datavorous/entrance-exam-dataset huggingface dataset:
For a reaction \( \mathrm{X}+\mathrm{Y} \rightarrow \mathrm{Z} \), rate \( \propto[\mathrm{X}] . \) What is (i) molecularity and (ii) order of reaction?

Options:
A (i) \( 2 \), (ii) \( 1 \)
B (i) \( 2 \), (ii) \( 2 \)
C (i) \( 1 \), (ii) \( 1 \)
D (i) \( 1 \), (ii) \( 2 \)

The correct answer is:
(i) \( 2 \), (ii) \( 1 \)

Solution:
X+Y→Z
Rate ∝X ; hence order of reaction = 1
Molecularity of reaction = 2

Topic: Chemistry, Chemical Kinetics, JEE Main
For a reaction,
N2g+3H2(g)⟶2NH3(g), identify di-hydrogen (H2) as a limiting reagent in the following reaction mixtures.

Options:
A 28 g of N2+6 g of H2
B 35 g of N2+8 g of H2
C 56 g of N2+10 g of H2
D 14 g of N2+4 g of H2

The correct answer is:
56 g of N2+10 g of H2

Solution:
N2g+3H2g→2NH3g

moles of N2stoichiometry of N2≠mole of H2stoichiometry of H2
Whose ratio is less called limiting reagent.

(a) mole of N2=2828=1
molesstiochiometry ratio of N2=11=1
mole of H2=62=3
molesstiochiometry ratio of H2=33=1
there is no limiting reagent.

(b) mole of N2=3528=1.25
molesstiochiometry ratio of N2=1.251 limiting reagent
mole of H2=82=4
molesstiochiometry ratio of H2=43=1.33

(c) mole of N2=5628=2
molesstiochiometry ratio of N2=21=2
mole of H2=102=5
molesstiochiometry ratio of H2=53=1.66 limiting reagent.

(d) mole of N2=1428=0.5
molesstiochiometry ratio of N2=0.51=0.5 limiting reagent
mole of H2=42=2
molesstiochiometry ratio of H2=23=1.66

∴  Answer is 3 .

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Main, JEE Main 2019 (09 Apr Shift 1)
For a reaction of order n, the unit of the rate constant is :

Options:
A mol1-nL1-ns
B mol1-nL2ns-1
C mol1-nLn-1 s-1
D mol1-nL1-ns-1

The correct answer is:
mol1-nLn-1 s-1

Solution:
Rate =k[A]n
comparing units
(mol/ℓ)sec=kmolℓn
⇒k=mol(1-n)ℓ(n-1)s-1

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2021 (27 Jul Shift 1)
For a reaction of type $\mathrm{A}+\mathrm{B} \rightarrow$ products, it is observed that doubling concentration of $\mathrm{A}$ causes the reaction rate to be four times as great, but doubling amount of $\mathrm{B}$ does not affect the rate. The unit of rate constant is

Options:
A $\mathrm{s}^{-1}$
B $\mathrm{s}^{-1} \mathrm{~mol} \mathrm{~L}^{-1}$
C $\mathrm{s}^{-1} \mathrm{~mol}^{-1} \mathrm{~L}$
D $\mathrm{s} \mathrm{s}^{-1} \mathrm{~mol}^{-2} \mathrm{~L}^{2}$

The correct answer is:
$\mathrm{s}^{-1} \mathrm{~mol}^{-1} \mathrm{~L}$

Solution:
Let the initial rate be $R$ and order with respect to $A$ be $x$ and $B$ be $y$. Thus, rate lawcan be written as, rate, $R=[A]^{x}[B]^{y}$...(i)
After doubling the concentration of $A$, rate becomes $4 R$
$$
4 R=[2 A]^{x}[B]^{y}...(ii)
$$
After doubling the concentration of $B$, rate remains $R$,
$R=[A]^{x}[2 B\}$...(iii)
From Eq. (i) and (ii), we get
$$
\frac{R}{4 R}=\left(\frac{1}{2}\right)^{x} \Rightarrow\left(\frac{1}{2}\right)^{2}=\left(\frac{1}{2}\right)^{x}
$$
So, $x=2$
From Eq. (i) and (iii), we get
$$
\frac{R}{R}=\left[\frac{1}{2}\right]^{y} \Rightarrow\left(\frac{1}{1}\right)^{0}=\left(\frac{1}{2}\right)^{y}
$$
So, $Y=0$
Hence, the rate lawis, rate $\mathrm{R}=[A]^{2}[B]^{0}$ This clearly shows that the order of this reaction is 2 and for second order reaction units of rate constant are $\mathrm{mol}^{-1} \mathrm{Ls}^{-1}$.

