datavorous/entrance-exam-dataset huggingface dataset:
For a differentiable function fx, if f'2=2 and f'3=1, then the value of limx→0fx2+x+2−f2fx2−x+3−f3 is equal to

Options:
A 2
B 1
C -2
D -1

The correct answer is:
-2

Solution:
Given, limx→0fx2+x+2−f2fx2−x+3−f3

limx→0fx2+x+2−f2x2+x+2-2×x2+x+2-2fx2−x+3−f3x2−x+3-3×x2−x+3-3

= f ′ 2 f ′ 3 lim x→0 x x+1 x x−1 =− 2 1 =−2

Topic: Mathematics, Continuity and Differentiability, JEE Main
For a dilute solution containing $2.5 \mathrm{~g}$ of a non-volatile non-electrolyte solute in $100 \mathrm{~g}$ of water, the elevation in boiling point at 1 atm pressure is $2^{\circ} \mathrm{C}$. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure ( $\mathrm{mm}$ of $\mathrm{Hg}$ ) of the solution is (take $K_{b}=0.76 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ )

Options:
A 724
B 740
C 736
D 718

The correct answer is:
724

Solution:
From Raoult's law,

$\frac{p^{\circ}-p}{p^{\circ}}=\frac{\text { No. of moles of solute }}{\text { No. of moles of solvent }+\text { No. of moles of solute }}$

When the concentration of solute is much lower than the concentration of solvent,

$\begin{array}{l}

\frac{p^{\circ}-p}{p^{\circ}}=\frac{\text { No. of moles of solute }}{\text { No. of moles of solvent }} \\

\frac{760-p}{760}=\frac{2.5 / m}{100 / 18}...(i)

\end{array}$

From elevation in boiling point, $\Delta T_{b}=K_{b} \times m$

$\begin{array}{l}

2=0.76 \times m \\

m=\frac{2}{0.76}...(ii)

\end{array}$

From(i) and (ii), $p=724 \mathrm{~mm}$

Topic: Chemistry, Solutions, JEE Main
For a dimerization reaction,

2A(g)→A2(g)

at 298K,ΔU-=-20kJmol-1,ΔS-=-30JK-1mol-1,then the ΔG- will be …..J.

Options:

The correct answer is:
13538

Solution:
From ΔH=ΔU+ΔngRT

Δ=−20×1000−1×8.314J/mol.K×298K

=−22477.572J

.ΔG=ΔH−TΔS

=−13537.572J

=−13538J

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2020 (05 Sep Shift 2)
For a first order reaction A→B, the rate constant, k=5.5×10-14 s-1. The time required for 67% completion of reaction is x×10-1 times the half life of reaction. The value of x is Nearest integer) (Given : log3=0.4771)

Options:

The correct answer is:
16

Solution:
For first order reaction, t=2.303klogA0At

Now, t67%=2.303klogA0.33A

& t50=2.303klogA0.5A

t67%=1kln11-0.67=t1/2ln2×ln11-23

⇒t67%=1.585×t1/2

X×10-1=1.585

⇒X=15.85

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2022 (28 Jun Shift 1)
For a first order reaction $(A \rightarrow B)$, the temperature $(T)$ dependent rate constant $(k)$ in $\mathrm{s}^{-1}$ was found to follow the equation :
$\log k=\left(-\frac{20}{T}\right)+4$. The activation energy $\left(E_a\right)$ and pre-exponential factor $(A)$ respectively, are

Options:
A $46.06 \mathrm{cal} \mathrm{mol}^{-1}$ and $10^{-4} \mathrm{~s}^{-1}$
B $92.12 \mathrm{cal} \mathrm{mol}^{-1}$ and $10^4 \mathrm{~s}^{-1}$
C $46.06 \mathrm{cal} \mathrm{mol}^{-1}$ and $10^4 \mathrm{~s}^{-1}$
D $9.212 \mathrm{cal} \mathrm{mol}^{-1}$ and $10^{-4} \mathrm{~s}^{-1}$

