datavorous/entrance-exam-dataset huggingface dataset:
Following figure shows two processes A and B for a gas. If ∆QA and ∆QB are the amount of heat absorbed by the system in two cases, and ∆UA and ∆UB are changes in internal energies, respectively, then:

Options:
A ∆QA< ∆QB; ∆UA< ∆UB
B ∆QA= ∆QB; ∆UA= ∆UB
C ∆QA> ∆QB; ∆UA= ∆UB
D ∆QA> ∆QB; ∆UA> ∆UB

The correct answer is:
∆QA> ∆QB; ∆UA= ∆UB

Solution:
⇒ ∆UA=∆UB as change in internal energy depends on initial and final state and not the path.
⇒ By comparing area under the curve,
WA>WB
From Q=U+W
We can say, QPA>QPB

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2019 (09 Apr Shift 1)
Following four solutions are prepared by mixing different volumes of $\mathrm{NaOH}$ and $\mathrm{HCl}$ of different concentrations, $\mathrm{pH}$ of which one of them will be equal to 1 ?

Options:
A
$55 \mathrm{~mL} \frac{\mathrm{M}}{10} \mathrm{HCl}+45 \mathrm{~mL} \frac{\mathrm{M}}{10} \mathrm{NaOH}$

B
$75 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{HCl}+25 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{NaOH}$

C
$100 \mathrm{~mL} \frac{\mathrm{M}}{10} \mathrm{HCl}+100 \mathrm{~mL} \frac{\mathrm{M}}{10} \mathrm{NaOH}$

D
$60 \mathrm{~mL} \frac{\mathrm{M}}{10} \mathrm{HCl}+40 \mathrm{~mL} \frac{\mathrm{M}}{10} \mathrm{NaOH}$

The correct answer is:
$75 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{HCl}+25 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{NaOH}$

Solution:
$75 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{HCl}+25 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{NaOH}$ $25 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{NaOH}$ will neutralise $25 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{HCl}$ $75-25=50 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{HCl}$ will remain. Total volume will be $75+25=100 \mathrm{~mL}$ $50 \mathrm{~mL} \frac{\mathrm{M}}{5} \mathrm{HCl}$ is diluted to $100 \mathrm{~mL}$ $\left[\mathrm{H}^{+}\right]=[\mathrm{HCl}]=\frac{\mathrm{M}}{5} \times \frac{50}{100}=\frac{\mathrm{M}}{10}$
$$
\mathrm{pH}=-\log _{10}\left[\mathrm{H}^{+}\right]=-\log _{10} \frac{\mathrm{M}}{10}=1
$$

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
Following gaseous reaction is undergoing in a vessel $\mathrm{C}_2 \mathrm{H}_4+\mathrm{H}_2 \Rightarrow \mathrm{C}_2 \mathrm{H}_6 . \Delta \mathrm{H}=-32.7 \mathrm{Kcal}$

Options:
A Increase of temperature
B By reducing temperature
C By removing some hydrogen
D By adding some $\mathrm{C}_2 \mathrm{H}_6$

The correct answer is:
By reducing temperature

Solution:
Exothermic reaction, favcured by low temperature.

Topic: Chemistry, Chemical Equilibrium, JEE Main
Following gates section is connected in a complete suitable circuit.
$\text { For which of the following combination, bulb will glow (ON) : }$

Options:
A $\mathrm{A}=0, \mathrm{~B}=0, \mathrm{C}=0, \mathrm{D}=1$
B $\mathrm{A}=0, \mathrm{~B}=1, \mathrm{C}=1, \mathrm{D}=1$
C $\mathrm{A}=1, \mathrm{~B}=0, \mathrm{C}=0, \mathrm{D}=0$
D $\mathrm{A}=1, \mathrm{~B}=1, \mathrm{C}=1, \mathrm{D}=0$