Topic: Chemistry, Chemical Kinetics, VITEEE, VITEEE 2010
For a reaction scheme A→ k1 B→ k2 C, if the net rate of formation of B is set to be zero then the concentration of B is given by:

Options:
A k1+k2 [A]
B k1k2 [A]
C k1k2 [A]
D k1-k2 [A]

The correct answer is:
k1k2 [A]

Solution:
dBdt= Rate of formation of B- Rate of dissociation of B

dBdt=k1A-k2B

Given; dBdt=0

k1A-k2B=0

B=k1Ak2

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2019 (08 Apr Shift 2)
For a reaction scheme \( \mathrm{A} \stackrel{\mathrm{k}_{1}}{\longrightarrow} \mathrm{B} \stackrel{\mathrm{k}_{2}}{\longrightarrow} \mathrm{C} \), if the net rate of formation of \( \mathrm{B} \) is set to be zero then the concentration of \( \mathrm{B} \) is given by:

Options:
A \( \left(\mathrm{k}_{1}+\mathrm{k}_{2}\right)[\mathrm{A}] \)
B \( \mathrm{k}_{1} \mathrm{k}_{2}[\mathrm{~A}] \)
C \( \left(\frac{k_{1}}{k_{2}}\right)[\mathrm{A}] \)
D \( \left(\mathrm{k}_{1}-\mathrm{k}_{2}\right)[\mathrm{A}] \)

The correct answer is:
\( \left(\frac{k_{1}}{k_{2}}\right)[\mathrm{A}] \)

Solution:
dBdt= Rate of formation of B- Rate of dissociation of B

dBdt=k1A-k2B

Given; dBdt=0

k1A-k2B=0

B=k1Ak2

Topic: Chemistry, Chemical Kinetics, JEE Main
For a reaction taking place in three steps at same temperature, overall rate constant K=K1K2K3. If Ea1,Ea2 and Ea3 are 40,50 and 60 kJ/mol respectively, the overall Ea is _________ kJ/mol.

Options:

The correct answer is:
30

Solution:
According to Arrhenius concept,

K=Ae-Ea/RT

Given that,

K=K1.K2/K3

Using Arrhenius equation we can expand above equation as

e-Ea(net)/RT=e-Ea1/RT×e-Ea2/RTe-Ea3/RT

Hence,

Ea(net)=Ea1+Ea2-Ea3⇒Ea(net)=40+50-60=30kJ/mol

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2024 (29 Jan Shift 1)
For a reaction X+Y=2Z,1.0mol of X,1.5mol of Y and 0.5mol of Z were taken in a 1L vessel and allowed to react. At equilibrium, the concentration of Z was 1.0molL-1. the equilibrium constant of the reaction is …………x15. The value of x is …….

Options:

The correct answer is:
16

Solution:
x + y= 2 z

t=0 1mol 1.5mol 0.5mol

Since moles of Z are increased at equilibrium therefore reaction goes in forward direction to attain the
equilibrium.

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2020 (05 Sep Shift 2)
For a reactions $A+B \rightarrow$ product, it was found that rate of reaction increases four times if concentration of ' $A$ ' is doubled, but the rate of reaction remains unaffected. If concentration of ' $B$ ' is doubled. Hence, the rate law for the reaction is

Options:
A rate $=k[A][B]$
B rate $=k[A]^2$
C rate $=k[A]^2[B]^1$
D rate $=k[A]^2[B]^2$

The correct answer is:
rate $=k[A]^2$

Solution:
Greater are the concentrations of the reactants, faster is the reaction. Conversely, as the concentrations of the reactants decreases, the rate of reaction also decreases.

