{2}+\frac{1}{2} \mathrm{I} \omega^{2}-\mathrm{m}_{1} \mathrm{gh}-\mathrm{m}_{2} \mathrm{gh}
\end{array}
$$
Also $\omega=\frac{\mathrm{v}}{\mathrm{R}}$
$$
\begin{array}{l}
\left(\mathrm{m}_{1}-\mathrm{m}_{2}\right) \mathrm{gh}=\frac{1}{2}\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right) \mathrm{v}^{2}+\frac{1}{2} \mathrm{I}\left(\frac{\mathrm{v}}{\mathrm{R}}\right)^{2} \\
\left(\mathrm{~m}_{1}-\mathrm{m}_{2}\right) \mathrm{gh}=\frac{\mathrm{v}^{2}}{2}\left[\mathrm{~m}_{1}+\mathrm{m}_{2}+\frac{\mathrm{I}}{\mathrm{R}^{2}}\right] \\
\mathrm{v} \propto \sqrt{\frac{\mathrm{m}_{1}-\mathrm{m}_{2}}{\mathrm{~m}_{1}+\mathrm{m}_{2}+\frac{\mathrm{I}}{\mathrm{R}^{2}}}}
\end{array}
$$

Topic: Physics, Rotational Motion, JEE Main
Consider two npn transistors as shown in figure. If 0 Volts corresponds to false and 5 Volts correspond to true then the output at C corresponds to :

Options:
A
A NAND B

B
A OR B

C
AAND B

D
ANOR B

The correct answer is:
A NAND B

Solution:
The output at $\mathrm{C}$ corresponds to A NAND
$$
\mathrm{B} \text { or } \overline{\mathrm{A} \cdot \mathrm{B}}=\mathrm{C}
$$

Topic: Physics, Semiconductors, JEE Main, JEE Main 2013 (09 Apr Online)
Consider two nuclei of the same radioactive nuclide. One of the nuclei was created in a supernova explosion 5 billion years ago. ihe other was created in a nuclear reactor $5 \mathrm{~min}$ ago. The probability of decay during the next time is

Options:
A different for each nuclei
B nuclei created in explosion decays first
C nuclei created in the reactor decays first
D independent of the time of creation

The correct answer is:
independent of the time of creation

Solution:
The decay's constant $(\lambda)$ and half-life period of radioactive nuclei remains constant and independent of time of creation of radioactive nuclei.
Hence, the probability of decay during the next time : independent of the time of creation.

Topic: Physics, Nuclear Physics, COMEDK, COMEDK 2017
Consider two physical quantities A and B related to each other as E=B−x2At where E, x and t have dimensions of energy, length and time respectively. The dimension of AB is

Options:
A L−2M1T0
B L2M-1T1
C L−2M-1T1
D L0M-1T1

The correct answer is:
L2M-1T1

Solution:
From the dimensional checking of the equation, we can say the dimensions of B=x2=L2

And dimensions of A=x2tE=L2TML2T−2=1MT−1

Hence, A=M−1T

Required value, AB=L2M−1T1

Topic: Physics, Units and Dimensions, JEE Main, JEE Main 2024 (31 Jan Shift 2)
Consider two polaroids A and B as shown. Unpolarised light is incident on polaroid A. Now, both the polaroids are rotated simultaneously by 180° in the same sense of rotation such that at every instant, their pass (transmission) axes always remain parallel to each other. During the rotation, intensity of transmitted light through polaroid B

Options:
A increases continuously
B first increases then decreases
C remains same
D decreases continuously

The correct answer is:
remains same

Solution:
Intensity after passing through polaroid A is
IA=I02
=Intensity of upolarised light2
Intensity after passing through polaroid B is
IB=IAcos2â¡Î¸
Here, Î¸= angle between pass axes of A and B.
Here, pass (transmission) axes always remain parallel to each other i.e. Î¸=0 .
â´ IB=IA=I02=constant
Hence, during the rotation, the intensity of transmitted light through polaroid B remains the same.