The correct answer is:
$\mathrm{A}=1, \mathrm{~B}=0, \mathrm{C}=0, \mathrm{D}=0$

Solution:
Bulb will glow if bulb have potential drop on it. One end of bulb must be at high (1) and other must be at low (0).
Option (3) satisfy this condition

Topic: Physics, Semiconductors, JEE Main, JEE Main 2024 (05 Apr Shift 1)
Following Hund's rule which element contains six unpaired electron

Options:
A $\mathrm{Fe}$
B $\mathrm{Co}$
C $\mathrm{Ni}$
D $\mathrm{Cr}$

The correct answer is:
$\mathrm{Ni}$

Solution:
According to the Aufbau's principle, electron will be first enters in those orbital which have least energy.

Topic: Chemistry, Structure of Atom, JEE Main
Following is the diagrammatic representation of the sectional view of cochlea. Select the correct option regarding it.

Options:
A A - Scala vestibuli-responsible for maintenance of balance of the body and posture.
B B - Organ of Corti-filled with perilymph that ends at the oval window.
C C - Scala tympani - terminates at the round window which opens to the middle ear.
D D - Reissner's membrane - nerve impulses are generated against it.

The correct answer is:
C - Scala tympani - terminates at the round window which opens to the middle ear.

Solution:
No Solution Available

Topic: Biology, Neural Control and Coordination, AIIMS, AIIMS 2018 (26 May)
Following is the graph between logâ t 1 2 and loga ( a= initial concentration of reactant) for a given reaction at 27oC. Hence, order of the reaction is

(Here, t 1/2 is half-life)

Options:
A 0
B 1
C 2
D 3

The correct answer is:
0

Solution:
t 1 2 ∝ 1 a n−1 for nth order reaction.
t 1 2 =k a 1−n
log t 1 2 = 1−n loga+logk
Thus, graph between log t 1 2 and loga is linear slope.
1−n =tan 45 o
n=0

Topic: Chemistry, Chemical Kinetics, JEE Main
Following Kjeldahl's method, 1 g of organic compound released ammonia, that neutralised 10 mL of 2MH2SO4. The percentage of nitrogen in the compound is _______ %.

Options:

The correct answer is:
56

Solution:
H2SO4+2NH3→NH42SO4

Equivalent of NH3=Equivalent of H2SO4

moles of NH3nNH3×1=2×10×21000

=4×10-2 moles

Moles of nitrogen=4×10-2 moles

Weight of Nitrogen=4×10-2×14=56×10-2 gm

Percent of nitrogen=56×10-2weight of the compound×100=56×10-21×100=56%

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (01 Feb Shift 2)
Following statements are given
(1) The average kinetic energy of a gas molecule decreases when the temperature is reduced.
(2) The average kinetic energy of a gas molecule increases with increase in pressure at constant temperature.
(3) The average kinetic energy of a gas molecule decreases with increases in volume.
(4) Pressure of a gas increases with increase in temperature at constant volume.
(5) The volume of gas decreases with increase in temperature.

Choose the correct answer from the options given below :

Options:
A (1) and (4) only
B (1), (2) and (4) only
C (2) and (4) only
D (1), (2) and (5) only

The correct answer is:
(1) and (4) only

Solution:
Average kinetic energy of a gas molecule is given by KEavg =32KT.

Pressure of a gas in terms of rms speed is given by P=13ρvrms2.

As KEavg∝T, so 1 is correct, 2 is incorrect, statement 3 can not be said.

Note : Statement (4) is correct only if we consider it at constant volume and not constant pressure. Ideally, this question must be bonus but most appropriate answer is option (A)

Topic: Physics, Kinetic Theory of Gases, JEE Main
Following statements are given
(1) The average kinetic energy of a gas molecule decreases when the temperature is reduced.
(2) The average kinetic energy of a gas molecule increases with increase in pressure at constant temperature.
(3) The average kinetic energy of a gas molecule decreases with increases in volume.
(4) Pressure of a gas increases with increase in temperature at constant volume.
(5) The volume of gas decreases with increase in temperature.