The correct answer is:
$92.12 \mathrm{cal} \mathrm{mol}^{-1}$ and $10^4 \mathrm{~s}^{-1}$

Solution:
Given, $\log k=\left(-\frac{20}{T}\right)+4$
Since, $\log k=\log A-\frac{E_a}{2.303 R T}$
Here, $\quad \log A=4$ and $-\frac{E_a}{2.303 R}=20$
So, $\quad A=10^{-4} \mathrm{~s}^{-1}$ and $E_a=2.303 \times 2 \times 20$
$E_a=92.12 \mathrm{cal} \mathrm{mol}^{-1}$

Topic: Chemistry, Chemical Kinetics, JEE Main
For a first order reaction, $\mathrm{A} \rightarrow \mathrm{P}, \mathrm{t}_{1 / 2}$ (half-life) is 10 days. The time required for $\frac{1}{4}^{\text {th }}$ conversion of $\mathrm{A}$ (in days) is: $(\ln 2=0.693, \ln 3=1.1)$.

Options:
A
$3.2$

B
$2.5$

C
$4.1$

D
5

The correct answer is:
$4.1$

Solution:
The half life $\mathrm{t}_{1 / 2}=10$ days The decay constant
$$
\mathrm{k}=\frac{0.693}{\mathrm{t}_{1 / 2}}=\frac{0.693}{10 \text { days }}=0.0693 \text { days }^{-1}
$$
The time required for one fourth conversion
$$
\begin{aligned}
&\mathrm{t}=\frac{2.303}{\mathrm{k}} \log _{10} \frac{\mathrm{a}}{\mathrm{a}-\mathrm{x}} \\
&=\frac{2.303}{0.0693 \text { day }^{-1}} \log _{10} \frac{1}{1-(1 / 4)}=4.1 \text { days }
\end{aligned}
$$

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
For a first order reaction A→P, the temperature (T) dependent rate constant k was found to follow the equation log⁡k=-20001T+6.0. The pre-exponential factor A and the activation energy Ea, respectively, are

Options:
A 1.0Ã106 s-1 and 9.2 kJmol-1
B 6.0Â s-1 and 16.6Â kJmol-1
C 1.0Ã106Â s-1 and 16.6Â kJmol-1
D 1.0Ã106 s-1 and 38.3 kJmol-1

The correct answer is:
1.0Ã106 s-1 and 38.3 kJmol-1

Solution:
logk=logA− E a 2.303RT          ...(1)
Also given logk=6.0−( 2000 ) 1 T …(2)
On comparing equations, (1) and (2) logA=6.0⇒A= 10 6 s −1
and E a 2.303R =2000;
⇒ E a =2000×2.303×8.314
=38.29kJmol −1

Topic: Chemistry, Chemical Kinetics, JEE Main
For a first order reaction, $(A) \rightarrow$ products, the concentration of $A$ changes from $0.1 \mathrm{~M}$ to $0.025 \mathrm{~M}$ in $40$ minutes. The rate of reaction when the concentration of $A$ is $0.01 \mathrm{~M}$ is :

Options:
A
$1.73 \times 10^{-5} \mathrm{M} / \mathrm{min}$

B
$3.47 \times 10^{-4} \mathrm{M} / \mathrm{min}$

C
$3.47 \times 10^{-5} \mathrm{M} / \mathrm{min}$

D
$1.73 \times 10^{-4} \mathrm{M} / \mathrm{min}$

The correct answer is:
$3.47 \times 10^{-4} \mathrm{M} / \mathrm{min}$

Solution:
$\begin{aligned}
& \mathrm{k}=\frac{2.303}{40} \log \frac{0.1}{0.025} \\
& \mathrm{k}=\frac{0.693}{20}
\end{aligned}$
For a F.O.R., rate $=k[\mathrm{~A}]$;
rate $=\frac{0.693}{20} \times 10^{-2}=3.47 \times 10^{-4} \mathrm{~M} / \mathrm{min}$.