Topic: Physics, Wave Optics, JEE Main
Consider two reactions having same Arrhenius factor A, but different energy of activation.
(i) AâB; Ea1=20 kJ
(ii) CâD; Ea2=30 kJ
Both are at temperature 25Â°C
If temperature in both reaction is increased slightly in such a way that change in temperature in both case is same than choose the correct options.

Options:
A The second reaction is faster
B The second reaction is more sensitive towards temperature variation
C If temperature increases, rate of first reaction increase more sharply
D All the above are correct

The correct answer is:
The second reaction is more sensitive towards temperature variation

Solution:
Arrhenius equation;
dlnk=EaRT2dT
Here, T and dT same and positive
dlnkαEa
lnk2-lnk1αEa
lnK2K1αEa
So, if temperature increases, then rate of reaction having more Ea increases more sharply.

Topic: Chemistry, Chemical Kinetics, JEE Main
Consider two rods of the same length and different specific heats $\left(S_1, S_2\right)$, conductivities $\left(K_1, K_2\right)$ and area of cross sections $\left(A_1, A_2\right)$ and both having temperature $T_1$ and $T_2$ at their ends. If the rate of loss of heat due to conduction is equal then:

Options:
A $K_1 A_1=K_2 A_2$
B $\frac{K_1 A_1}{S_2}=\frac{K_2 A_2}{S_1}$
C $K_2 A_1=K_1 A_2$
D $\frac{K_2 A_1}{S_2}=\frac{K_1 A_2}{S_1}$

The correct answer is:
$K_1 A_1=K_2 A_2$

Solution:
Rate of conduction of heat is given as
$\frac{\Delta \mathrm{Q}}{t}=\frac{K A}{L}\left(T_2-T_1\right)$
According to the question
$\begin{aligned}
& \frac{\Delta Q_1}{\Delta t}=\frac{\Delta Q_2}{\Delta t} \\
& \Rightarrow \frac{K_1 A_1}{L}\left(T_1-T_2\right)=\frac{K_1 A_1}{L}\left(T_1-T_2\right) \\
& \Rightarrow K_1 A_1=K_2 A_2
\end{aligned}$

Topic: Physics, Thermal Properties of Matter, JEE Main
Consider two satellites S1 and S2 with periods of revolution 1hr and 8hr respectively revolving around a planet in circular orbits. The ratio of angular velocity of satellite S1 to the angular velocity of satellite S2 is:

Options:
A 8:1
B 2:1
C 1:4
D 1:8

The correct answer is:
8:1

Solution:
T1 T2=18

2π/ω12π/ω2=18

ω1ω2=81

Topic: Physics, Gravitation, JEE Main, JEE Main 2021 (24 Feb Shift 1)
Consider two spheres A and B of the same size and made of the same material. The sphere A is hollow while the sphere B is solid. If both the spheres are heated to the same temperature, then

Options:
A A expands more than B
B A expands less than B
C both the spheres expand equally
D data is insufficient.

The correct answer is:
both the spheres expand equally

Solution:
A cavity in a material expands in exactly the same way as if the cavity were filled with material. Thus, both spheres will expand by the same amount.

Topic: Physics, Thermal Properties of Matter, JEE Main
Consider two thin identical conducting wires covered with very thin insulating material. One of the wires is bent into a loop and produces magnetic field $\mathrm{B}_1$, at its centre when a current $\mathrm{I}$ passes through it. The ratio $\mathrm{B}_1: \mathrm{B}_2$ is:

Options:
A
1:1

B
$1: 3$

C
$1: 9$

D
$9: 1$

The correct answer is:
$1: 3$

Solution:
For loop $B=\frac{\mu_0 \mathrm{nI}}{2 \mathrm{a}}$ where, $a$ is the radius of loop.
Then, $B_1=\frac{\mu_0 I}{2 a}$
Now, for coil $\mathrm{B}=\frac{\mu_0 \mathrm{I}}{4 \pi} \cdot \frac{2 \mathrm{nA}}{\mathrm{x}^3}$ at the centre $x=$ radius of loop
$$
\begin{aligned}
\mathrm{B}_2 &=\frac{\mu_0}{4 \pi} \cdot \frac{2 \times 3 \times(\mathrm{I} / 3) \times \pi(\mathrm{a} / 3)^2}{(\mathrm{a} / 3)^3} \\
&=\frac{\mu_0 \cdot 3 \mathrm{I}}{2 \mathrm{a}} \\
\therefore & \frac{\mathrm{B}_1}{\mathrm{~B}_2}=\frac{\mu_0 \mathrm{I} / 2 \mathrm{a}}{\mu_0 \cdot 3 \mathrm{I} / 2 \mathrm{a}} \\
\mathrm{B}_1: \mathrm{B}_2=1: 3
\end{aligned}
$$

Topic: Physics, Magnetic Effects of Current, JEE Main, JEE Main 2014 (12 Apr Online)
Consider two uniform discs of the same thickness and different radii R1=R and R2=αR made of the same material. If the ratio of their moments of inertia I1 and I2, respectively, about their axes is I1:I2=1:16 then the value of α is :

Options:
A 22
B 2
C 2
D 4

The correct answer is:
2

Solution:
Moment of inertia of disc is given by I=MR22=ρπR2tR22

I α R4

I2I1=R2R14

161=α4

α=2

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2020 (04 Sep Shift 2)
Consider two uniform discs of the same thickness and different radii \( R_{1}=R \) and \( R_{2}=\alpha R \) made of the same material. If the ratio of their moments of inertia \( I_{1} \) and \( I_{2} \), respectively, about their axes is \( I_{1}: I_{2}=1: 16 \) then the value of \( \alpha \) is:

Options:
A \( 2 \sqrt{2} \)
B \( \sqrt{2} \)
C \( 2 \)
D \( 4 \)

The correct answer is:
\( 2 \)

Solution:
Moment of inertia of disc is given by I=MR22=ρπR2tR22

I α R4

I2I1=R2R14

161=α4

α=2

Topic: Physics, Rotational Motion, JEE Main
Consider two vectors a→=3i^-2j^+4k^ and b→=j^+2k^. If c→ is a unit vector and k be the maximum value of a·b→×c→, then the value of k2-50 is equal to

Options:

The correct answer is:
59

Solution:
a→×b→=i^j^k^3-24012=i^-8-j^6+k^3
=-8i^-6j^+3k^
a → b → c → = a → × b → ⋅ c → ≤ a → × b → c →
⇒ maximum value of a→b→c→=a→×b→c→

⇒k=64+36+9.1=109
⇒k2-50=109-50=59

Topic: Mathematics, Vector Algebra, JEE Main
Consider two vectors \( \vec{a}=3 \hat{i}-2 \hat{j}+4 \hat{k} \) and \( \vec{b}=\hat{j}+2 \hat{k} \). If \( \vec{c} \) is a unit vector and \( k \) be the maximum value of
\( a \cdot(\vec{b} \times \vec{c}) \), then the value of \( k^{2}-50 \) is equal to

Options:

The correct answer is:
59

Solution:
a→×b→=i^j^k^3-24012=i^-8-j^6+k^3

=-8i^-6j^+3k^

a → b → c → = a → × b → ⋅ c → ≤ a → × b → c →

⇒ maximum value of a→b→c→=a→×b→c→

⇒k=64+36+9.1=109

⇒k2-50=109-50=59

Topic: Mathematics, Vector Algebra, JEE Main
Considering a group of positive charges, which of the following statements is correct?