Choose the correct answer from the options given below :

Options:
A (1) and (4) only
B (1), (2) and (4) only
C (2) and (4) only
D (1), (2) and (5) only

The correct answer is:
(1) and (4) only

Solution:
Average kinetic energy of a gas molecule is given by KEavg =32KT.

Pressure of a gas in terms of rms speed is given by P=13ρvrms2.

As KEavg∝T, so 1 is correct, 2 is incorrect, statement 3 can not be said.

Note : Statement (4) is correct only if we consider it at constant volume and not constant pressure. Ideally, this question must be bonus but most appropriate answer is option (A)

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2022 (25 Jul Shift 1)
Following statements are the characteristics of "X". Read the statements carefully and conclude "X" stands for.

I. "X" should not show too much variation in productivity from year to year.
II. "X" must be either resistant or resilient to occasional disturbances.
III. "X" must also be resistant to invasions by alien species.

Options:
A Stable species
B Stable community
C Less species diversity
D Stable biomass

The correct answer is:
Stable community

Solution:
Various groups of interacting organisms in a particular region are included in community. A community is said to be stable when its populations size of all species remains same over a period of time. In other words, the population is not showing much change or variation and is able to withstand any hardships or changes in the environment. A stable community is not affected by invasion of other organisms including alien species. It resists loss of biodiversity.
Species is a group of similar organisms which breed true to produce fertile offsprings. Biomass is the total dry weight of an organism.

Topic: Biology, Biodiversity and its Conservation, NEET
Following statements regarding the periodic trends of chemical reactivity of the alkali metals and the halogens are given. Which of these statements gives the correct picture?

Options:
A
The reactivity decreases in the alkali metals but increases in the halogens with increase in atomic number down the group

B
In both the alkali metals and the halogens the chemical reactivity decreases with increase in atomic number down the group

C
Chemical reactivity increases with increase in atomic number down the group in both the alkali metals and halogens

D
In alkali metals the reactivity increases but in the halogens it decreases with increase in atomic number down the group

The correct answer is:
In alkali metals the reactivity increases but in the halogens it decreases with increase in atomic number down the group

Solution:
No Solution Available

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Main, JEE Main 2006
Food preservatives prevent microbial growth and avoid spoilage of food. Which is the commonly used preservative

Options:
A Salts of sorbic acid and propanoic acid .
B Vegetable oils and sodium benzoate (C6H5COONa) .
C Table salt and sugar
D All of these

The correct answer is:
All of these

Solution:
The commonly used preservatives added in the food are table salt, sugar, vegetable oils and sodium benzoate, C6H5COONa . Sodium benzoate is used in limited quantities and is metabolised in the body. Salts of sorbic acid and propanoic acid are also used as preservatives.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
Food preservatives prevent spoilage of food due to microbial growth. The most commonly used preservatives are

Options:
A table salt, sugar
B vegetable oils and sodium benzoate
C C6H5COONa
D all of the above

The correct answer is:
all of the above

Solution:
Table salt (NaCl), sugar, sodium nitrite, spices, vinegar, and alcohol have been used to preserve foods for centuries. Sodium benzoate, calcium propionate, and potassium sorbate are used to prevent microbial growth that causes spoilage and to slow changes in color, texture, and flavor.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
For 0<θ<π/2, if the eccentricity of the hyperbola x2−y2cosec2θ=5 is 7 times eccentricity of the ellipse x2cosec2θ+y2=5, then the value of θ is:

Options:
A π6
B 5π12
C π3
D π4

The correct answer is:
π3

Solution:
Given:

The equation of hyperbola is, x2-y2cosec2θ=5

⇒x25-y25sin2θ=1

And the equation of ellipse is x2cosec2θ+y2=5

⇒x25sin2θ+y25=1 here a<b as sin2θ≤1

So, eH=1+sin2θ & eE=1-sin2θ

Also given, eH=7eE

⇒1+sin2θ =71-sin2θ

⇒1+sin2θ =7-7sin2θ

⇒8sin2θ =6

⇒sinθ =32

⇒θ=π3

Topic: Mathematics, Hyperbola, JEE Main, JEE Main 2024 (01 Feb Shift 1)
For α,β∈0,π2 let 3sin(α+β)=2sin(α-β) and a real number k be such that tanα= tanβ. Then the value of k is equal to

Options:
A -5
B 5
C 23
D -23

The correct answer is:
-5

Solution:
Given:

3sinα+β=2sinα-β

⇒sinα+βsinα-β=23

⇒sinα+β+sinα-βsinα+β-sinα-β=2+32-3

⇒2sinαcosβ2cosαsinβ=-5

⇒tanαtanβ=-5

⇒tanα=-5tanβ

⇒k=-5

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2024 (30 Jan Shift 2)
For 0<a<1, the value of the integral ∫0πdx1-2acosx+a2 is :

Options:
A π2π+a2
B π2π-a2
C π1-a2
D π1+a2

The correct answer is:
π1-a2

Solution:
Given: I=∫0πdx1-2acosx+a2 ...i

⇒I=∫0πdx1-2acosπ-x+a2

⇒I=∫0πdx1+2acosx+a2 ...ii

Adding i and ii,

⇒2I=∫0πdx1-2acosx+a2 +∫0πdx1+2acosx+a2

⇒2I=∫0π11-2acosx+a2 +11+2acosx+a2dx

⇒2I=∫0π1+2acosx+a2+1-2acosx+a21+a22-2acosx2dx

⇒2I=∫0π21+a21+a22-2acosx2dx

⇒2I=2∫0π221+a21+a22-4a2cos2xdx

⇒I=∫0π221+a2·sec2x1+a22·sec2x-4a2dx

⇒I=∫0π221+a2·sec2x1+a22+1+a22tan2x-4a2dx

⇒I=∫0π221+a2·sec2x1+a2-2a2+1+a22tan2xdx

⇒I=∫0π22·1+a2·sec2x1+a22·tan2x+1-a22dx

⇒I=∫0π22·sec2x1+a2·dxtan2x+1-a21+a22

Let, tanx=t

⇒sec2xdx=dt

⇒I=21+a2∫0∞dtt2+1-a21+a22

⇒I=21+a2×11-a21+a2tan-1t1-a21+a20∞

⇒I=21-a2tan-1∞-tan-10

⇒I=21-a2π2-0

⇒I=π1-a2

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2024 (27 Jan Shift 2)
For $\lambda>0$, let $\theta$ be the angle between the vectors $\vec{a}=\hat{i}+\lambda \hat{j}-3 \hat{k}$ and $\vec{b}=3 \hat{i}-\hat{j}+2 \hat{k}$. If the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are mutually perpendicular, then the value of (14 cos $\theta)^2$ is equal to

Options:
A 50
B 40
C 25
D 20

The correct answer is:
25

Solution:
$\begin{aligned} & (\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=0, \lambda>0 \\ & |\vec{a}|^2-|\vec{b}|^2=0 \rightarrow 1+\lambda^2+9=9+1+4 \\ & \therefore \lambda=2, \cos \theta=\frac{\vec{a}-\vec{b}}{|\vec{a}| \cdot|\vec{b}|}=\frac{3-\lambda-6}{\sqrt{14} \cdot \sqrt{14}} \\ & 14 \cos \theta=3-8=-5 \\ & \therefore(14 \cos \theta)^2=25\end{aligned}$

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2024 (04 Apr Shift 2)
For θ∈[0,π], let f(θ)=sin(cosθ) and g(θ)=cos(sinθ). Let a=max0≤θ≤πfθ,b=min0≤θ=πfθ,c=max0≤9≤πgθ and d=min0<θ<πgθ. The correct
inequalities satisfied by a, b, c, d are