Topic: Chemistry, Chemical Kinetics, JEE Main
For a reactions A+B→ product, it was found that rate of reaction increases four times if concentration of 'A' is doubled. But the rate of reaction remains unaffected, if concentration of 'B' is doubled. Hence, the rate law for the reaction is

Options:
A rate=k[A][B]
B rate=k[A]2
C rate=k[A]2[B]1
D rate=k[A]2[B]2

The correct answer is:
rate=k[A]2

Solution:
Let the rate of reaction depends on xth power of [A] . Then
r1=k[A]x and r2=k[2A]x
∴r1r2=[A]x[2A]x=14=122(∵r2=4r1)
∴x=2 . As the reaction rate does not depend upon the concentration of B. Hence, the correct rate law will be rate=K[A]2[B]o or =K[A]2

Topic: Chemistry, Chemical Kinetics, NEET
For a real number $x$, let $[x]$ denote the largest integer less than or equal to $x$, and let $\{x\}=x-[x] .$ The number of solutions $x$ to be equation $[x]\{x\}=5$ with $0 \leq x \leq 2015$ is

Options:
A 0
B 3
C 2008
D 2009

The correct answer is:
2009

Solution:
$$
\begin{array}{l}
\{x\}=x-[x] \\
{[x]\{x\}=5 \quad f \neq 0} \\
\quad \text { I } f=5 \\
\quad 0 < f < 1
\end{array}
$$
Possible solutions
$$
\left(6+\frac{5}{6}\right),\left(7+\frac{5}{7}\right), \ldots \ldots .,\left(2014+\frac{5}{2014}\right)=2009
$$

Topic: Mathematics, Functions, JEE Main
For a regular polygon, let $\mathrm{r}$ and $\mathrm{R}$ be the radii of the inscribed and the circumscribed circles. A false statement among the following is

Options:
A
There is a regular polygon with $\frac{r}{R}=\frac{1}{\sqrt{2}}$

B
There is a regular polygon with $\frac{r}{R}=\frac{2}{3}$

C
There is a regular polygon with $\frac{r}{R}=\frac{\sqrt{3}}{2}$

D
There is a regular polygon with $\frac{r}{R}=\frac{1}{2}$

The correct answer is:
There is a regular polygon with $\frac{r}{R}=\frac{2}{3}$

Solution:
$$
\mathrm{r}=\frac{\mathrm{a}}{2} \cot \frac{\pi}{\mathrm{n}}
$$
'a' is side of polygon.
$$
\begin{aligned}
& R=\frac{a}{2} \operatorname{cosec} \frac{\pi}{n} \\
& \frac{r}{R}=\frac{\cot \frac{\pi}{n}}{\operatorname{cosec} \frac{\pi}{n}}=\cos \frac{\pi}{n} \\
& \cos \frac{\pi}{n} \neq \frac{2}{3} \text { for any } n \in N
\end{aligned}
$$

Topic: Mathematics, Properties of Triangles, JEE Main, JEE Main 2010
For a reversible process, necessary condition is

Options:
A In the whole cycle of the system, the loss of any type of heat energy should be zero
B That the process should be too fast
C That the process should be slow so that the working substance should remain in thermal and mechanical equilibrium with the surroundings
D The loss of energy should be zero and it should be quasi static

The correct answer is:
The loss of energy should be zero and it should be quasi static

Solution:
For a reversible process $\int \frac{d Q}{T}=0$

Topic: Physics, Thermodynamics, JEE Main
For a reversible reaction A⇌B, the ∆H forward reaction =20 kJ mol–1. The activation energy of the uncatalyzed forward reaction is 300 kJ mol–1. When the reaction is catalysed keeping the reactant concentration same, the rate of the catalysed forward reaction at 27∘C is found to be same as that of the uncatalyzed reaction at 327∘C. The activation energy of the catalysed backward reaction is _______ kJ mol–1.

Options:

The correct answer is:
130

Solution:
As per the question

To determine the activation energy of the catalyzed backward reaction, we need to use the Arrhenius equation and the given information.

The Arrhenius equation is given by:

k=Ae(-Ea/RT)

Ae-300×103600×R=Ae-Ea300×R

⇒1032=Ea300

⇒Ea=150×103J mol–1

⇒Ea=150 kJ mol–1

∴ Activation energy of catalysed backward reaction

Energy of activation for backward reactionEb=Ea-∆H

=150-20=130 kJ mol–1

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2023 (15 Apr Shift 1)
For a reversible reaction, if the concentration of reactants is increased, the equilibrium constant of the reaction

Options:
A increases
B remains constant
C decreases
D depends on amount of reactant

The correct answer is:
remains constant

Solution:
Equilibrium constant is independent of change in concentration of reactants or products. It depends only on temperature. Hence, the correct option is (b).