Options:
A Net potential of the system cannot be zero at a point but net electric field can be zero at that point
B Net potential of the system at a point can be zero but net electric field can't be zero at that point
C Both the net potential and the net field can be zero at a point
D Both the net potential and the net electric field cannot be zero at a point

The correct answer is:
Net potential of the system cannot be zero at a point but net electric field can be zero at that point

Solution:
As we know, net potential will be algebraic sum of potential due to each charge, so we can write Vnet=∑kQiri.

As given in the question, Qi and ri are positive, so

∴ Vnet>0

But electric field is a vector quantity, so it can be zero for a given arrangement of charges. For example, electric field at the centre of the two equal charges at certain distance will be zero as electric field due to both charges will be in opposite direction and hence will cancel each other.

Topic: Physics, Electrostatics, JEE Main, JEE Main 2023 (31 Jan Shift 2)
Considering acetic acid dissociates in water, its dissociation constant is \(6.25 \times 10^{-5}\). If \(5 \mathrm{~mL}\) of acetic acid is dissolved in 1 litre water, the solution will freeze at \(-x \times 10^{-2}{ }^{\circ} \mathrm{C}\), provided pure water freezes at \(0^{\circ} \mathrm{C}\).
\(x=\) ______ . (Nearest integer)
Given :
\(\begin{aligned}
& \left(\mathrm{K}_f\right)_{\text {water }}=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} \text {. } \\
& \text { density of acetic acid is } 1.2 \mathrm{~g} \mathrm{~mol}^{-1} \text {. } \\
& \text { molar mass of water }=18 \mathrm{~g} \mathrm{~mol}^{-1} \text {. } \\
& \text { molar mass of acetic acid=60 } \mathrm{g} \mathrm{mol}^{-1} \text {. } \\
& \text { density of water }=1 \mathrm{~g} \mathrm{~cm}^{-3}
\end{aligned}\)
Acetic acid dissociates as \(\mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{\ominus}+\mathrm{H}^{\oplus}\)

Options:

The correct answer is:
19

Solution:
$\begin{aligned} & \text { Mass of } \mathrm{CH}_3 \mathrm{COOH}=\mathrm{V} \times \mathrm{d} \\ & =5 \mathrm{ml} \times 1.2 \mathrm{~g} / \mathrm{ml} \\ & =6 \mathrm{gm} \\ & \mathrm{n}_{\mathrm{CH}_3 \mathrm{COOH}}=\frac{6}{60}=0.1 \mathrm{~mol} \\ & \mathrm{~m}_{\mathrm{CH}_3 \mathrm{COOH}} \approx \mathrm{M}_{\mathrm{CH}_3 \mathrm{COOH}}=\frac{0.1}{1}=0.1 \mathrm{M} \\ & \mathrm{CH}_3 \mathrm{COOH} \rightleftharpoons \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}^{+}\end{aligned}$
C
$\begin{aligned} & \mathrm{C}-\mathrm{C} \alpha \quad \mathrm{C} \alpha \quad \mathrm{C} \alpha \\ & \mathrm{K}_{\mathrm{a}}=\frac{\mathrm{C} \alpha^2}{1-\alpha} \\ & 1-\alpha \approx 1 \Rightarrow \mathrm{K}_{\mathrm{a}}=\mathrm{C} \alpha^2 \\ & \alpha=\sqrt{\frac{\mathrm{Ka}}{\mathrm{C}}}=\sqrt{\frac{6.25 \times 10^{-5}}{0.1}}=25 \times 10^{-3}\end{aligned}$
$\begin{aligned} & \text { V.f. }(\mathrm{i})=1+\alpha(\mathrm{n}-1)=1+\alpha(2-1)=1+\alpha \\ & =1+25 \times 10^{-3}=1.025 \\ & \Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \mathrm{K}_{\mathrm{f}} \mathrm{m} \\ & =(1.025)(1.86)(0.1) \\ & =0.19 \\ & =19 \times 10^{-2}\end{aligned}$

Topic: Chemistry, Solutions, JEE Main, JEE Main 2024 (05 Apr Shift 2)
Considering entropy (S) as a thermodynamic parameter, the criterion for the spontaneity of any process is:

Options:
A $\Delta \mathrm{S}_{\text {system }}+\Delta \mathrm{S}_{\text {surrounding }}>0$
B $\Delta \mathrm{S}_{\text {system }}-\Delta \mathrm{S}_{\text {surrounding }}>0$
C $\Delta \mathrm{S}_{\text {System }}>0$ only
D $\Delta \mathrm{S}_{\text {Surroundings }}>$ only

The correct answer is:
$\Delta \mathrm{S}_{\text {system }}+\Delta \mathrm{S}_{\text {surrounding }}>0$

Solution:
According to Second Law of Thermodynamics, for a spontaneous process, total change in entropy (system + surrounding) should be positive.
Related Theory
(Ref. to Fig. 28.1)
For the latter two cases, the temperature at which the spontaneity changes will be determined by the relative magnitudes of $\triangle S$ and $\Delta H$.

Topic: Chemistry, Thermodynamics (C), JEE Main
Considering $\mathrm{H}_2 \mathrm{O}$ as a weak field ligand, the number of unpaired electrons in $\left[\mathrm{Mn}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ will be (At. No. of $M n=25$ )

Options:
A Two
B Four
C Three
D Five

The correct answer is:
Five

Solution:
$M n^{25} \longrightarrow 3 d^5+4 s^2$
$\mathrm{Mn}^{2+} \longrightarrow 3 d^5$ \(\begin{array}{|l|l|l|l|l|}
\hline 1 & 1 & 1 & 1 & 1 \\
\hline
\end{array}\)
In presence of weak ligand field, there will be no pairing of electrons. So it will form a high spin complex. i.e. the number of unpaired electrons $=5$.

Topic: Chemistry, Coordination Compounds, JEE Main
Considering only the principal values of inverse trigonometric functions, the number of positive real values of x satisfying tan-1(x)+tan-1(2x)=π4 is :

Options:
A More than 2
B 1
C 2
D 0

The correct answer is:
1

Solution:
Let,

fx=tan-1x+tan-12x

Now, differentiating the above function we get,

⇒f'x=11+x2+21+4x2

⇒f'x>0; ∀x∈R

⇒limx→∞fx=π and limx→-∞fx=-π

Hence, fx is monotonically increasing for x∈R and the range of fx is -π,π.

Hence, fx=π4 has exactly one solution for x>0.

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2024 (27 Jan Shift 2)
Considering only the principal values of the inverse trigonometric functions, the domain of the function fx=cos-1x2-4x+2x2+3 is

Options:
A -∞,14
B -14,∞
C -13,∞
D -∞,13

The correct answer is:
-14,∞

Solution:
Given,

fx=cos-1x2-4x+2x2+3

We know that domain of cos-1 is -1,1

So, x2+4x+2x2+3≤1

⇒x2-4x+22≤x2+32

⇒x2-4x+22-x2+32≤0

⇒2x2-4x+5-4x-1≤0

Since 2x2-4x+5>0 as its discriminant D<0

So, -4x-1≤0⇒x≥-14

⇒x∈[-14,∞)

Topic: Mathematics, Functions, JEE Main, JEE Main 2022 (28 Jul Shift 1)
Considering that Δ0>P, the magnetic moment (in BM) of RuH2O62+ would be

Options:

The correct answer is:
0

Solution:
Ru2*=4d6=t2g2,2.2eg0,0 since Δ0>P
No. of unpaired electrons = zero
Magnetic Moment =0

Topic: Chemistry, Coordination Compounds, JEE Main, JEE Main 2020 (05 Sep Shift 2)
Considering the above chemical reaction, identify the product X:

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main, JEE Main 2021 (18 Mar Shift 1)
Considering the above reaction, the major product among the following is:

Options:
A
B
C
D

The correct answer is:

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main, JEE Main 2021 (26 Feb Shift 2)