Options:
A b<d<c<a
B d<b<a<c
C b<d<a<c
D b<a<d<c

The correct answer is:
b<d<a<c

Solution:
f(θ)=sin(cosθ)

g(θ)=cos(sinθ)

f'(θ)=cos(cosθ)(-sinθ)<0∀θ∈[0,π]

∴f(θ) decreases monotonically

∴a=maxf(θ)=f(0)=sin1

b=minf(θ)=f(π)=-sin1

g'(θ)=-sin(sinθ)cosθ

g(θ)=1;g(π)=1;gπ2=cos1

∴c=maxg(θ)=1

d=ming(θ)=cos1

∴b<d<a<c

Topic: Mathematics, Application of Derivatives, JEE Main
For $0 \leq x \leq \frac{\pi}{2}$, the value of
$$
\int_0^{\sin ^2 x} \sin ^{-1}(\sqrt{t}) d t+\int_0^{\cos ^2 x} \cos ^{-1}(\sqrt{t}) d t \text { equals : }
$$

Options:
A
$\frac{\pi}{4}$

B
0

C
1

D
$-\frac{\pi}{4}$

The correct answer is:
$\frac{\pi}{4}$

Solution:
Consider
$$
\int_0^{\sin ^2 x} \sin ^{-1}(\sqrt{t}) d t+\int_0^{\cos ^2 x} \cos ^{-1}(\sqrt{t}) d t
$$
Let $\mathrm{I}=f(x)$ after integrating and putting the limits.
$$
\begin{aligned}
&f^{\prime}(x)=\sin ^{-1} \sqrt{\sin ^2 x}(2 \sin x \cos x)-0 \\
&\quad+\cos ^{-1} \sqrt{\cos ^2 x}(-2 \cos x \sin x)-0 \\
&\therefore f^{\prime}(x)=0 \Rightarrow f(x)=\mathrm{C} \quad \text { (constant) }
\end{aligned}
$$
Now, we find $f(x)$ at $x=\frac{\pi}{4}$
$$
\begin{aligned}
\therefore \mathrm{I} &=\int_0^{1 / 2} \sin ^{-1} \sqrt{t} d t+\int_0^{1 / 2} \cos ^{-1} \sqrt{t} d t \\
&=\int_0^{1 / 2}\left(\sin ^{-1} \sqrt{t}+\cos ^{-1} \sqrt{t}\right) d t \\
&=\int_0^{1 / 2} \frac{\pi}{2} d t=\frac{\pi}{4}=\mathrm{C} \\
\therefore \quad f(x)=\frac{\pi}{4} \\
\therefore \text { Required integration }=\frac{\pi}{4}
\end{aligned}
$$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2013 (25 Apr Online)
For $0 \leq x \leq \pi$, the area bounded by $y=x$ and $y=x+\sin x$, is

Options:
A $2$
B $4$
C $2 \pi$
D $4 \pi$

The correct answer is:
$2$

Solution:
The curves $y=x$ and $y=x+\sin x$ intersect at $(0,0)$ and $(\pi, \pi)$. Hence area bounded by the two curves
$\begin{aligned}& =\int_0^\pi(x+\sin x) d x-\int_0^\pi x d x=\int_0^\pi \sin x d x \\
& =[-\cos x]_0^\pi=-\cos \pi+\cos 0=-(-1)+(1)=2\end{aligned}$

Topic: Mathematics, Area Under Curves, JEE Main
For $1 \mathrm{~mol}$ of an ideal gas at a constant temperature $T$, the plot of $(\log P)$ against $(\log V)$ is a ( $P$ : Pressure, $V$ : Volume)

Options:
A
Straight line parallel to $x$-axis.

B
Straight line with a negative slope.

C
Curve starting at origin.

D
Straight line passing through origin.

The correct answer is:
Straight line with a negative slope.

Solution:
According to Boyle's law, $P V=$ constant
$\therefore \log P+\log V=$ constant
$\log P=-\log V+$ constant
Hence, the plot of $\log P$ vs $\log V$ is straight line with negative slope.