Topic: Chemistry, Chemical Equilibrium, AP EAMCET, AP EAMCET 2020 (21 Sep Shift 2)
For a reversible reaction, if the concentration of the reactants is reduced to half, the equilibrium constant will be.........

Options:
A doubled
B halved
C reduced to one-fourth
D remains same

The correct answer is:
remains same

Solution:
The value of equilibrium constant \(K_{\mathrm{eq}}\) is independent of initial concentrations of reactants and products. So, options (a), (b) and (c) are not correct, i.e. so, for a reversible reaction, if the concentration of the reactants is reduced to half, the equilibrium constant will be remains same.

Topic: Chemistry, Chemical Equilibrium, JEE Main
For a reversible reaction, if the concentration of the reactants is reduced to half, the equilibrium constant will be.........

Options:
A doubled
B halved
C reduced to one-fourth
D remains same

The correct answer is:
remains same

Solution:
The value of equilibrium constant \(K_{\mathrm{eq}}\) is independent of initial concentrations of reactants and products. So, options (a), (b) and (c) are not correct, i.e. so, for a reversible reaction, if the concentration of the reactants is reduced to half, the equilibrium constant will be remains same.

Topic: Chemistry, Chemical Equilibrium, AP EAMCET, AP EAMCET 2020 (18 Sep Shift 1)
For a sample size of 10 observations x1, x2, ...x10, if ∑i=110xi-52=350 and ∑i=110xi-2=60, then the variance of xi is

Options:

The correct answer is:
26

Solution:
∑i=110xi-2=60⇒∑xi=80
∑i=110xi2-10xi+25=350
⇒∑i=110xi2=350+800-250=900
Variance is 90010-82=90-64=26

Topic: Mathematics, Statistics, JEE Main
For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is

Options:
A $2$
B $\frac{1}{2}$
C $\frac{1}{\sqrt{2}}$
D $\sqrt{2}$

The correct answer is:
$\frac{1}{2}$

Solution:
For a moving satellite kinetic energy $=\frac{G M m}{2 r}$
Potential energy $=\frac{-G M m}{r}$
$\therefore \frac{\text { Kineticenergy }}{\text { Potentialenergy }}=\frac{1}{2}$

Topic: Physics, Gravitation, JEE Main
For a screw gauge least count of the main scale is 1 mm. Find the minimum number of divisions on the circular scale so that it can measure a diameter of 5âÎ¼m

Options:
A 500
B 100
C 50
D 200

The correct answer is:
200

Solution:
Least count of screw gauge =5 μm
L.C=pitchno. of div on circular  scale
5 μm=1 mmN
N=200

Topic: Physics, Experimental Physics, NEET
For a second order reaction, 2A→ Products, a plot of log t1/2 vs log a (where, 'a' is initial concentration) will give an intercept equal to which one of the following?

Options:
A 1k
B log⁡12k
C log⁡1k
D log⁡k

The correct answer is:
log⁡k

Solution:
Kinetic equation for second order reaction is
k=1t1(a-x)-1a
when x=a2
t=t1/2
So t1/2=1k×1a
log⁡t1/2=-log⁡k-log⁡a
Here intercept = log k

Topic: Chemistry, Chemical Kinetics, JEE Main
For a sequence tn , if Sn=52n-1 then tn=…

Options:
A 52n+1
B 5×2n4
C 52n-1
D 2×2n-15

The correct answer is:
52n-1

Solution:
We have, Sn=52n-1
We know that an=Sn-Sn-1
=52n-1-52n-1-1
=52n-2n-1
=52n-1

Topic: Mathematics, Sequences and Series, JEE Main
For a series LCR circuit with R=100 Ω, L=0.5 mH and C=0.1 pF connected across 220 V-50 Hz AC supply, the phase angle between
current and supplied voltage and the nature of the circuit is:

Options:
A 0°, resistive circuit
B ≈90°, predominantly inductive circuit
C 0°, resonance circuit
D ≈90°, predominantly capacitive circuit

The correct answer is:
≈90°, predominantly capacitive circuit

Solution:
R=100 Ω

XL=ωL=50π×10-3

XC=1ωC=1011100π

XC≫XL

and XC-XL≫R

Topic: Physics, Alternating Current, JEE Main, JEE Main 2021 (20 Jul Shift 2)
For a set of five true or false questions, no student has written the all correct answers and no two students have given the same sequence of answers. The maximum number of students in the class for this to be possible is