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2012 (Offline)
For a first order reaction at $27^{\circ} \mathrm{C}$, the ratio of time required for $75 \%$ completion to $25 \%$ completion of reaction is

Options:
A $3.0$
B $2.303$
C $4.8$
D $0.477$

The correct answer is:
$4.8$

Solution:
For a first order reaction,
$$
t=\frac{2.303}{\lambda} \log _{10} \frac{a}{a-x}
$$
Let initial amount of reactant is 100 .
$$
\begin{aligned}
\frac{t_{1}}{t_{2}} &=\frac{\log \frac{100}{100-75}}{\log \frac{100}{100-25}}[\because \lambda \text { remains constant }] \\
&=\frac{\log \frac{100}{25}}{\log \frac{100}{75}}=\frac{\log 4}{\log 4 / 3} \\
&=\frac{\log 4}{\log 4-\log 3} \\
&=\frac{2 \times 0.3010}{2 \times 0.3010-0.4771} \\
&=\frac{0.6020}{0.1249}=4.81
\end{aligned}
$$

Topic: Chemistry, Chemical Kinetics, JEE Main
For a first order reaction at $27^{\circ} \mathrm{C}$, the ratio of time required for $75 \%$ completion to $25 \%$ completion of reaction is

Options:
A $3.0$
B $2.303$
C $4.8$
D $0.477$

The correct answer is:
$4.8$

Solution:
For a first order reaction,
$$
t=\frac{2.303}{\lambda} \log _{10} \frac{a}{a-x}
$$
Let initial amount of reactant is 100 .
$$
\begin{aligned}
\frac{t_{1}}{t_{2}} &=\frac{\log \frac{100}{100-75}}{\log \frac{100}{100-25}}[\because \lambda \text { remains constant }] \\
&=\frac{\log \frac{100}{25}}{\log \frac{100}{75}}=\frac{\log 4}{\log 4 / 3} \\
&=\frac{\log 4}{\log 4-\log 3} \\
&=\frac{2 \times 0.3010}{2 \times 0.3010-0.4771} \\
&=\frac{0.6020}{0.1249}=4.81
\end{aligned}
$$

Topic: Chemistry, Chemical Kinetics, VITEEE, VITEEE 2009
For a first order reaction, if the time taken for completion of 50% of the reaction is 't' second, the time required for completion of 99.9% of the reaction is n 't'. Find the value of n?

Options:

The correct answer is:
10

Solution:
t=0.693k for first order reaction
k=2.303timelogâ¡aa-x
0.693 t 1/2 â=â 2.303 time âlogâ 100 100â99.9
time= 6.903 0.693 t 1/2 ≈ 10t 1/2

Topic: Chemistry, Chemical Kinetics, JEE Main
For a first order reaction, the half-life is 10 mins. How much time in minutes will it take to reduce the concentration of reactant to 25% of its original concentration?

Options:

The correct answer is:
20

Solution:
It takes two half-lives to reduce the concentration of reactant to 25% of its original concentration.
Hence, total time taken =10+10
=20 minutes.

Topic: Chemistry, Chemical Kinetics, JEE Main
For a first order reaction, the half-life period is

Options:
A dependent on the square of the initial concentration
B dependent on first power of initial concentration
C dependent on the square root of initial concentration
D independent on initial concentration

The correct answer is:
independent on initial concentration

Solution:
$t_{1 / 2}=\frac{1}{(a)^{n-1}}$
where, $n=$ order of reaction $a=$ initial concentration For first order reaction,
$$
\begin{aligned}
n &=1 \\
t_{1 / 2} &=\frac{1}{a^{1}-1} \\
a &=\frac{1}{a^{0}}=0
\end{aligned}
$$
Thus, for a first order reaction, $t_{1 / 2}$ is independent on initial concentration.