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2012 (07 May Online)
For 1 molal aqueous solution of the following compounds, which one will show the highest freezing point?

Options:
A CoH2O3Cl3.3H2O
B CoH2O6Cl3
C CoH2O5ClCl2.H2O
D CoH2O4Cl2Cl.2H2O

The correct answer is:
CoH2O3Cl3.3H2O

Solution:
Depression in freezing point ΔTf=i.kf.m
For one molal soloution ; ΔTf∝i
i is least for  [Co(H2O)3Cl3].3H2O ; So ΔTf is least but freezing point is high.

Topic: Chemistry, Solutions, JEE Main, JEE Main 2018 (08 Apr)
For -π2≤x≤π2, the number of points of intersection of curves y=cos⁡x and y=sin⁡3x is

Options:
A 0
B 1
C 2
D 3

The correct answer is:
3

Solution:
y=cos⁡x; y=sin⁡3x
⇒cos⁡x=cosπ2-3x⇒x=2nπ±π2-3x
⇒4x=2nπ+π2, -2x=2nπ-π2
⇒x=nπ2+π8, x=kπ+π4
⇒x=4n+1π8, x=4k+1π4
n=0⇒x=π8, k=0⇒x=π4
n=-1⇒x=-3π8
⇒x=-3π8,π8 and π4

Topic: Mathematics, Trigonometric Equations, JEE Main
For (2n+1) observations $x_1,-x_1, x_2,-x_2, \ldots \ldots x_n,-x_n$ and 0 where $x$ 's are all distinct. Let S.D. and M.D. denote the standard deviation and median respectively. Then which of the following is always true

Options:
A S.D. $ \lt $ M.D.
B S.D. $\gt$ M.D.
C $S.D. = M.D.$
D Nothing can he said in general about the relationship

The correct answer is:
S.D. $\gt$ M.D.

Solution:
On arranging the given observations in ascending order, we get
All negative terms $\underbrace{O}_{(n+1)^n t e r m}$ All positive terms
The median of given observations $=(n+1)^{\text {th }}$ term $=0$ $\therefore \mathrm{S} . \mathrm{D} .\gt\mathrm{M}$ D.

Topic: Mathematics, Statistics, JEE Main
For $a>0, t \in\left(0, \frac{\pi}{2}\right)$, let $x=\sqrt{a^{\sin ^{-1} t}}$ and $y=\sqrt{a^{\cos ^{-1} t}}$, Then, $1+\left(\frac{d y}{d x}\right)^2$ equals :

Options:
A
$\frac{x^2}{y^2}$

B
$\frac{y^2}{x^2}$

C
$\frac{x^2+y^2}{y^2}$

D
$\frac{x^2+y^2}{x^2}$

The correct answer is:
$\frac{x^2+y^2}{x^2}$

Solution:
$$
\begin{aligned}
& \text { Let } x=\sqrt{a^{\sin ^{-1} t}} \\
& \Rightarrow \quad x^2=a^{\sin ^{-1} t} \Rightarrow 2 \log x=\sin ^{-1} t \cdot \log a \\
& \Rightarrow \quad \frac{2}{x}=\frac{\log a}{\sqrt{1-t^2}} \cdot \frac{d t}{d x} \\
& \Rightarrow \quad \frac{2 \sqrt{1-t^2}}{x \log a}=\frac{d t}{d x}
\end{aligned}
$$
Now, let $y=\sqrt{a^{\cos ^{-1} t}}$
$$
\begin{aligned}
& \Rightarrow 2 \log y=\cos ^{-1} t \cdot \log a \\
& \Rightarrow \quad \frac{2}{y} \cdot \frac{d y}{d x}=\frac{-\log a}{\sqrt{1-t^2}} \cdot \frac{d t}{d x}
\end{aligned}
$$
$\Rightarrow \frac{2}{y} \cdot \frac{d y}{d x}=\frac{-\log a}{\sqrt{1-t^2}} \times \frac{2 \sqrt{1-t^2}}{x \log a} \quad$ (from(1)
$$
\Rightarrow \frac{d y}{d x}=-\frac{y}{x}
$$
Hence, $1+\left(\frac{d y}{d x}\right)^2=1+\left(\frac{-y}{x}\right)^2=\frac{x^2+y^2}{x^2}$