Options:
A 30
B 31
C 32
D 16

The correct answer is:
31

Solution:
Each of the five questions can be solved in two ways.
$\therefore$ Maximum number of wrong answers.
$$
=(2 \times 2 \times 2 \times 2 \times 2)-1=31
$$

Topic: Mathematics, Permutation Combination, JEE Main
For a simple harmonic motion in a mass spring system shown, the surface is frictionless. When the mass of the block is 1 kg, the angular frequency is ω1. When the mass block is 2 kg the angular frequency is ω2. The ratio ω2ω1 is :

Options:
A 2
B 12
C 2
D 12

The correct answer is:
12

Solution:
In case of SHM in spring-block system, angular frequency is given by, ω=km.

Therefore,

ω2ω1=m1m2=12

Topic: Physics, Oscillations, JEE Main, JEE Main 2023 (30 Jan Shift 2)
For a simple pendulum, a graph is plotted between its kinetic energy \( (K . E .) \) and potential energy \( (P . E .) \) against its
displacement \( d \). Which one of the following represents these correctly?

Options:
A
B
C
D

The correct answer is:

Solution:
In simple harmonic motion, kinetic energy is given by,

K.E.=12kA2-d2.

So, the graph is a downward parabola.

Here, A=amplitude of simple harmonic motion

and d=displacement from mean position in simple harmonic motion
and

P.E.=12kd2.

So, the graph is an upward parabola.

At mean, d=0⇒K.E.=12KA2

and potential energy is, P.E.=0.

At extreme positions,

d=A⇒K.E.=0.

Topic: Physics, Oscillations, JEE Main
For a simple pendulum, a graph is plotted between its kinetic energy (K.E.) and potential energy (P.E.) against its displacement d. which one of the following represents these correctly? (graphs are schematic and not drawn to scale)

Options:
A
B
C
D

The correct answer is:

Solution:
For simple pendulum performing the simple harmonic motion, displacement, y=Asinωtv=dydt=ωAcosωt=ωA2-y2⇒K.E.=12mv2,
This represents a downward parabola, Y=±A, K.E.=12ω2m A2-A2=0. P.E.=12mω2y2. This represents an upward parabola. On plotting graphs of potential energy & kinetic energy.

Topic: Physics, Oscillations, JEE Main, JEE Main 2015 (04 Apr)
For a singular matrix A=aij2Ã2 and the matrix B=1121, if AB=BA, then 3a11a124 is equal to

Options:

The correct answer is:
12

Solution:
A=a11a12a21a22 and AB=BA (given)
i.e. a11a12a21a221121=1121a11a12a21a22
⇒a11+2a12=a11+a21 …i
a11+a12=a12+a22 …ii
a21+2a22=2a11+a21 …iii
and a21+a22=2a12+a22 …iv
From ii,a11=a22
From iv,a21=2a12
Since, A=0⇒a11a22-a21a12=0

⇒a112=2a122⇒a11a122=2

Hence, 3a11a124=12

Topic: Mathematics, Determinants, JEE Main
For a solid rod, the Young's modulus of elasticity is 3.2×1011 N m-2 and density is 8×103 kg m-3. The velocity of longitudinal wave in the rod will be

Options:
A 145.75×103 m s-1
B 3.65×103 m s-1
C 18.96×103 m s-1
D 6.32×103 m s-1

The correct answer is:
6.32×103 m s-1

Solution:
The velocity of longitudinal wave in the rod will be, v=Yρ=3.2×10118×103

=2×10310

≈6.32×103 m s-1

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2023 (31 Jan Shift 2)
For a solid with the following structure, the coordination number of the point B is

Options:
A 3
B 4
C 5
D 6

The correct answer is:
6

Solution:
It is evident from figure that B occupies octahedral voids and thus, coordination number is 6.