Topic: Chemistry, Chemical Kinetics, JEE Main
For a first order reaction, the ratio of the time for 75% completion of a reaction to the time for 50% completion is ___________ . (Integer answer)

Options:

The correct answer is:
2

Solution:
T=2.303klog100100-x%
T75%=2.303klog10025
T50%=2.303klog10050
T75% T50%=log4log2=2

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2021 (31 Aug Shift 1)
For a first order reaction, the time required for completion of 90% reaction is 'x' times the half life of the reaction. The value of 'x' is

(Given: ln10=2.303 and log2=0.3010)

Options:
A 1.12
B 2.43
C 3.32
D 33.31

The correct answer is:
3.32

Solution:
ln2t1/2t90%=ln10010=ln10t90%×0.693t1/2=2.303t90%=2.3030693×t1/2=3.32t1/2

Hence, x=3.32

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2022 (24 Jun Shift 2)
For a function fx=2x2+1x (where . denotes the greatest integer function), if 1≤x<4, then

Options:
A the range of f is 4,172
B f is bijective function
C the maximum value of fx is 343
D the minimum value of fx is 4

The correct answer is:
the minimum value of fx is 4

Solution:
We have, fx=2x2+1x
When x∈1,2, then fx=2x2+1⇒Rf=4,10
When x∈2,3, then fx=x2+1⇒Rf=5,10
When x∈3,4, then fx=2x2+13⇒Rf=203,343
∴Rf=4,343
Clearly, for x∈1,2 and x∈2,3 range of fx has common elements
⇒fx is many-one
⇒fx is not bijective

Topic: Mathematics, Functions, JEE Main
For a gas CP-CV=R in a state P and CP-CV=1.10R in a state Q,TP and TQ are the temperatures in two different states P and Q, respectively. Then

Options:
A TP=TQ
B TP<TQ
C TP=0.9TQ
D TP>TQ

The correct answer is:
TP>TQ

Solution:
CP-CV=R for ideal gas and gas behaves as ideal gas at high temperature, so TP>TQ.

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2021 (25 Jul Shift 1)
For a given alkyl group the densities of the halides follow the order

Options:
A $R I \lt R B r \lt R C l$
B $R I \lt R C l \lt R B r$
C $\mathrm{RBr} \lt \mathrm{RI} \lt \mathrm{RCl}$
D $R C I \lt R B r \lt R I$

The correct answer is:
$R C I \lt R B r \lt R I$

Solution:
Density of alkyl halide increases as the size of halogen atom increases $R F \lt R C l \lt R B r \lt R I$

Topic: Chemistry, Haloalkanes and Haloarenes, JEE Main
For a given chemical reaction A→B at 300 K the free energy change is -49.4 kJ mol-1 and the enthalpy of reaction is 51.4 kJ mol-1. The entropy change of the reaction is _________JK-1 mol-1.

Options:

The correct answer is:
336

Solution:
Given chemical reaction:

A⟶T=300 KB[ΔG]P,T=-49.4 kJ/mol

ΔHrxn=51.4 kJ/mol

ΔSrxn=?

⇒From the relation [ΔG]P,T=ΔH-TΔS

⇒ΔSrxn=ΔHrxn-[ΔG]P,TT

=[51.4-(-49.4)]×1000300JmolK

⇒ΔSrxn=336JmolK

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2021 (20 Jul Shift 2)
For a given distribution of marks mean is 35.16 and its standard deviation is $19.76$. The co-efficient of variation is

Options:
A $\frac{35.16}{19.76}$
B $\frac{19.76}{35.16}$
C $\frac{35.16}{19.76} \times 100$
D $\frac{19.76}{35.16} \times 100$

The correct answer is:
$\frac{19.76}{35.16} \times 100$

Solution:
Coefficient of variation $=\frac{\text { S.D. }}{\text { Mean }} \times 100=\frac{19.76}{35.16} \times 100$.