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2013 (22 Apr Online)
For a $1^{\text {st }}$ order reaction if concentration is doubled then rate of reaction becomes

Options:
A doubles
B half
C four times
D remains same.

The correct answer is:
doubles

Solution:
For a first order reaction, rate $=k[A]$, when concentration of $A$ is doubled, the rate becomes double.

Topic: Chemistry, Chemical Kinetics, AIIMS, AIIMS 2010
For a âABC, if Bâ¡3,0, Câ¡-3,0 and the angle BAC is 90Â°, then the locus of the centroid of the triangle is

Options:
A x2+y2=1
B x2+y2=2
C 9x2+y2=1
D 9x2+y2=4

The correct answer is:
x2+y2=1

Solution:
Let, Aa,b and Gh,k.

Now, A,G,O are collinear

âh=2.0+a3âa=3h and similarily b=3k.
Now, a,b lies on the circle x2+y2=9âa2+b2=1

Hence, locus of the centroid is x2+y2=1

Topic: Mathematics, Straight Lines, JEE Main
For a ∆ABC the vertices are A0,3, B0,12 and Cx,0. If the circumcircle of ΔABC touches the x-axis, then the area (in sq. units) of the ∆ABC is

Options:
A 36
B 27
C 30
D 24

The correct answer is:
27

Solution:
OAÃOB=OC2
â3Ã12=x2
âx=6
Centre C'=6,152
â AC'B=2â ACB=2Î¸
Area of âACB=12Ã9Ã6=27 sq. units

Topic: Mathematics, Circle, JEE Main
For $\mathrm{a}, \mathrm{b}>0$, let $f(x)=\left\{\begin{array}{cc}\frac{\tan ((\mathrm{a}+1) x)+\mathrm{b} \tan x}{x}, & x < 0 \\ 3, & x=0 \\ \frac{\sqrt{\mathrm{a} x+\mathrm{b}^2 x^2}-\sqrt{\mathrm{a} x}}{\mathrm{~b} \sqrt{\mathrm{a}} x \sqrt{x}}, & x>0\end{array}\right.$
be a continous function at $x=0$. Then $\frac{\mathrm{b}}{\mathrm{a}}$ is equal to :

Options:
A 6
B 4
C 5
D 8

The correct answer is:
6

Solution:
$\begin{aligned} & \lim _{x \rightarrow 0} f(x)=f(0)=3 \\ & \lim _{x \rightarrow 0^{+}} \frac{\sqrt{a x+b^2 x^2}-\sqrt{a x}}{b \sqrt{a} x \sqrt{x}}=3 \\ & \lim _{x \rightarrow 0^{+}} \frac{a x+b^2 x^2-a x}{b \sqrt{a} x^{3 / 2}\left(\sqrt{a x+b^2 x^2}+\sqrt{a x}\right)} \\ & \lim _{x \rightarrow 0^{+}} \frac{b^2}{b \sqrt{a}\left(\sqrt{a+b^2 x}+\sqrt{a}\right)} \\ & \frac{b}{\sqrt{a} \cdot 2 \sqrt{a}} \Rightarrow \frac{b}{2 a}=3 \Rightarrow \frac{b}{a}=6\end{aligned}$

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2024 (08 Apr Shift 2)
For A+B⇌C+D , if initially the concentration of A and B are equal but at equilibrium concentration of D will be twice of that of A then what will be the equilibrium constant of C+D⇌A+B reaction ?