Topic: Chemistry, Solid State, JEE Main
For a somatic cell with 2n = 4. Which of the following is true?
(Note:- G1 - growth phase 1, G2 - growth phase 2, M- Metaphase, P - Prophase and T - Telophase)

Options:
A Number of chromatids in G2 = 4, Number of chromosome in G1 = 4
B Number of chromatids in G1 = 8, Number of sister chromatids in T = 8
C Number of chromatids in P = 8, Number of chromosomes in G2 = 4
D Number of Chromatids in G2 = 4, Number of Chromosomes in M = 8

The correct answer is:
Number of chromatids in P = 8, Number of chromosomes in G2 = 4

Solution:
Somatic cells undergo mitotic cell division so, there is no reduction division and chromosome number remain the same. At the beginning of mitosis, there are 4 chromosomes, and each chromosome replicates during S phase and form sister chromatids. So the number of chromosomes is 4, and chromatids are 4 x 2 = 8. Hence, during G1 phase
Chromatids number - 4
Chromosomes number- 4
During S phase
Chromatids number - 8
Chromosome number – 4
During the G2 phase
Chromatids number - 8
Chromosomes number - 4

Topic: Biology, Cell Cycle and Cell Division, NEET
For a sparingly soluble salt $A_{p} B_{q}$, the relationship of its solubility product $L s \rightarrow K_{s p}$ with its solubility (S) is

Options:
A $\quad \mathrm{Ls} \rightarrow \mathrm{K}_{\mathrm{sp}}=\mathrm{S}^{\mathrm{pq}}(\mathrm{pq})^{\mathrm{P}+\mathrm{q}}$
B $\mathrm{Ls}=\mathrm{S}^{\mathrm{p}+\mathrm{q}} \cdot \mathrm{p}^{\mathrm{p}} \mathrm{q}^{\mathrm{q}}$
C $\quad \mathrm{Ls} \rightarrow \mathrm{K}_{\mathrm{sp}}=\mathrm{S}^{\mathrm{p}+\mathrm{q}} \cdot \mathrm{p}^{\mathrm{q}} \mathrm{q}^{\mathrm{p}}$
D $\quad \mathrm{Ls} \rightarrow \mathrm{K}_{\mathrm{sp}}=\mathrm{S}^{\mathrm{pq}} \mathrm{p}^{\mathrm{P}} \mathrm{q}^{\mathrm{q}}$

The correct answer is:
$\mathrm{Ls}=\mathrm{S}^{\mathrm{p}+\mathrm{q}} \cdot \mathrm{p}^{\mathrm{p}} \mathrm{q}^{\mathrm{q}}$

Solution:
$\mathrm{ApBq} \rightleftharpoons \mathrm{pA}_{\mathrm{Sp}}^{+}+\mathrm{qB}_{\mathrm{Sq}}$

Let the solubility be $\mathrm{S} \mathrm{mol} /$ liter Thus,

\begin{aligned}

\mathrm{K}_{\mathrm{sp}} &=\left[\mathrm{A}^{+}\right]^{\mathrm{p}}\left[\mathrm{B}^{-}\right]^{\mathrm{q}} \\

&=[\mathrm{Sp}]^{\mathrm{p}}[\mathrm{Sq}]^{\mathrm{q}}=\mathrm{p}^{\mathrm{p} \cdot} \mathrm{q}^{\mathrm{q}}(\mathrm{S})^{\mathrm{p}+\mathrm{q}}

\end{aligned}

Topic: Chemistry, Ionic Equilibrium, JEE Main
For a sparingly soluble salt \(\mathrm{AB}_2\), the equilibrium concentrations of \(\mathrm{A}^{2+}\) ions and \(B^{-}\)ions are \(1.2 \times 10^{-4} \mathrm{M}\) and \(0.24 \times 10^{-3} \mathrm{M}\), respectively. The solubility product of \(\mathrm{AB}_2\) is :

Options:
A \(6.91 \times 10^{-12}\)
B \(0.276 \times 10^{-12}\)
C \(27.65 \times 10^{-12}\)
D \(0.069 \times 10^{-12}\)

The correct answer is:
\(6.91 \times 10^{-12}\)

Solution:
$\begin{aligned} & \mathrm{AB}_{2(\mathrm{~s})} \rightleftharpoons \mathrm{A}_{(\mathrm{aq})}^{+2}+2 \mathrm{~B}_{(\mathrm{aq})}^{-} \\ & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{A}^{+2}\right]\left[\mathrm{B}^{-}\right]^2 \\ &=1.2 \times 10^{-4} \times\left(2.4 \times 10^{-4}\right)^2 \\ &=6.91 \times 10^{-12} \mathrm{M}^3\end{aligned}$

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2024 (09 Apr Shift 2)
For a specific wavelength 670 nm of light coming from a galaxy moving with velocity v, the observed wavelength is 670.7 nm. The value of v is

Options:
A 3×108 m s-1
B 3.13×105 m s-1
C 3×1010 m s-1
D 4.48×105 m s-1

The correct answer is:
3.13×105 m s-1

Solution:
The apparent change in the wavelength is given by

Δλλ=vc

⇒v=cΔλλ=3×108×670.7-670670=3.13×105 m s-1

Hence, velocity of the planet is 3.13×105 m s-1.