Topic: Mathematics, Statistics, JEE Main
For a given electric current the drift velocity of conduction electrons in a copper wire is $v_d$ and their mobility is $\mu$. When the current is increased at constant temperature

Options:
A $v_d$ increases, $\mu$ remains the same
B $v_d$ remains the same, $\mu$ increases
C $v_d$ decreases, $\mu$ remains the same
D $v_d$ remains the same, $\mu$ decreases

The correct answer is:
$v_d$ increases, $\mu$ remains the same

Solution:
We know that, mobility of electron,
$\mu=\frac{e \tau}{m}$
$\Rightarrow \quad \mu \propto \tau \quad[\tau \rightarrow$ relaxation time $]$
Thus, for constant temperature, $\mu$ remains the same.
Since, $I=n e A v_d$
$\Rightarrow \quad v_d \propto I$
Hence, when current is increased, then drift velocity $\left(v_d\right)$ also increases.

Topic: Physics, Current Electricity, KCET, KCET 2023
For a given equilibrium reaction, addition of inert argon gas at constant volume can shift the equilibrium in
$\mathrm{N}_2(g)+3 \mathrm{H}_2(g) \rightleftharpoons 2 \mathrm{NH}_3(g)$

Options:
A forward direction
B backward direction
C remain unaffected
D initially forward and then backward direction

The correct answer is:
remain unaffected

Solution:
The expression to $K_p$ of an equilibrium state at
a given temperature for addition of inert gases is
$K_p=K_n\left(\frac{p}{\Sigma n}\right)^{\Delta n_g}
$where, volume is constant, $\left(\frac{p}{\Sigma n}\right)$ remains constant for any value of $\Delta n_g$.
So, the given equilibrium reaction remains unaffected, if addition of inert Ar gas at constant volume.

Topic: Chemistry, Chemical Equilibrium, TS EAMCET, TS EAMCET 2020 (14 Sep Shift 1)
For a given one mole of ideal gas kept at 6.5 atm in a container of capacity 2.463 L, the Avogadro proportionality constant for the hypothesis is (see figure)

Options:
A 0.406
B 2.46
C 22.4
D none of these

The correct answer is:
2.46

Solution:
According to Avogadro hypothesis
Vâ n
V=nRTP
Proportionality constant=RTP
Using PV = nRT
â´ RT P = V n = 2.463 1

Topic: Chemistry, States of Matter, JEE Main
For a given reaction, ΔH=35.5 kJmol-1 and ΔS=83.6 Jk-1mol-1 . The reaction is spontaneous at
(Assume that ΔH and ΔS do not vary with tempearature)

Options:
A T > 425 K
B All temperatures
C T > 398 K
D T < 525 K

The correct answer is:
T > 425 K

Solution:
As Both ΔH and ΔS are positive so to keep ΔG=-ve High Temp is favourable so that process become spontaneous.
ΔG=ΔH-T.ΔS
T=ΔHΔS=35.5×100083.6>424.6K
T>425K

Topic: Chemistry, Thermodynamics (C), JEE Main
For a given reaction $t_{1 / 2}=\frac{1}{K a}$. . The order of the reaction is

Options:
A 1
B 0
C 3
D 2

The correct answer is:
2

Solution:
$t_{1 / 2}=\frac{1}{K a}$
for second order reactions.

Topic: Chemistry, Chemical Kinetics, JEE Main
For a given series LCR circuit it is found that maximum current is drawn when value of variable capacitance is $25 \mathrm{nF}$. If resistance of $200 \Omega$ and $100 \mathrm{mH}$ inductor is being used in the given circuit. The frequency of ac source is ______ $\times 10^3 \mathrm{~Hz}$. (given $\pi^2=10$ )

Options:

The correct answer is:
10

Solution:
$\begin{aligned} & \mathrm{f}_0=\frac{1}{2 \pi \sqrt{\mathrm{L} \times \mathrm{C}}} \\ & \mathrm{f}_0=\frac{1}{2 \pi \sqrt{100 \times 10^{-3} \times 25 \times 10^{-9}}} \\ & =\frac{1}{2 \pi \sqrt{25 \times 10^{-10}}} \\ & =\frac{1}{2 \pi \times 5} \times 10^5 \times \sqrt{10} \mathrm{~Hz} \\ & =\frac{100}{10} \times 10^3 \mathrm{~Hz} \\ & \mathrm{f}_0=10 \times 10^3 \mathrm{~Hz}\end{aligned}$

Topic: Physics, Alternating Current, JEE Main, JEE Main 2024 (06 Apr Shift 2)
For a good quality cement, the ratio of lime to the total of the oxides of Si,Al and Fe should be as close as to

Options:
A 4
B 1
C 2
D 3

The correct answer is:
2

Solution:
Cement is made by strongly heating a mixture of lime stone and clay in a rotatory kiln. Lime stone and clay are finely powdered, and a little water is added. For a good quality cement, the ratio of silica (SiO2) to alumina (Al2O3) should be between 2.5 and 4 and the ratio of lime (CaO) to the total of the oxides of silicon (SiO2) aluminium (Al2O3) and iron (Fe2O3) should be as close as possible to 2. If lime is in excess, the cement cracks during setting. On the other hand, if lime is less than the required, the cement is weak in strength. Therefore, a proper composition of cement must be maintained to get cement of good quality.

Topic: Chemistry, s Block Elements, JEE Main, JEE Main 2023 (08 Apr Shift 2)
For a G.P., if Sn=4n-3n3n, then, t2= ________

Options:
A 19
B 29
C 79
D 49

The correct answer is:
49

Solution:
Given Sn=4n-3n3n
Tn=Sn-Sn-1
T2=S2-S1=42-3232-4-33
=79-13⇒49

Topic: Mathematics, Sequences and Series, JEE Main
For a group of 50 male workers, the mean and the standard deviation of their daily wages are Rs. 630 and Rs. 90 respectively and for a group of 40 female workers these are Rs. 540 and Rs. 60 respectively. Then, the standard deviation of all these 90 workers is

Options:
A 60
B 70
C 80
D 90

The correct answer is:
90

Solution:
n1=50, x¯1=630, σ1=90
n2=40, x¯2=540, σ2=60
x ¯ = 50×630+40×540 50+40 = 3150+2160 9
= 5310 9 =590
Now, d1=x¯1−x¯, d2=x¯2−x¯
d 1 =630−590=40
d 2 =540−590=−50
S.D. of the 90 workers
= n1(σ12+d12)+n2(σ22+d22)n1+n2
= 50(8100+1600)+40(3600+2500)90
= 5(9700)+4(6100)9
= 48500+244009
= 729009=8100=90

Topic: Mathematics, Statistics, JEE Main
For a group of 100 observations, the arithmetic mean and standard deviation are 8 and $\sqrt{10.5}$ respectively. The mean and standard deviation of 50 items selected from these 100 observations are 10 and 2 respectively. Then the standard deviation of the remaining 50 observation is

Options:
A 2
B 3
C 3.5
D 4

The correct answer is:
3

Solution:
Let $\sigma_1$ and $\bar{x}_1$ are the standard deviation and mean of 50 observation and $\sigma_2$ and $\bar{x}_2$ are standard deviation one mean of another 50 observation respectively
$$
\begin{gathered}
\bar{x}_1=10, \sigma_1=2, n_1=50 \\
\bar{x}=\text { Mean of } 100 \text { observation }=8 \\
\sigma=\text { SD of } 100 \text { observation }=\sqrt{10.5} \\
\bar{x}=\frac{50 \bar{x}_1+50 \bar{x}_2}{50+50} \\
800=500+50 \bar{x}_2 \\
\bar{x}=6 \\
\sigma^2=\frac{n_1\left(\sigma_1\right)^2+n_2\left(\sigma_2\right)^2}{n_1+n_2}+\frac{n_1 n_2\left(\bar{x}_1-\bar{x}_2\right)^2}{\left(n_1+n_2\right)^2} \\
10.5=\frac{50(4)+50\left(\sigma_2\right)^2}{100}+\frac{2500(10-6)^2}{(100)^2} \\
10.5=\frac{200+50\left(\sigma_2\right)^2}{100}+\frac{25 \times 16}{100} \\
\Rightarrow \quad 1050=600+50\left(\sigma_2\right)^2 \\
\Rightarrow \quad \sigma_2^2=\frac{1050-600}{50}=9 \Rightarrow \sigma_2=3
\end{gathered}
$$