Options:
A 94
B 97
C 14
D 4

The correct answer is:
14

Solution:
x=2 aâx
3x=2a
x=2a3
K c xâx aâx aâx = 4 a 2 9Ã a 3 Ã a 3 =4
C + D⇌A + B; K c = 1 4

Topic: Chemistry, Chemical Equilibrium, JEE Main
For a,b∈Z and |a-b|≤10, let the angle between the plane P: a x+y-z=b and the line L: x-1=a-y=z+1 be cos-113 If the distance of the point (6,-6,4) from the plane P is 36, then a4+b2 is equal to

Options:
A 32
B 85
C 25
D 48

The correct answer is:
32

Solution:
Given,

For a,b∈Z and |a-b|≤10, let the angle between the plane P: a x+y-z=b and the line L: x-1=a-y=z+1 be cos-113

Now rewriting the equation of line we get,

L: x-11=y-a-1=z+11

Now let, θ=cos-113=sin-1223

Now we know that,

Angle between plane and line is given by,

cos90°-θ=a+1-1+-11a2+2·3

⇒a-2a2+2=223

⇒3a-22=8a2+2

⇒3a2+4-4a=8a2+2

⇒5a2+12a+4=0

So, a=-25,a=-2

Hence, a=-2 (∵a∈Z)

So, the equation of plane will be,

P:-2x+y-z-b=0

Now given, distance of point 6,-6,4 from plane is=36

⇒-12-6-4-b22+12+12=36

⇒|22+b|=18

⇒b=-4, -40

So, b=-4 (∵|a-b|≤10)

Hence, a4+b2=16+16=32

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2023 (08 Apr Shift 2)
For a binomial distribution $B(n, p), \eta p=4$ and variance $n p q$ $=4 / 3 .$ What is the probability $P(x \geq 5)$ equal to?

Options:
A $(2 / 3)^{6}$
B $(1 / 3)^{6}$
C $(1 / 3)^{6}$
D $Q^{8} / 3^{6} y$

The correct answer is:
$Q^{8} / 3^{6} y$

Solution:
Given, $n p=4$ and $n p q=\frac{4}{3}$
$\therefore \quad 4 q=\frac{4}{3} \Rightarrow q=\frac{1}{3}$
$\therefore \quad p=1-\frac{1}{3}=\frac{2}{3}$
$\Rightarrow \quad n=\frac{4 \times 3}{2}=6$
Now, $P(X \geq 5)={ }^{6} C_{5}(p)^{5}(q)^{1}+{ }^{6} C_{6} p^{6} q^{0}$
$={ }^{6} C_{5}\left(\frac{2}{3}\right)^{5}\left(\frac{1}{3}\right)+{ }^{6} C_{6}\left(\frac{2}{3}\right)^{6}$
$=\frac{6 \times 32}{3^{6}}+\frac{64}{3^{6}}=\frac{256}{3^{6}}=\frac{2^{8}}{3^{6}}$

Topic: Mathematics, Probability, JEE Main
For a body executing S.H.M. :
(a) Potential energy is always equal to its K.E.
(b) Average potential and kinetic energy over any given time interval are always equal.
(c) Sum of the kinetic and potential energy at any point of time is constant.
(d) Average K.E. in one time period is equal to average potential energy in one time period.
Choose the most appropriate option from the options given below :

Options:
A (c) and (d)
B only (c)
C only (b)
D (b) and (c)

The correct answer is:
(c) and (d)

Solution:
The formula of kinetic energy in S.H.M. is, K.E.=12mω2A2-x22, where A is the amplitude of the S.H.M. the formula of potential energy is, P.E.=12mω2x2, so the net energy is defined as, T.E.=P.E.+K.E.=12mω2A2 so the total energy is always constant. Since, energy changes from potential to kinetic and vice-versa the average K.E. in one time period is equal to the average P.E. in one time period.

Topic: Physics, Oscillations, JEE Main, JEE Main 2021 (31 Aug Shift 2)