Topic: Physics, Waves and Sound, JEE Main, JEE Main 2022 (26 Jun Shift 2)
For a spontaneous change, free energy change $\Delta G$ is

Options:
A Positive
B Negative
C Zero
D Can be positive or negative

The correct answer is:
Negative

Solution:
For spontaneous change $\Delta G=-v e$.

Topic: Chemistry, Thermodynamics (C), JEE Main
For a spontaneous reaction the $\Delta \mathrm{G}$, equilibrium constant $(\mathrm{K})$ and $\mathrm{E}_{\mathrm{cell}}^{\circ}$ will be respectively

Options:
A
-ve, >1, +ve

B
+ve, >1, -ve

C
-ve, < 1, -ve
D
-ve, >1, -ve

The correct answer is:
-ve, >1, +ve

Solution:
No Solution Available

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2005
For a strong electrolyte, a plot of molar conductivity against (concentration) \({ }^{1 / 2}\) is a straight line, with a negative slope, the correct unit for the slope is

Options:
A \(\mathrm{S} \mathrm{cm}{ }^2 \mathrm{~mol}^{-3 / 2} \mathrm{~L}^{-1 / 2}\)
B \(\mathrm{S} \mathrm{cm}^2 \mathrm{~mol}^{-3 / 2} \mathrm{~L}^{1 / 2}\)
C \(\mathrm{S} \mathrm{cm}^2 \mathrm{~mol}^{-3 / 2} \mathrm{~L}\)
D \(\mathrm{S} \mathrm{cm}^2 \mathrm{~mol}^{-1} \mathrm{~L}^{1 / 2}\)

The correct answer is:
\(\mathrm{S} \mathrm{cm}^2 \mathrm{~mol}^{-3 / 2} \mathrm{~L}^{1 / 2}\)

Solution:
$\Lambda_{\mathrm{m}}=\Lambda_{\mathrm{m}}^{\circ}-\mathrm{A} \sqrt{\mathrm{C}}$
Units of $\mathrm{A} \sqrt{\mathrm{C}}=\mathrm{S} \mathrm{cm}^2 \mathrm{~mole}^{-1}$
Uits of $\mathrm{A}=\mathrm{S} \mathrm{cm}^2 \mathrm{~mole}^{-3 / 2} \mathrm{~L}^{1 / 2}$

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2024 (04 Apr Shift 2)
For a substance the average life for Î± - emission is 1620 years and for Î² emission is 405 years. After how much time the 14th of the material remains by simultaneous emission?

Options:
A 648 years

B 324 years

C 449 years

D 810 years

The correct answer is:
449 years

Solution:
Taug= 1/ Î»
1/ Î» Î± = 1620 â Î» Î± = 1 1620 years
1/ Î» Î² = 405 â Î» Î² = 1 405 years
N = No e - Î» Î± + Î» Î² t
No 4 = No e - Î» Î± + Î» Î² t
2ln2 = 1 1620 + 1 405 t = t 324
t = 648 ln2 = 449 years

Topic: Physics, Nuclear Physics, NEET
For a suitably chosen real constant a, let a function, f : R--a→R be defined by fx=a-xa+x. Further supposed that for any real number x≠-a,and f(x)≠-a, fofx=x. Then f-12 is equal to :

Options:
A 13
B -13
C -3
D 3

The correct answer is:
3

Solution:
fof(x)=a-fxa+f(x)=x

⇒a-ax1+x=f(x)

⇒a(1-x)1+x=a-xa+x (∵a=1)

so f(x)=1-x1+x

f-12=3

Topic: Mathematics, Functions, JEE Main, JEE Main 2020 (06 Sep Shift 2)
For a system to be in equilibrium, the torques acting on it must balance. This is true only if the torques are taken about

Options:
A the centre of the system
B the centre of mass of the system
C any point on the system
D any point on the system or outside it

The correct answer is:
the centre of the system

Solution:
For a system to be in equilibrium torques on the system must be equal to zero This is true only if the torques are taken about centre of mass as the whole mass of the object seams to be situated at centre of mass

Topic: Physics, Rotational Motion, JEE Main