Topic: Mathematics, Statistics, TS EAMCET, TS EAMCET 2019 (04 May Shift 2)
For a hypothetical H-like atom, which follows Bohr model, some spectral lines are observed as shown. If it is known that line E belong to visible region., then the line, which possibly belong to U.V. region is/are? (n1 is not necessarily the ground state)

Options:
A B and D
B D only
C C only
D A only

The correct answer is:
A only

Solution:
Line E belong to visible so n1 will be 2.
For E, ΔE=+13.6×z214-19=13.6×z2×0.138
For D, ΔE=13.6×z219-136=13.6×z2×0.083
For C, ΔE=13.6×z2116-136=13.6×z2×0.034
For A, ΔE=13.6×z214-125=13.6×z2×0.21
For B, ΔE=13.6×z219-125=13.6×z2×0.07
For ΔE for A is larger than that of E, so it means line A belong to U.V. region.

Topic: Chemistry, Structure of Atom, JEE Main
For a∈ℂ, let A={z∈ℂ:Re(a+z¯)>Im(a¯+z)} and B={z∈ℂ:Re(a+z¯)<Im(a¯+z)}. Then among the two statements:
(S1) : If Re(a),Im(a)>0, then the set A contains all the real numbers
(S2) : If Re(a),Im(a)<0, then the set B contains all the real numbers,

Options:
A Only S2 is true
B only S1 is true
C Both are true
D Both are false

The correct answer is:
Both are false

Solution:
Let a=x1+iy1, a¯=x1-iy1

And z=x2+iy2, z¯=x2-iy2

Now,

a+z¯=x1+x2+iy1-y2

and a¯+z=x1+x2+iy2-y1

Now according to the question,

Re(a+z¯)=x1+x2 and

Im(a¯+z)=y2-y1

A=z:x1+x2>y2-y1=z:x1+y1+x2>y2
B=z:x1+x2<y2-y1=z:x1+y1+x2<y2

If y2=0 and x1,y1>0 then

A={z:x2>-x1+y1}

A covers a part of negative real axis and therefore, does not contain whole real axis

Similarly if y2=0, and x1,y1<0, then
B={z:x2<-x1+y1}
∴ B covers part of positive real axis and therefore does not cover whole real axis.

Hence, both are false.

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2023 (11 Apr Shift 2)
For a linear plot of log xm versus log p in a Freundlich adsorption isotherm, which of the following statements is correct? (k and n are constants)

Options:
A Only 1n appears as the slope.
B log⁡1n appears as the intercept.
C Both k and 1n appear in the slope term.
D 1n appears as the intercept.

The correct answer is:
Only 1n appears as the slope.

Solution:
According to Freundlich isotherm,

xm=k.p1n {Taking log}

log⁡xm=log⁡k+1nlog⁡P

y=c+mx (Straight line equation)

So, intercept is log k and slope is=1n.

Topic: Chemistry, Surface Chemistry, JEE Main, JEE Main 2016 (03 Apr)
For a liquid the vapour pressure is given by
log10⁡P=-400T+10
Vapour pressure of the liquid is 10 x  mm Hg at 400 K . The value of x will be?

Options:

The correct answer is:
9

Solution:
log10â¡P=-400T+10=-400400+10=9
P= 10 9 âmmâHg
x = 9 .

Topic: Chemistry, Solutions, JEE Main
For a material medium, the values of refractive index for violet and red colours are given as nv=1.56 and nr=1.44. The dispersive power of a prism made out of this material is

Options:
A 0.06
B 0.24
C 0.03
D none of these

The correct answer is:
0.24

Solution:
Dispersive power of a prism is given as,

ω=nv−nr(nv+nr2)−1=1.56−1.44(1.56+1.442)−1=0.24

Topic: Physics, Ray Optics, JEE Main
For a \( \mathrm{d}^{4} \) metal ion in an octahedral field, the correct electronic configuration is :

Options:
A \( \mathrm{t}_{2_{\mathrm{g}}}^{3} \mathrm{e}_{\mathrm{g}}^{1} \) when \( \Delta_{\mathrm{O}} < \mathrm{P} \)
B \( \mathrm{t}_{2_{\mathrm{g}}}^{3} \mathrm{e}_{\mathrm{g}}^{1} \) when \( \Delta_{\mathrm{O}}>\mathrm{P} \)
C \( \mathrm{t}_{2_{\mathrm{g}}}^{4} \mathrm{e}_{g}^{0} \) when \( \Delta_{\mathrm{O}} < \mathrm{P} \)
D \( \mathrm{e}_{\mathrm{g}}^{2} \mathrm{t}_{2_{\mathrm{g}}}^{2} \) when \( \Delta_{\mathrm{O}} < \mathrm{P} \)

The correct answer is:
\( \mathrm{t}_{2_{\mathrm{g}}}^{3} \mathrm{e}_{\mathrm{g}}^{1} \) when \( \Delta_{\mathrm{O}} < \mathrm{P} \)

Solution:
For d4 configuration if Δ0<P the electronic configuration is t2g3, eg1.

Topic: Chemistry, Coordination Compounds, JEE Main
For a matrix A, A=aij3×3≠0 . If each element aij is multiplied by ki-j, we obtain matrix B, such that αA+βB=0, then the value of α+β is equal to

Options:

The correct answer is:
0

Solution:
Let, A=a11a12a13a21a22a23a31a32a33, then

B=a11a12ka13k2ka21a22a23kk2a31ka32a33
B=a11a12ka13k2ka21a22a23kk2a31ka32a33
B=1k3k2a11ka12a13k2a21ka22a23k2a31ka32a33=a11a12a13a21a22a23a31a32a33=A
⇒-A+B=0
On comparison, α+β=0

Topic: Mathematics, Determinants, JEE Main
For a matrix $A, A I=A$ and $A A^{\top}=1$ is true for

Options:
A If $A$ is a square matrix
B If $\mathrm{A}$ is a non singular matrix
C If $A$ is a symmetric matrix
D If $A$ is any matrix

The correct answer is:
If $A$ is a square matrix

Solution:
For a matrix $\mathrm{A}, \mathrm{AI}=\mathrm{A}, \mathrm{AA}^{\mathrm{T}}=\mathrm{I}$
Then, A is square matrix.

Topic: Mathematics, Matrices, JEE Main
For a matrix A, if A2=A and B=I-A, then AB+BA+I-I-A2 is equal to (where, I is the identity matrix of the same order of matrix A)

Options:
A B
B A
C AB
D I

The correct answer is:
A

Solution:
AB+BA+I-I-A2
=AI-A+I-AA+I-I-2A+A2
=A-A2+A-A2+I-I+2A-A2
=A-A+A-A+I-I+2A-A=A

Topic: Mathematics, Matrices, JEE Main
For a metal ion, the calculated magnetic moment is 4.90 BM. This metal ion has _______ number of unpaired electrons

Options:

The correct answer is:
4

Solution:
The magnetic moment of the metal ion can be calculated using

μ=nn+2BM

n=number of unpaired electrons.

μ=4.90 BM

4.90BM=(n)(n+2) BM

24 = (n) (n + 2)

n = 4

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2023 (10 Apr Shift 2)